TheSpeedMathBible

writtenbyYamadaTakumi,withthespecialcollaborationofDaniloLapegna

Transformyourbrainintoanelectroniccalculatorandmasterthemathematicalstrategiestotriumphineverychallenge!

"The101bibles"series

I-TheSpeedMathBible

I'mquitesurewecouldallagreeaboutafact: the traditionalwayto teachmathematicshasa lotofstructural problems, and in most of cases it doesn't really help students to get confident with thesubject.Infact,toomanystudentsfinishtheirschoolingstillhavingareal"mathematicalilliteracy",unitedwithaburninghatredofeverythingconcerningnumbersandoperations.

Inparticular,talkingaboutclassicmethodofteachingmathematics,I'mstronglyoftheopinionthat:

It's reallypoor incentive for individual creativity: in fact, toomany times schoolwill teachyou that the proper method for performing a set of calculations is "rigidly" one, and thateverythingshould always bemade in the sameway. This obviously can't domuchmore thanboringeverystudentandgeneratingthefeelingthatthematteritself,ratherthanimprovingone'smentalskills,actuallyshrinksthem,gradually transforminghim/her intosomething that'smoresimilartoanindustrialmachine.Thisbook,however, isdesignedtogofarbeyondthisrestrictedvisionandwill teachyouthatthe classic approach is not the only possible approach and that every set of mathematicalcalculationscanbetransformedintoadeeplycreativechallenge.

The ideaofmathematical"trick" isunjustly"demonised":very few people know that theycould perform very complex calculations just by using extraordinarily rapid and effectivenumericaltricks.And although it's widely accepted that learning the right balance between tricks and rigour,creativity and structure, quick solutions and harder solutions would be ideal for any student,schoolsusuallycontinuetopreferfollowinga"complexityatallcosts"thatofcoursedoesnotnothingbutalienatingpeoplefromthematter.

It'spromotedverylittleself-expression:toomanypeoplefeelthattheyhave"nothingtoshare"with some mathematical concepts even because the rigidity of the taught method preventseveryone from expressing himself according to his talent, his wishes and his naturalpredispositions.Butthisbookpresentsacompletelydifferentapproach,sinceeverythingamongitspageswillbe

deeply self-expression oriented, and the methods shown here will teach you not just one,unquestionablemethod to solve everything, butwill give you the freedom to act according towhatyoufeeltobeeasierandmorecompatiblewithyournaturalattitude.

It'spromotedverylittlecuriosityand"researchingspirit":math,astaughtfromsomeverybadteachers,ispresentedasagreyandsquaredworld,madeofendlessrepetitionsandveryfewinteresting things. But thanks to this book you will start an original journey through thesingularities and the curiosities of the numbers world and you'll soon find out how manyinterestingandbeautifulthingsarehiddenintoitsdeepharmony.

They speakvery little about the realutilityofmathematics: beyond the ability to calculateyourrestaurantbillinaninstant,theenormousadvantageyouwillgainineveryofficialtestorexam, or the awesomeprogressionof your logical anddeductive faculties, youwill discoverthatmathematics has extraordinary strategic and creative applications thatwill help you veryoften inyour everydaychallenges,givingyou that extraoomph in the administrationofyourpersonalfinance,yourwork,yourstudies,yourhealth,yourself-confidenceandyourstrategicintelligencewhendealingwithanykindofchallenge.Math is also vital engine and fuel for every little technologicalwonder in themodernworld.ComputersandtabletscouldneverexisttodayiftwomathematicianslikeJohnvonNeumannandAlanTuringdidn'twritethemathematicalprinciplesbehindthefirstcalculators.Internetwouldneverhaveexisted ifnoonehaddevelopedthemathematicalprinciples thenetworks theory isbasedon.Andthesearchenginesandsocialnetworkscouldneverbebornwithouttheequationsand algorithms allowing each user to track, retrieve and organize documents,web pages andprofiles directly from a set of chaotic data. Moreover, the inventors of Google andrevolutionaries of the modern world, Larry Page and Sergey Brin, are both graduates inmathematics.What I'm trying to tellwith this?Well, I'mnotsure thatevery readerof thisbookwillendupfoundingthenewGoogle,butI'msurethatagreatermathematicscompetenceinaworldtunedonthesefrequenciescanhelpeachofustobemoreaprotagonistandlessabarespectatorofit.

Inotherwords,thisbookwillnotsimplybeasetofstrategiesforimpressingsomeoneorincreasingyouracademicperformance,butwillletyoustartajourneythroughapleasantandintriguingpathofpersonalgrowth,alongwhichyou'lllearntobemoreeffective,creative,confidentand,whynot,moreintelligent.

Inadditiontothat,Iwouldliketogiveyoualastadvice:donotimmediatelytrytolearneverymethodexplained here, but go slowly, make notes, select the techniques you like most and train yourselfcalmlyandalways takingyour time.Thiswillhelpyourmind to learneverythingwithmuchmoreeaseandlesseffort.

Enough said: now I can't do anything butwishing you all the best and, of course, to enjoy yourreading!

YamadaTakumi

Abouttheauthors:

YamadaTakumiandDaniloLapegnaaretwoSoftwareEngineersworkinginLondonassoftwaredevelopersandfreelancewriters.Theirseriesofbooks,"The101bibles",hasbeenagreatsalessuccessintheself-publishingsectorinItaly,engagingthousandsofelectronicreadersevenina

momentofeconomicinstabilityfortheItalianmarket.Nowthey'reconstantlyexpandingtheirserieswithnewbooksaboutpersonalandprofessionalself-developmentandtheirworkrepresentsan

importantreferencepointinthemarket.

Foranykindofquestions,helprequestsor,mostofall,feedbacks,youcanwritethematdanilolapegna-101bibles@yahoo.it

Wealsoconstantlyimproveourworkwithyourfeedbacks,sodon'thesitatetosendusanythingaboutpossiblemistakes,typosorimprovementssuggestions.Asalways,we'llseriouslytakeinto

considerationtheopinionsofourreadersandwe'llusethemtoenrichournexteditions,addingaspecialmentionforyourhelpaswell,ifyoulikeso.TurnevenontheautomaticKindleUpdatesfor

yourdevicetoautomaticallyreceiveournextupdatesforthisebook!

Thankyouforbuying"TheSpeedMathBible",andthankyouforbecomingpartofthe"101bibles"family!

"Themovingpowerofmathematicalinventionisnotreasoning,butimagination."

(ADeMorgan)

II-5mathematicalstrategiesthatwillseriouslyimproveyourlife

Weconstantlycountandmakemeasurementsinoureverydaylife:howmanyhoursmustwesleeptofeel really refreshed, what's our ideal body fat percentage, how much time do we need to finishexaminingthosedocuments,afterhowmanymilesthegastankofourcarwillbeempty...theseareallexamplesofeverydayproblemsweconstantlyhavetodealwith,andthatobviouslywouldrequireasolutionasmoreeffectiveaspossible.Atthesametime,however,wetoooftenevaluatethefactorsinvolvedinourproblemsfromapurelyqualitativepointofviewratherthanaquantitativeone,givingthempurelyempiricalsolutions.Thatis,forexample,insteadoftryingtounderstandtheproperamountofsleepinghoursforourbody,wetoooftentrytosleepenoughtofeelrelativelyfit.Ratherthancalculatingourbodyfatpercentageandadjusting our diet accordingly, we prefer some kind of homemade diet that does not really payattention toourrealphysiologicalneeds.Andwedo thatdespite it is intuitive thataddressing theseissues from a quantitative point of view, and so measuring, evaluating, calculating the quantitiesinvolved,wouldofferagreaterefficiencytoouractions,andsowouldgiveustheabilitytoproducemuchmorewithmanylessexpensesintermsoftime,moneyandresources.Wait:thisdoesnotmeanthatyoushouldrunintosomeweirdobsessionforrationalassessmentsorformeasuring every single aspect of reality.Your existence of course needs even impulsively doneactions,lessonslearntfromyourmistakesandthecoexistencewiththeinevitableunknown.However, when you actually feel the need to get more results in a specific field, a strongermathematical"grip"onitcouldbeagreatstrategytoenhanceyourpersonalimprovement.Afterall,themostsuccessfulcompaniesareexactlythosedoingthebestinretrievingandanalysingdataabouttheconsequencesoftheirbehaviour.Buthowcouldwedothat?Here'ssomeadviceswecouldfollowforthepurpose:

Play givingmarksmore often. No, this has nothing to dowith school. If, for example, youreally enjoy doing something, like, I don't know, eating a particular type of food or goingtravelingsomewhere,trygivingtotheseactions,ortotheadvantageyougetfromdoingthem,amark.From0 to10, from0 to100, itdoesn'tmatter!Thepoint isplayingmakingquantitativeconceptsoutofqualitativeones,andstartingmakingbetteranalysisontheirbasis.Forexample,ifdedicatingyourselftojogginginthemorninghasforyouahighermarkthangoingtogym,youcouldsimplystarttodecideto...gojoggingmoreoften!

Quantityinmostofcaseshelpsyoutobetterfocusonthemostimportantthings,andcleansyourthoughtsfromtheunnecessarystuff!

Try to make measurements where nobody else would do it. "Not everything that can becountedreallycounts,andnoteverythingthatcountscanbereallycounted",saidAlbertEinstein.ThisphrasenotonlyconfirmswhatIearliersaidaboutthefactthatmathematicsandcalculationssometimesmustsimplybeputaside(luckily),butcanhelpustounderstandthat,toomuchoften,themostcommonlyusedmeasurementsystemscandeceiveus,andcandivertourattentionfromthethingsthatreallymatter.AnexampleIlovedoingaboutthisisintheschoolreportcards:theirobviouspurposeshouldbegivingthemeasurementof theperformanceofastudent,butweoftenforget that,ontheotherside,therecouldbeateacherthatsimplyhasanawfulteachingmethod.Andthat'swhereweshouldstartto"makemeasurements"wherenobodyelsewoulddoit: lettingthestudentsevaluatethequalityandthegoodnessoftheirteacher'sworkcouldpusheverybodyintodoingbetterhis/herwork,andconsequentlyintoproducingbetterresultsforeverybody.Makingmeasurements,evaluating,focusingonthethingsthatotherpeopletendtoignore,givesyouthatextraoomphthatwillproduceasignificantadvantageinanykindofsituation!

Increase thequantityand improve thequalityofyourmeasurement instruments. Imagine:it'salazyeveningandyoudecidetogowatchingamovietothecinema.Now,youhavetodecidewhichmoviebuyingtheticketsfor,butyou'rehesitatingbetweenthatnewactionmoviewithalotof old celebrities and that new horror Japanesemovie about a deadwomanwho comes backfrom the grave. Now you could take your smartphone, read the movie reviews from somefamouswebsite and simply choose themoviewith the highest rate. But of course youwouldbetter estimate the quality of thosemovies after increasing the quantityof yourmeasurementinstrumentsand, forexample, readingmovie ratings frommanydifferentwebsites instead thanfromonlyoneofthem.Nowthere isaconcept thatcanreallycome inhandy: themathematicalmean that, although isquitewidelyknown,we'llexplaininafewwords,justtomakesurethateverybodyknowswhatarewetalkingabout.If you have multiple quantities, multiple measurements, multiple values (ex. 3, 4 and 5 starsratingforamovieonthreedifferentwebsites),youcansumthosenumbers(ex.3+4+5=12)andthendividetheresultbythenumberofvaluesyouconsidered(threeratingsinthiscase,so12/3=4).This will give you the mean of those quantities, which is universally considered as a verypreciousvaluesince,asinthemeasurementstheory,asintheprobabilityone,it'ssaidtobeveryclosetothe"real"valueofsomething(ifarealvaluecanphilosophicallyorscientificallyexist,ofcourse).Butnow,let'saddsomethingandlet'stalkaboutqualityofmeasurementinstruments:howmanytimeshappenedthatamulti-awardwinningmoviejustleftyouasleepinyourchair?Thatsimplymeansthatthosemeasurementinstruments(the"official"moviereviewsweusedtoestimatethegoodnessoftheproduct)weren'tasreliableastheyweresupposedtobe.Andthisreallyhappensalotoftimes,doesn'tit?Butwhat'sthesolution?Howcouldweimprovethequalityofourinstruments?Well,wecould,forexample,browsejustamongtheopinionsofthefriendsofourswhousually

havesimilartastetoours.Afteradoptingthisstrategywe'llautomaticallynoticethatevenfewer"measurements"willbeproventobemuchmoretruthful,andsoagainthatsometimesquantityisapoorconcept(weshouldnevertrustanopiniononlybecausesharedbymostofpeople,afterall)andqualityshouldhaveaheavierweightinoureverydayreasoning.

If numbers rule over a situation, you must ally with them or you'll definitely bedefeated.Sun-Tzu,themostimportantstrategicbookofalltimes,repeatedlyaffirmsthatvictorycanonlybeachievedifoneharmonizesitselfwiththenaturalconstants.So,ifwemonthlyspendmorethanhowmuchweearn,bankruptwillsoonbemuchmorethananabstractconcept,andthereareno"qualitativefactors"thatcanchangethis.Ifwebuildashelvingunitwhoseshelvesarestructurallymadetosustainaheavierweightthanthedecorativeobjectwejustbought,weshouldsimplyavoidputtingitthere.Allthefieldsstrictlyruledbyphysical,economic,naturalrulesareclearexamplesofcontextsinwhich you just can't escape from "numbers supremacy". So, making extra measurements foryourshelf,checkingyourmonthlyexpenses,orbeingsurethatyourself-employingprojectwillsustain your rent cost in the long-term are allnecessary actions, that not onlywill definitelyimprovethequalityofyourwork,butwillalsohelpyoutoavoiddisasters.Yes, thismaysoundtrivialsometimes,butat thesametimeweshouldpayattention to thefactthattoooftenwetendtosufferforignoringexactlythemostobviousthings.

Doubtful?Use thepowerof thegain-lossratio!Suppose thatyou just receivedmultiple joboffers,withdifferentproandcons,andsupposetobetotallydoubtfulaboutwhichonecouldbethebestchoiceforyourcareerprogression.Ormaybeyou'relookingforanewhouse,youhavevariousoptionsandyouwanttounderstandwhichonewillbemoresuitableforyourneeds.Well,inthesecasesyoucouldputyourselfintothehandsofthegain-lossratio:trytomakealistof theconsof eachchoice, andgive to eachdisadvantageadamage ratingbetween1 (almostannoying)and10 (death).Andbecareful:eachchoicemusthaveat leastadisadvantagewithadamageratinggreaterthan0.Forexample:

Flat1:Veryexpensive:7(prettybigdeal)Nosmokingallowed:5(annoyingbutalmostacceptable)Nopetsallowed:8(thisisgoingtobeaseriousproblemifIwanttobuyadog)

Flat2:Verydirty:9Veryfarfromthecitycentre:8

Now,foreachchoice,calculatethemeanofthedamageratingsbyitsdisadvantages,andcalltheresult"damagerating"ofthatspecificchoice.Inourexampleitwouldbe:

Flat1:7+5+8=20/3=Damagerating:6.66

Flat2:

9+8=17/2=Damagerating:8.5

Nowit'stimetostartconsideringthebrightsideofthemedal:makealistoftheprosforeachchoice and give to each advantage a "convenience rating". Now, calculate the "conveniencerating" of every specific choice by calculating the averages of the convenience ratings of itsadvantages.Inourdecisionaboutwhichapartmentgoingtolivein,thiscouldbe:

Flat1:Verybright:7Roomsareprettyclean:7Veryneartomyworkplace:8Conveniencerating:7+7+8=7.3

Flat2:Verycheap:8Petsallowed:8Conveniencerating:8

You're almost done. Now, for each choice, just divide the convenience rating by the damagerating, and you'll have your gain-loss ratio: an approximate measurement of which choice is(possibly)thebesttotake:

Flat1:Gain-lossratio:7.3/6.66=1.096

Flat2:Gain-lossratio:8/8.5=0.94

Flat1wins!

Ofcoursethisdoesn'thavetobetakenastheultimatetruth,butitcanbereallyusefultoshowyouabrighterway inyourdecisionmakingprocess. In addition to this, remember thatmorehonest you will be with yourself in your rating, and more trustworthy will be this index inshowingyouthebestchoicetotake.Last important thing: when you're dealing with a situation in which fortuity has a stronginfluence, this indexlosesmostof itsusefulness,andin thatcase thebest thingtodois takingyourdecisionwiththehelpofotherfactorswe'llexplaininChapterXIX.

Thischapterendshere,butyourjourneyintothepracticalandstrategicapplicationsofmathematicsjust started.Mostof calculation techniques shown in thenextpages, in fact,will often comealongwith theirmoreuseful, and funnyapplications for everyday life, likeprobability theory applied togamblingorgametheoryappliedtopoker.So,ifyoucontinuereading,youwillbepageafterpagemoresurprisedandwonderingwhyyourteachersonlytaughtyouthepoorestsideofthismatter.

III-4basicstepstoboostyourcalculationskills

HereI'llstartexplainingyoufourhelpfulfundamentalsforimprovingthespeedofyourmathematicalskills that you'll absolutely need to learn before starting to practise with any creative calculationstrategy.Onceyoulearntthem,the80%ofthehardworkisdone,so...let'sstart!

1-Taketherightattitude

Yes, influence of the mental attitude in math is something as underestimated as powerful. Ifyousubconsciouslykeeprepeatingyourselfthatmathisnotpartofyourworld,thatit'stoodifficultforyouandthatyouwillnevertrulyuseitinyoureverydaylife,youcanbesurethatyouwillneverdoanyprogressaswell.Rather:

Remember that, whatever is your opinion about your "innate predisposition" to mathematics,you'renotobligatedtofollowit.Yes,definitelytherearepeoplethatarecutoutformathematicsandlogicsincetheywereborn,andpeoplethathadapanicattackinfrontoftheirfirstnumberwrittenonablackboard,butournaturedoesn'thavetobeourcurse.ForexampleGertMittring,whowonthegoldmedalattheMSOmentalcalculationfornineconsecutiveyears,atschoolwasone of theworstmath student in its class. And, before him,many other geniuses likeAlbertEinsteinorThomasAlvaEdison,achievedawfulmarksexactly in thesamesubjects todayareconsideredasreferencepointsfor.Edisonwasevencalleda"mentallyretardedboy"byoneofhisteachers.So,1)Ourschoolgradeswillneverbearealmeasurementofourrealskillsand,2)Fromtheadaptivenatureofourbraininevitablyfollowsthatwe,andonlywecandecidewhichpartofourmindtocultivate,andwhichparttoleavewitheringanddecaying.AndIcouldbetanyamountofmoneyonthefactthat,ifyouproperlyworkwiththestrategiesI'llshowyou,youwilldevelopskillsyouthoughtyouneverhad!

Keep yourself motivated! Motivation is the foundation of every personal growth and self-improvementpath!Infact,ifwekeepourselvesmotivated,everybarrierbecomesnothingbutanopportunitytolearn,buildandimproveourtalentandcreativity.Buthowdowekeepmotivatedin front of a potentially tedious pathmade up of numbers and theorems?By learning how toconverteveryarithmeticalproblemintoadeeplycreativechallenge!Byconstantlyrememberingourselves thatbackingout fromamental challengemeansgradually losingourbrightnessand

cleverness!Byunderstandingthatmathematicscansharpenandempowerallthose"mindtools"thatoverandoveragainwillmakethedifferenceinourlifebetweenprofitandloss,happinessandunhappiness,successand failure.And, lastbutnot least,bybeingaware that an improvedabilityto"play"withnumberscanevenimpressthepeopleweknow,aswe'llbetterseelater.Thiswillinevitablyhelpustobuildourself-confidenceandwillimproveasourrelationshipwiththeothers,asourconnectionwithourinnerself.

Trainyourself.Stopusingyour smartphoneeven toperforma "7x6"anduseyoureverydaychallenges to train your brainwithmathematics! For example, try tomentally calculate yourrestaurant bill or your change after buying your daily newspaper! This advice may soundobvious, but at the same time is probably the most effective of all. After all, we should justrememberwhatAristotle, theancientGreekphilosopher, said:"Wearewhatwerepeatedlydo.Excellence,then,isnotanact,butanhabit!".

Don't be stubborn! Yes, keep yourself motivated and trained, but don't stubbornly insist onmathematicalproblemsthatyouapparentlycannotsolve,becausesometimesthesolutionscomejustasyouthinkofsomethingelse.Infact,evenwhenweabandonaproblemsolvingprocedure,ourbrainoftencontinuesworking"inbackground",expands itsperspectiveand then itsuddenlybecomesable togiveus thebestsolutions to our problems. Famous,moreover,was the expression "Eureka" thatArchimedes,accordingto legend,shoutedafterfindingagoodsolutionforcalculatingthevolumeofsolidobjects while relaxing during a bath. Legend or not, in any case, when the solution to amathematicalproblemorarithmeticjustdoesnotseemtoarrive,trytotakeabreakandmoveon.Theactofcomingback lateronhardproblemshelpsus toapproach themfromanew,widerperspectiveand,althoughitmayseemcounterintuitive,it'sthebestwaytohelpourbraintodoitsjob.

2-Respectyourbrain

Onethingweoftenforgetisthatbrainandbodyusuallygiveusbacknothingmorethanwhatwegavethem.

Sowe should just stop looking at them asbare tools to squeeze the best possible result from, andinsteadweshouldalwaysseethemasimportantelementsofourselftobefedasmuchaspossibleinordertoletthemgivetheirbestfruitsinreturn.

Sowhat?Agoodbrain,andthereforegoodcalculationskillscomesfromtherespectforyourbody,yourhealthandyourownbiologicalrhythmsfirst.Inparticular:

Makephysicalactivityahabit.Someconstantworkout,infact,oxygenatesourbodyandbrain,improvingourreasoningskillsandmakingevenmorecomplexthoughtseasiertoprocess.No,youdon't have to engageyourself in anything strenuous and evenone hour a daywalking inordertoboostyourmetabolismwillbefine.Oh,andalwayskeepinmindthatDr.MarilynAlbert,aresearcherinthefieldofbrainfunction,hasfoundthatadultsaremorelikelytokeepagoodbraininoldageiftheyarephysicallyactive

andkeepagoodcardiovascularhealth.So...makeaworkoutplantokeepyoualwaysyoung!

Eatproperly.Therearedozensoftheoriesaboutwhichtheexactmeaningof"eatingproperly"shouldbeand,inadditiontothis,thehabitsandbodyofeachofusandmakeusbettersuitableforspecifickindofdiets,sofindingawaythroughthedifferentvoicesisalwaysaquitedifficultchallenge.Well, without necessarily having to consult a nutritionist, my simple advice here is to neverabandon some elementary common sense rules: variety, don't overdo, always have a properbreakfast,haveagooddailyamountofantioxidants,don'tomitorexaggeratewithrefinedsugarand, most of all, don't forget to insert into your diet a good amount of foods containingphosphorousandB-complexvitamins,greatsubstancestoletyourbrainworkatitsbest.Someexamplesoffoodscontainingthesenutritionalsubstances?Cereals,fish,nuts.

Coffee?Yes,but in the right amount.Yeah, coffee, canned drinks, black and green tea, andanything containing caffeine certainly enhances our cognitive skills, boosts our attention andimprovesthespeedofourneurologicalreactions.However, if taken in highdoses, thesedrinks just endupbeing counterproductive, since theirhyper-stimulatingeffectstartscompromisingourabilitytoreasonasbestaswecan.Sobecareful,andbeawareofwhatshouldbeyourdailylimitofcaffeine.Ofcourse, this limitcanbedifficulttocalculatebecauseitchangesdependingonfactorssuchasourgenderandage;however,payingattentiontothesignalsofyourbodycanbereallyhelpful,sojusttrytonoticeafter how many cups of coffee, tea or soft drinks you start to feel confused, distracted andnervous.

Relax.Stresscausesouradrenalglandstoproduceexcessiveamountsofcortisol,aneurotoxicsubstance thatcandamageoursynapses (theconnectionsamong themassesofneurons inourbrain).So if you reallywant to improve your brain skills you always have to leave yourselfsometimetoworkonyourconcerns,toenjoy the thingsyoureally likeand toeliminate (oratleastlearntohandle)asmoresourcesofanxietyandstressaspossible.

Sleep.Wecananywherereadadvicesaboutsleeping8,9or10hourseverynight,butthetruthisthat each of us can have different needs, depending on our physiology and state of fatigue orstress.Sodonothingbutpayingattentiontoyourbodysignalsandgiveareasonableprioritytoyoursleeptime,sinceitwillgreatlyboostyourbrainskills,yourabilitytofocusandevenyourproblemsolvingandstressmanagementabilities.

3-Understandtheimportanceofyourmemory

Anycalculationprocessisbasicallymadeupoftwodifferentphases:thefirstoneisthecalculationitself, while the other one is the memorization of the previous results. And the latter probablyrepresentsacrazilycriticalpointformostofpeople.

Imagineyourself trying to solvewithout anypenorpaper a longmultiplicationbetween three-or-more-figurenumbers:probablythehardestthingforyouwillbeexactlytryingtoretainallthedigitscomingoutofthepartialmultiplications.

So,ontheonehandyou'llseethatthe"secret"behindmanycalculationstrategiesinthisbookisinthe

fact that they'rebuilt inorder to letusoptimize theuseofourshort-termmemory,andon theotherhand it follows that your memory represents a key-skill to train if you want to improve yourmathematicalabilities.

So,trytofollowtheseadvices:

Keep yourself focused:This may sound obvious, but focusing on the tasks you're dedicatingyourself to will seriously improve your ability to memorize everything involved. So, try toisolate yourself from any potential disturbance source, always force yourself to completelyfocusonthe"hereandnow"andyourmemorywillgetanawesomeboost!

Try tomemorizeonly theessential things. If for exampleyouarementallyperforminga longaddition,don'trepeatyourselfanydigitofthenewaddendsyou'regetting,butjustmemorizethepartial results you obtain. Same thing with subtraction, multiplication or division: memorizeonly thebareminimumamount of elements necessary to go ahead in your calculation.You'llrealizebyyourselfthat,afteradoptingthiswayofthinking,everyarithmeticprocedurewillbemuchfasterandlesstiring.

Know yourself. Somebody, for example, is capable to retain something much better into hismemory after visualizing it, somebody else after listening to it instead, and so on. In otherwords,trytounderstandhow,whenandwhereyourbrainusestostoreinformationinafasterandmoreefficientway,andthen...alwaysdoyourbesttohelpittodoitswork!Imagineabig,coloured picture of the numbers you're working with if you prefer visualization, and repeatyourselftheirnameifyouarean"auditive"person:thiswillhelpyoutorememberthemmuchmoreeasily!

Divide.Yourbrainworks lesshardwhile trying tomemorizemanycombinationsmadeupoffew elements, than while trying tomemorize one, single set made up of a lot of things. So,especially if you try to operate with very large numbers, try to split them in many shorternumbersmadeupofafewdigits:rememberingthemlaterwillbeaseasyastheproverbialpie.

Usethemnemonicmajorsystem:Themnemonicmajorsystemisprobablythemostfamousandmostcommonlyusedmnemonictechnique,duetoitssimplicityandpower.Infact,thankstoit,youwillbeabletomemorizeevenareallylongnumberbyturningitintoawordoraphrase,whichofcourseisdefinitelymucheasiertorememberthanasequenceofdigits.Thistechniqueismadeupofthreephases:1.Converteachdigitinyournumberintoaconsonantusingaspecifictable.2.Mixtheconsonantsyougotwiththevowelsyouneed (orsome"unassigned"consonants), inordertocreateakey-wordorakey-phrasetoremember.3.When you need to have your number again, you will just have to take your key-phrase,removethevowels,andre-useyourtableinreverse.Oh,andbytheway,thisisyourtable:

ButsinceIthinkthatexamplesarealwaysthebestwaytoexplainsomething,let'simagineyouhavetorememberthenumber3240191.YouhaveM-N-R-S-T-P(orb)-D(ort).Aftermixingthoseconsonantswithsomevowelsyoucouldobtain"Menarestupid".Andmorestupidorstrangeistheresultingphrase,andeasieritwillbeforyoutoremember!Onthecontrary,let'simagineyourkey-phraseis"Blackhorse".YouhaveB-L-hardC-R-S:yournumberis95640!Yes, this technique may sound difficult at the beginning, but a little training can give youawesomeresults.Andconsiderthatit'saloneworththecostoftheentirebook,sinceitwon'tgiveyouonlytheabilitytomakeanycalculationwithoutpenorpaper,butwillcomeinhandyeverytimeyou'llneedtorememberdatesortelephonenumbersaswell!

Now you have in your "mental toolbox" a lot of extraordinary instruments for improving your"numerical"memory. Treasure those which better work for you, and sharpen them to boost yourmathematicalskillsattheirmaximumpower!

4-Strengthenyourbasis

Mathematics is likea"Lego"buildingstructure: thesimplestconceptscanbecombined together inorder toshapemorecomplex theories, those theoriescanbecombined in turn togiverise toevenmorecomplexideas,andsoon.

Soit'sobviousthat,ifthestarterbricksaretin-pot,thefinalstructureswillalwaysbecrumblingandunstable.Inotherwords,ifyourmathematicalbasisisnotwellsetinyourmind,allthethinkingthatlaysonthatbasiswillbeuncertain,slowandunreliable.Andthisisahardlycriticalpoint,becauseit'squiteunderestimatedby99%ofpeople.This,forexample,remindsmeaboutalotofengineers-to-beI was studying with, who very often were reporting bad results in calculus and algebra tests, notbecausetheyactuallywereignorantaboutcalculusoralgebra,butbecausevery,veryoftentheywerejustmakingbanalarithmeticalmistakes.

It'salsointuitivethatstrengtheningyourbasisisdefinitelyakey-skilltoaccelerateyourcalculationmaking:if,forexample,youtrainyourselftoinstantlyrecallfromyourmemoryalltheresultsofthebasicsumsfrom1+1to9+9,youwon'tloseanymoreasinglesecondwhiletryingtoperforma"7+8"

youneedtocarryoutduringaseven-figuresum.

So, my advice here is to work on reinforcing the following concepts in your mind, as banal orobvioustheymaysound:

Thebasicnumberproperties.Thebasicnumberpropertieswillbeoftenused in thisbook in"creative"waysinordertobuildalotofspecialspeedmathstrategies,soyoucouldfinditveryusefultobrushuponthem:

○Commutativepropertyof additionandmultiplication: changing theorderof theoperands inanyadditionormultiplicationdoesn'tchangethefinalresult.Andso,forexample:3+4+5=5+4+3=5+3+4=12.

○ Associative property of addition and multiplication: changing the order you perform youradditionormultiplicationwith,doesn'tchangethefinalresult.So,forexample,(3+4)+5which,afterlookingatthebrackets,impliesdoing3+4firstandthenadding5,won'tbedifferentfrom3+(4+5),whichinsteadimpliescalculating4+5firstandthenadding3.

○Distributivepropertyofmultiplicationoveraddition.Thatis:ifIhavean"ax(b+c)",theresultwillbethesameas"(axb)+(axc)".

○Invariantivepropertyofdivision:givenana/bdivision,ifyoumultiplyordividebothaandbforthesamequantity,thefinalresultwon'tchange.Let'smakeasimpleexampleandlet'simagineyouhavea30/10toperform:afterdividingbothnumbersby10you'llhave3/1,whichresultobviouslyisthesameas30/10andequals3.Samethingifyoumultipliedbothnumbersby2:60/20infactstillequals3.Themultiplication has a very similar property, butwith one very significant difference: afterperformingany"axb"multiplication,theresultdoesn'tchangeifyoumultiply"a"byagenericquantityanddivide"b"bythesamequantity,orviceversa.So,givenforexamplea16x2,youcanhalve16,butthentokeeptheexactresultyou'llhavetodouble2.So16x2=8x4.Samethingwith,forexample,100x9:itwillgivethesamequantityasaresultas300x3,or900x1.

○Identityproperty:Addingorsubtracting0fromanumberdoesn'tchangeit.Thesamehappensafterdividingormultiplyingitby1.

○Zeroproperty:Any number, ifmultiplied by 0will still equal 0.And, in the sameway, theresultofamultiplicationcanneverbe0ifneitherofthefactorsis0.Itmaybealsointerestingtorememberthatthisruleimpliesthatyouwillneverbeabletodivideany number by 0. Let's take in fact a non-null number, like 14. Division is the inverse ofmultiplication,andsotryingtodivide14by0meansaskingyourself:"Whichnumberequals14when multiplied by 0?". And the zero property obviously tells us that this question has noanswers.

Dividingormultiplyingby10,100,1000oranyotherpowerof10:This is a pretty simple

calculation as well, but brushing up on it is still a good practice, even because some fastmultiplication strategies we'll see in the next pages are exactly based on the concept oftransformingan"ordinary"multiplicationintoamultiplicationbyapowerof10.So:○Tomultiplyanintegernumber"a"byapowerof10,justaddto"a"asmanytrailingzeroesasthepowerof10has.However, if "a" has a decimal point, thenyouhave to shift the point on the right by asmanypositionsasthetrailingzeroesofthepowerof10are.So:5x1000=5000(justwritethreetrailingzeroes)1.28x1000=1280(shiftthepointbytwopositionstotherightandyouget128.But1000hasthreetrailingzeroes,soyouhavetowriteonemorezerototheresultinordertocompletethemultiplication)567.3x1000=567300(shiftthepointbyonezerototherightandyouget5673.But1000hasthreezeroes,soyouhavetowritetwomoretrailingzeroestotheresultinordertocompletethemultiplication)

○Todivideanumber "a"byapowerof 10, just shift its decimal point to the left by asmanypositionsasthepowerof10has.Ifthenumberisaninteger,justthinkasifithada".0"afteritsrightmostdigit.Andif,duringyouroperation,thepointpassesoverthedigitontheextremeleft,thenwritea0onitsleft.So,forexample:345/1000=0.345(shiftthepointtotheleftbythreepositions.Thepointpassesoverthe3,soputa0onitsleft)2/1000=0.002(shiftthepointtotheleftbythreepositions.Youwillhave.2atfirst,whichwillbecomea0.2.Then,byshiftingagain,itwillbe.02=0.02.Andsoon.)14300/1000=14.300=14.3700000/1000=700.000=700Andavery interesting thing is thatmastering this techniquemakesyour lifeeasierevenwhenyouhavetoperformanykindofmultiplicationordivisionbetweennumbershavingoneormoretrailingzeroes.In particular, in themultiplicationcase, you can just remove any trailing zeroes, perform themultiplication,andthenmultiplytheresultbyapowerof10havingthesameamountofzeroesasthosethatwereremoved.So, for example, 5677 x 300 = 5677 x 3, whichmust then bemultiplied by 100 (two zeroesremoved).And, similarly, 350 x 2000 = 35 x 2, whichmust then bemultiplied by 10 000 (four zeroesremoved).Thedivisioncaseisalittlebitmoredifficultthough.Inthiscase,infact,youmuststillremoveanytrailingzeroes,but thenyouhavetomultiply the resultbyapowerof10having thesameamountofzeroesas thoseremovedfromthedividend,anddivide the resultbyapowerof10havingthesameamountofzeroesasthoseremovedfromthedivisor.So, if forexampleyoumustcalculate90 /6,youcanremove the0fromthe90firstand thencalculate9/6=1.5.Youjustremovedazerofromthedividend,sonowyouhavetomultiplytheresultby10=15.Ifyouhavetocalculate9/60instead,youmustremovethe0from60first,thenperformagain9/6=1.5,andattheenddividetheresultby10,gettinga0.15asaresult.

Last case: let's imagineyoumust calculate900 /60. In this caseyoumust remove twozeroesfrom900andonezerofrom60,so,aftercalculatingagainyourgoodold9/6=1.5,you'llhavetomultiplytheresultby100andthendivideitby10.Thisobviouslyisthesameasmultiplyingitby10,andsothefinalresultis15onceagain.

Multiplicationandadditiontablesfrom0to9:probablyeverybodyknowsthesetables,butatthe same time very few people are able to recall them from memory instantly and withoutmakinganymistakes.Sotrytobrushuponthem,andexamineindepthallthe"hardercells".Forexample,itisproventhatmajorityofpeoplehasproblemswithmultiplicationtablesfrom7to9,andthesamehappenswiththesumsbetweendigitsfrom6to9.

Additiontablefrom0+0to9+9

Multiplicationtablefrom0x0to9x9

So, this is where your "4 basic steps to improve your calculation skills" end. Just keep the rightattitude,respectyourbrain,trainyourmemoryandstrengthenyourbasis...you'llenhanceyourskillsinwaysyoucouldneverbelieve.Andthisisjustthebeginning!

IV-Thefingercalculator

Everybodyknowsthatonecanperformlittleadditionssimplybyusinghisorherfingers.It'soneofthefirstthingsthattheytaughtusintheschool,afterall,isn'tit?Butdidyouknowthat,byusingyourfingers,youcouldperformevensomevery...simplemultiplications?

If,forexample,youhaveashockingblackoutduringacomplexcalculation,andyoucan'trememberaspecificmultiplication table,youcanuse twoawesometricks thatwillmakeyouable to instantlyperformalotofpossiblesingle-digitmultiplications.

Butlet'sstartwiththefirst,reallystraightforwardtechnique,whichtakesadvantageofthefactthatifyou sum together the digits of any multiple of 9 (<90), you will always get 9 as a total. Morespecifically, it letsyou immediatelyperformanymultiplication from9x1 to9x10.Andhere'showyoucanuseit:

Turnbothyourpalmstowardsyourface,pointingtheextremityofyourfingersupwards.Nowassociatetoeachfingeranumberfrom1to10,correspondingtotheordertheyhave.So,yourleftthumbwillbe1,theleftindexwillbe2,andsoon,tilltherightthumb,whichwillbea10.Toperform9multipliedbya,simplylowerthe"a"finger.Thefirstdigitofyourresultwillbeequaltothenumberoffingersbeforetheloweredone.Theseconddigitofyourresultwillbeequaltothenumberoffingersaftertheloweredoneinstead.

So,ifforexampleyouwanttocalculate9x2,thenumber2isassociatedtoyourleftindex.Ifyoulowerit,youwillseethatthereis1fingerbeforetheindexand8fingersafterit:18!Thesameifyoudon'tremembertheresultof9x6:sixisassociatedtoyourrightpinkie,beforetherightpinkiethereare5fingers,andsothefirstdigitis5.Afteritthereare4fingersinsteadandinfact9x6=54.

But let's immediately look at the second technique too, useful for retrieving the result of anymultiplicationfrom6x6to10x10.Hereitis:

Turn again both your palms towards your face, but now point the extremity of your fingers

laterallyinwards,inordertoletthetwomiddlefingerstoucheachother.Forbothhands,associate6tothepinkie,7totheannular,8tothemiddlefinger,9totheindex,and10tothethumb.Ifyouwanttomultiplytwonumbers,lettheextremitiesofthecorrespondingfingerstoucheachother,makingsureyoukeepthepalmsorientationIjustdescribed.The fingersunder thepointofcontact, including theones in thepointofcontact itselfare the"under"fingers.Thefingersabovethepointofcontactwillbethe"above"fingers.Multiplythenumberofthe"under"fingersbyten.Callthisresult"a".Multiply"abovefingersontheleft"by"abovefingersontheright".Callthisresult"b".Sumaandbtogether.

Butsincethistechniquemayseemalittlebitmorecomplexthanthepreviousone(butstillveryeasytomanage,anyway),let'sintroducealittleexampleandlet'strytouseittocalculate6x7.Wewillactlikethis:

Letourleftpinkietouchourrightannular(orviceversa).Wewillhave3"under"fingers(thetwotouchingfingersplusthepinkie).3x10=30."Aboveright"equals3,while"Aboveleft"equals4.3x4=12.30+12=6x7=42

Secondexample,8x7:

Ourrightannularwilltouchourleftmiddlefinger(orviceversa)."Under"equals5.Multipliedby10equals50."Aboveright"equals2."Aboveleft"equals3instead.2x3=6.50+6=8x7=56

Here comes the fact that this technique can't be used properly if you cannot remember the lowermultiplication tables.But this probably itwon't be a big problem, considering thatmost of peoplehaveproblemsrightwithmultiplicationsbythedigitsverycloseto10,like7,8or9.

Throughyour fingersyoucanevenuse two little "mathematical" lifehacks that cancome in reallyhandyineverydaylife:

Measure the lengthof anyobject. It's quite simple:measure one of your fingerswith a rule,alwayskeepinmindtheresultandthenexttimeyouneedtomeasureanyobject,andyoudon'thavethepropertoolswithyou,youdon'thavetodoanythingbutcountinghowmanytimestheobjectislongerthanyourfinger,andthenmultiplyingthatresultbyyourfingerlength.

Measuretheheightofatree,abuilding,oranyotherbigobject.Thistechniquewasusedsinceancienttimes:putapersonyouknowtheheightofnexttothebigobjectyouwanttomeasure,placeyourfingernexttoyoureyeatsuchadistancethatthatit'sexactlyashighasthatperson...andapply theprevious technique!Themultiplicationhere isgoing tobedifficult toperform?Don'tworry;thetechniquesinthenextchapterswillexactlycometoyouraid!

Nowyoujustgotthefullbasistoenterintotheworldofthespeedmathandsoit'stimetointroduceavery fascinating topic, loved asmuchby themathematics fans asby theAsian spiritualityone: the

VedicMathematics.

"Mathematicsisagreatandvastlandscapeopentoallthinkingmenthatadverselyjoy,butnotverysuitableforthosewhodonotlikethetroubleofthinking."

(LazarusFuchs)

V-MathematicsandHinduwisdom

Was the beginning of the 20th century, while an influential researcher of Hindu philosophydiscovered into its personal copy of a sacredHindu book, theAtharvaveda, a summarise of veryoriginalmathematical rules.More specifically, in thatbookhe found sixteenSutras (aphorisms ofHinduwisdom)andsomecorollariespresentingacompletelyoriginalandmuchmorecreativeandflexibleapproachtovariousmathematicalproblems.

Sincethenthis"Vedicapproach",whichvascompletelydifferentfromanyotherapproachtaughtintheOccidentalworldsofar,startedspreadingallovertheworld.Forexampletodayit'sprettyrenownamongthemostimportantUSAschools,whereitevenrepresentsthemostprestigiouschapterintheirteachingplan.

ProbablythepopularityoftheVedicMathematicsisexactlyduetothefactthatitcommendseveryone'screativity and inner expression, making mathematics easier and more attractive for any kind ofstudent. In fact,many of themathematical stratagemswe'll see in the next chapterswill be exactlydrawnfromtheVedicSutras,ortheircorollaries,oftencombiningtheminordertogetfast,creativeandextraordinarilyeffectivecalculationstrategies.

But let's start immediately and let's explain themeaning and the purpose of the "Nikhilam Sutra",whichdeclaims:"Allfrom9andthelastfrom10".

What? Well, it's not cryptic as it sounds, actually: this Sutra reveals nothing but a very fast andefficientmethodtocalculatethedistance(ordifference)betweenanynumber"a"anditshigherpowerof10(whichis,intheunluckycaseyouforgotit,thatnumbermadeupofa"1"andasmanyzeroesasthedigitsof"a".So,forexample,willbe10for4,100for55,1000for768,andsoon).

Soitwillbasicallyhelpyoutorapidlyperformcalculationslike"1000-658","10000-4530","1000000-564324",andsoon.Butitspower,aswewillsoonsee,doesn'tendhere.

Inanycase,toapplytheSutrainitsessentialform,youcandolikethis:

Makesurethatthesubtrahend(thenumberaftertheminus)hasasmanydigitsasthezeroesofthepowerof10youwanttosubtractfrom.Forexample,youcanuseittocalculate100-67,1000-345,or10000-4589.

Startingfromleft,subtracteachdigitofthesubtrahendfrom9(so,becareful,calculate9minusthe subtrahend and not the otherway round), except the units digit,whichmust be subtractedfrom10.Then,putalltheresultsofthesesubtractionsinorderintothefinalresult.

Everytimeyouget"10"asaresult(like,forexample,ithappenswhentheunitsdigitis0),set0ascurrentresultdigitandadd1totheresultdigitonitsleft.

That'stheresultofyoursubtraction!

So,forexample,let'strytocalculate10000-4350.Wewillhavethat:

Aftersubtracting4from9thefirstdigitintheresult(ontheleft)willbe5

Aftersubtracting3from9,theseconddigitintheresultwillbe6

Aftersubtracting5from9,thethirddigitintheresultwillbe4

Aftersubtracting0from10,thefourthdigitintheresultwillbe10.Put0intheresultandcarry1tothethirddigitthatwillthenbecomea5.Finalresult:5650

Withtheclassicmethodwelearntatschoolasubtractionlikethiswouldhaverequiredamuchlongerandmore complex procedure.But here you can dealwith the problem just through a few, simplesteps,limitingsoeventheprobabilitytorunintoanerror.

Butlet'sextendtheSutranow,andlet'suseittoperformdifferentkindsofsubtractions,too.

Forexample,youcanuseittofasterperformsubtractionsbetweennumbershavingthesameamountofdigits,andwhoseminuend(thenumberbeforetheminus)ispositiveandmadeonlyupofzeroesexceptfortheleadingdigit.So,forexample,youcanapplyittosubtractionslike"500-388","6000-4567",or"70000-43200".Infactinthiscaseyoucan:

ApplytheSutra"All from9andthe last from10" inorder toobtain thedifferencebetweenthenumberanditsgreaterpowerof10.

SubtractfromtheresultoftheSutrathedifferencebetweenthatpowerof10andyourminuend.So,ifforexampleyourminuendwas3000,thenyouwillhavetosubtract7000fromtheSutratoobtainyourfinalresult.Ifyourminuendwas500,thenyouwillhavetosubtract500,andsoon.

ButanotherexampleofextensionofthepossibleapplicationsoftheSutracouldbeinusingittofindthedifferencebetweenanumberandthepowerof10havingonemorezerothanitsamountofdigits.So,forexampleyoucanuseittocalculate1000-28,10000-345or100000-6432.Inthatcase,infact,youwilljusthaveto:

ApplytheSutratothesubtrahend.

Addtotheresultanumbermadeupofa9andasmanytrailingzeroesastheamountofdigitsofthesubtrahenditself (andso,forexample,900if thesubtrahendhastwodigits,9000for threedigits,andsoon).

It'sprettysimple, isn't it?So, if forexampleyouwant tocalculateyour1000-28youwillhave to

applytheSutrato28,obtaining72,andthenbanallyadd900:972!

At thispointyoumayhavenoticed that those last twoexamplesofSutraextensionscamebasicallyfromthesamelineofreasoning,whichis,givenanykindofa-b,youcan:

ApplytheSutratob.

Findoutwhat'sthehigherpowerof10forb.

If"a"exceedsthatpowerbyaspecificamount,addthatamounttothefinalSutraresult.

If"a"islesserthanthatpowerbyaspecificquantityinstead,subtractthatquantityfromtheSutraresult.

So...1002-456?ApplytheSutrato456andadd2totheresult!

20004-3498?ApplytheSutrato3498andadd10004!

9998-4432?ApplytheSutrato4432andthensubtract2fromtheresult.

The first practical application of these strategies could consist in using them to rapidly calculateexpenses, starting from an initial budget, which in fact is very often a "rounded" number. Forexample, if before starting a project you had 7000 dollars on your bank account and now, after amonth,only1288dollars remain,youcanapply theSutraand thensubtract3000fromits result inordertorapidlycalculatehowmuchdidyouspend.So,inthiscase,theexpenseswere8712-3000=5712dollars.

ButthiswasjustanintroductiontothetopicandtheapplicationsoftheVedicmathematicsofcoursedon't end here. In fact, in the next pages, itwill come in handywhenwe'll start talking about fastmultiplication.

VI-Assemble,decomposeandBlackjack

This chapter will teach you three basic techniques for more rapidly performing additions andsubtractions.Then,at theendof theexplanations, itwill showyouadefinitely interestingpracticalapplicationforallofthem.

Inaddition,I'mprettysureyou'veusedsomeofthesestrategiesinpastyet,evenif inan"intuitive"and "automatic" way. But, anyway, making you fully aware of them will help you to exactlyunderstandwhentousethem,andsotounleashtheirfullpotential.

So,let'sstartexplainingthefirsttechnique,whichisverysimpleanduseful,especiallyifyouhavetoadd (or subtract) amongmany small numbers. It banally consists inpre-emptively summing up theaddends,orthesubtrahends,inordertogetmultiplesof10.

Forexample,let'ssupposethatyouhavetocalculatea3+4+5+6+7+2+5+1+3+4+3

Thefirstthingyoucandoiscombiningthefirst"three"andthefifth"seven",getting10+4+5+6+5+1+2+3+4+3.

Now we take the penultimate "four", the last "three", then the third last "three". After beingsummedtogethertheyequalten,sowe'llhave20+4+5+6+2+5+1.

Again,afteradding"four","five"and"one"Igetathird10,andthen30+6+2+5=43.

Apart from the obvious practical simplicity, this method has even the advantage to be essentially"creative",especiallyforchildren.The"Questfortheten",infact,canalsobefunandeasilyhelpthechildrentohavefunwiththisspecificstrategy.

But let's rapidlygo to thesecond technique,which isverysimilar to thepreviousone,canbeusedwithadditionsorsubtractionsofanykind,andconsists in"borrowing"anamountfromoneof theaddends(orsubtrahends)andaddingittoanotheroneoftheminordertotransformthelatterintoamultipleof10.

Let'sassume,forexample,wewanttocalculatea368+214.Onethingthatyoucanimmediatelydohereistosimplifythecalculationsbyborrowinga"2"from214andcarryingittothe368.Soyou

willgeta"370+212"thatcanbecalculatedintwoseconds.

Samethingifwehavea967-255.We"borrow"a3fromthe255,gettingamucheasier970-258.Asyoumayhavenoticed,sincethisisasubtraction,heretheruleschangealittlebit:infact,anyquantityborrowed from the subtrahend must then be added to the subtrahend itself, while any quantityborrowedfromtheminuendmustthenbesubtractedfromthesubtrahendinstead.So...becarefulinthiscase!

Offtopic:inordertoperform970-258youcoulduseaSutraasintroducedinthepreviouschapter,and thisproves that therealpointofstrengthof the"tools"contained into thisbook is thefact thatthey can be freely recombined in order to create appropriate calculation patterns for any kind ofsituation.

The "borrowing rule" also lets us build these easy simplification rules.They of coursemay seemobvioustosomebody,butit'salwaysagoodpracticetorefreshinyourmindevenwhat'sobvious,sothatyouwillbeabletoevenmorequicklyrecallitfromyourmemoryanytimeit'sneeded(exactlyaswesaidinChapterII).

Ifyouwanttoadd9,justadd10andthensubtract1Ifyouwanttoadd8,justadd10andthensubtract2Ifyouwanttoadd7,justadd10andthenremove3Ifyouwanttosubtract7,justsubtract10andthenadd3Ifyouwanttosubtract8,justsubtract10andthenadd2Ifyouwanttosubtract9,justsubtract10andthenadd1

Andofcoursethesamereasoningcanbeappliedtonumberslike19,90,80,900,800,andsoon.

So,let'sintroducethelasttechnique,that'sverysimpleaswell,andwillcomeinveryhandyanytimedealingwiththree-or-more-figurenumbers:justdecomposeoneoftheoperandsintoasumamongitsunitsplusitstens,plusitshundreds,andsoon...andthenoperateseparatelywiththeseparts!

So, when for example you have to calculate a 3456 - 1234, will be much easier for you if youdecompose1234into1000+200+30+4andthencalculate,insequence:

3456-1000=

2456-200=

2256-30=

2226-4=

2222

The humanmind in fact is muchmore rapidly and easily able to perform a longer set of simplecalculations than a single and more complex one. And, as we will see in future, this is the basicprinciplealotofspeedmathstrategiesarebasedon.

Now that we explained the three fundamental speed math strategies involving addition andsubtraction, let's introducea commonlyknownandquite interestingpractical application for them:theBlackjack"cardcounting".

Ifyoudon'tknowwhatI'mtalkingabout,let'sstartwithalittlepremise:everygamblinggameisbuiltinsuchawaythat,inthelongterm,wewillalwaysloseourmoneywhiletheCasinowillalwaystakeeverything dropping out of our pockets. This means that every gambling game has a negative"expectedvalue",aswewillexplainmoreindetailinChapterXIX.

Theonlyexception to this truth is in theBlakJack and in the fact that, if approachedwith the rightstrategy("countingcards",forexample),itcanhaveapositiveexpectedvalueandsoitcanleadustoaprogressiveenrichmentinthelongterm.

Furthermore,Blackjack is a very simple game: you, the dealer and the other players are given aninitialtwo-cardshand.It'seachplayeragainstthedealer.Facecardsarecountedastenpoints.Aceiscounted as one or eleven points depending on the player's choice and every other card keeps its"standard"value.

So,inturn,eachplayercanchoosewhetherto"stay"withhis/herhandortohaveanothercard,tryingtoget as closer as possible to 21points, butwithout exceeding this value. In fact,who exceeds21pointsautomaticallyloseshis/herbet,whichgoestothedealer.

So,aftereachplayerstill ingamedefinedhis/herscorebydecidingto"stay"withhis/herhand,it'sthedealer'sturn.Atthispointhe/shewillhavetomakethesameastheotherplayers,decidingcardafter card if continuing to draw or keeping the hand. But the only difference here is that he/she'sforcedtocontinuedrawinguntilhedoesn'tmakeascoreof17atleast.Andifhe/shewillexceed21,he/she'llhavetopaytoeveryplayerhis/herbet.Otherwisehe/shewillsimplyhavetopaythebettoanyplayerwithahigherscorethanhis/hers,andtakeitfromthosehavingalesserscore.

Now that I'm sure you know what I'm talking about, let's introduce some good news about cardcounting:

Countingcardsissimple,orbetter,we'lltalkaboutaverysimplecardcountingmethod:theso-called"High/Low".Countingcardsisperfectlylegal.If you have been traumatized by someHollywoodmovie inwhich the peoplewho have beendiscovered counting cards were hardly beaten up, you can sit and relax because it's justHollywoodfiction!

Andnow,thebadnews:

Even if anybodywill unlikely touchyouwhenhe/she findsout thatyou're countingcards, it'sseriouslypossiblethathe/shecouldkindlyinviteyoutoleavetheCasino.YouwillneverbeabletocountcardswhenplayingatsomevirtualCasinos:thevirtualdecksarevirtuallyandrepeatedlyshuffled,withnochancetobuildanywinningstrategy.In the "physical" Casinos they actually do everything in their power to reduce your positiveexpectedvaluetotheminimum.

So,thequestionherecouldbe:whyshouldIeverstillbeinterestedinthiskindofactivity?Well,Icanthinkabouttwoanswers:

Becausetrainingyourcalculationskillsbygamblingisinfinitelyfunnierthandoingitbyreadingbooks.

Becauseifyouflylowanddon'texceedinusingthistechnique, insomelittleCasinos it'sstillpossibletoearnsomelittlecashwithoutanyrisksortoomucheffort.

Let'sstart,then!Let'simagineyou'resittingattheBlackjacktablewithamojito(hopingit'syourfirstoneoryourmathematicalskillsdefinitelywon'tbeattheirbest).Wewillstartourcountfrom0and,aslongasweseethedealerdrawingthecardsfromthedeck,wewilldolikethis:

Whenyouseea2,3,4,5,ora6,thenadd1toyourcountWhenyouseea10,Jack,Queen,KingoranAce,thensubtract1fromyourcountWhenyousee7,8or9,thenleaveyourcountasitis

Asyoucanimmediatelynotice,sureit'saverysimplecount,butitmustnecessarilybedoneveryfast,andinordertoachievethisyoucouldusethefirsttwostrategiesintroducedinthischapter.

So, as far aswekeep counting,wemust consider that a high total (+6/+9) reveals that the deck isfreightwithhighcardsandfiguresand,asaconsequence,itmeansthatwehaveastrategicadvantageagainstthedealer(whoinfactwillmoreeasilyexceed21).Alowtotalinsteadrevealsthatwehaveahigherprobabilitytoloseagainsthim/her.

More in detail, this is an example of a both prudent and effective strategy. Decide how much areasonably"basic"betisand:

Countisnegativeorequalto0:don'tbetCountis+1:betyour"basic"betCountis+2or+3:betthedoubleofyourbasicbetCountis+4or+5:betthetripleofyourbasicbetCountis+6or+7:betthequadrupleofyourbasicbetCountis+8,+9orhigher:betthequintupleofyourbasicbet

So, here ends our digression about this eternally fascinating Casino game.Many more things ofcoursecouldbesaid,butunfortunately this isnot the rightplace.Anyway,never forget thatyou'reluckyenoughtobeborninthedigitalinformationage:justasimpleGooglesearchandyou'llbeableto find a lot of interesting tricks to maximize the probability to beat the dealer while playingBlackjack.Goodluck!

VII-Themagiccolumn

Let'srepeatsomethingveryimportantwesaidinChapterII:acalculationstrategybecomesasfasterandmoreefficientasit'sdesignedtoletthebrainstoreaslessinformationaspossible.

Let'ssuppose,forexample,wemustperformthisaddition:

989+724+102+670+112=

If we had to calculate it through the classic column method we traditionally learnt at school, weshould separately sum the units first, then the tens and the hundreds at last, trying to bear inmindeverypartial amountcarried toget the final result.And trying todo thatwithoutanypenorpaperwouldbetotallyfrustrating,whennotimpossible.

So,as inmathematicsas in life,whensomethingdoesn'tworkproperlywehave todonothingbutchangingourstrategy,andhereiswherethe"Magiccolumnstrategy"comesinourhelp.Buthowcanweapplyit?Let'sseeit:

Insteadofstartingfromtheunits,startfromthefirstdigitontheleftandsumallthenumbersontheleftcolumntogether,makingsureyoumentallyrepeat,whileyoucalculate,onlytheresultsofthepartialsums.Soforexample,ifyoulookattheabove-mentionedaddition,youshouldmentallyperform9+7,thentheresult+1,then+6,etc.,butonlysayinginyourmind"9,16,17,23,24"(andheretofasterperformallthepartialsumsyoucaneasilyusethetechniquesintroducedinthepreviouschapter).

Onceyoufinishedperformingthesumsontheleftcolumn,yourpartialresultis24.Nowmovetothenextcolumnontheright,takeitsfirstdigit,putitnexttoyourpartialresult(sonowyou

will have 24_8) and, without touching the result of the previous column, add to the secondnumberofyourpartial result theotherdigitsofyournewcolumn,exactly likeyoudid in thepreviousstep.So,repeatingthesameprocedureasbefore,youwillhave:"24_8,(afteradding2)24_10,(afteradding0)24_10,(afteradding7)24_17,24_18".Andonlyif,likeinthiscase,afterperformingyoursumsyougetanumberthat'sgreaterthan10,putinthenewpartialresultonlytheunitsandsumthetenstothenumberontheleft.So,inthiscaseyourpartialresultis24_18=25_8.

Moveagaintothenextcolumnontherightandputitsfirstdigitnexttoyourpreviousresult(inthiscase,so,youwillhave25_8_9).Then,behaveexactlyasyoudidintheprevioussteps:don'ttouchthepreviousresult,andsumthedigitsofthenewcolumnstothenumberontheright.Soyou will have: "25_8_9, 25_8_13, 25_8_15, 25_8_17 = 25_9_7 (since we removed the tensdigit)".Therearenomorecolumns,soyoucanremovetheunderscoresand2597isyourfinalresult.

Sowedefinitelygottheresultofalongsuminanextraordinarilyfastandefficientway,keepinginmindonlyaverysmallamountofdigitsandwithoutpayinganyattention to thepartialamounts tocarry.Ofcourse,incasewemustputincolumnnumbersmadeupofadifferentamountofdigitsonefromeachother,thenweshould:

Aligntheunitswiththeunits,thetenswiththetens,thehundredswiththehundreds,andsoon.Using the same technique as the empty spaces (that inevitably will be on the left side of thesmallernumbers)wereinsteadleadingzeroes.Wecouldalsowritethosezeroes,ifitmakesthethingseasierforus.

Butlet'shaveanotherexampleandlet'ssaywehavetocalculate1341+450+2451+888+9872.Asafirst thing,wemust align the numbers in a column and, tomake the things easier,we canwrite aleadingzeronexttoeachthree-figurenumber:

1341+0450+2451+0888+9872=

Startingfromtheleftwehave:"1,3,12".Soourpartialresultis12.Aftergoingtotherightwehave:"12_3,12_7,12_11,12_19,12_27".Soourpartialresultisnow147.Aftergoing to therightagainwehave:"147_4,147_9,147_14,147_22,147_29".So theotherpartialresultis1499.Afterworkingonthelastcolumnwehave:"1499_1,1499_2,1499_10,1499_12".Aftercarryingthe1twice,thefinalresultwillsobe15002.

Now,showthisnewabilitytoyourfriends,ofcoursewithoutexplainingthestrategybehind!I'mquitesurethat,ifyoutrainyourselftoperformanypartialsumrapidlyandwithoutmakinganyerrors,the"sceniceffect"willbeabsolutelyawesome!

Doyouwanttopasstothe"nextlevel"?Trainyourselfto:

WorkevenwithoutaligningthenumbersWorkontwoormorecolumnsatthesametimeinordertoperformevenfastercalculations.

Butlet'sexplainthesecondpoint.Let'simaginewehave:

561+343+912+134+451=

Weknowhowtooperatecolumnbycolumn.Butifwewanttomakethethingsevenfaster,wecould:

Startfromtheleftasbefore,butnow,asafirststep,sumdirectlythetwo-figurenumberstakenfromthefirsttwocolumns.Andsobeginningwith"56,(+34=)90,(+91=)181,194,239"Behaveasbeforeonthelastcolumn,calculatingso:"239_1,239_4,238_6,239_10,239_11".Thefinalresult,infact,is2401.

This of course is just the next step, and you'll be confident with it only after properly trainingyourselftorapidlyperformanytwo-figurenumbersums(eventhankstothestrategiesweintroducedinthelastchapter)andthe"single-column"versionofthistechniqueaswell.

"Mathematicsislikecheckers,inbeingsuitablefortheyoung,nottoodifficult,amusing,andwithoutperiltothestate."

(Plato)

VIII-Gametheorypills

Now, since we just learnt quite almost anything about fast addition and subtraction, let's start tointroduce an extremelyuseful, interesting and fascinatingpractical applicationof these operations:let'sstarttalkingaboutgametheory.Game theory, for thosewho don't know it, is nothing but amatter studying any kind of situationsinvolving people making decisions. It's an extremely large and complex argument, and so we'redefinitelynotgoingtohaveacompletedissertationaboutit.Anyway,givingyousomehintsaboutthetopicwillcomeyouinextremelyhandyanytimeyouwillfacesituationsinwhichyou'llconfrontwithotherpeoplemakingtheirownchoices,potentiallyinconflictwithyours.More in detail, I'll give you mostly some hints about the so called "2 x 2 games": all kinds ofsituationsinwhichthedecidersaretwo,andthesetwopeoplecanmaketheirdecisionsinarangeoftwospecificchoices.AndI'lldolikethatasbecauseofthesimplicityofthecase,asbecauseit'saquiterecurringcircumstance,aswe'llsoonseeinourexamples.Butlet'sstartintroducingsomegametheorynotation,takingasanexamplethecanonical"Prisoner'sdilemma":"Twomembersofacriminalgang,AandB,arechargedandarrestedbecauseofanimportantbankrobbery.Eachof themcan't speak to theotherone.Thepolice anywayhaven't enoughevidence toconvict both of them on the principal charge, and so the officers try persuading them to confess,makingbothofthemanoffer:"

Ifbothconfess,eachofthemservesjustoneyearinprison.IfAconfesses,butBdenies,AwillbesetfreeandBwillserve10yearsinprison.Ifbothdeny,eachofthemwillserve5yearsinprison.

Now,let'simaginewe're"A",andlet'strytoschematisethissituationthroughatable:

Aswecan immediatelynotice, in this table the rowsareourpotentialchoices, thecolumnsare thepotentialchoicesoftheotherperson,whilethecorrespondingcellindicatestheyearsofprisonwe'llserveincasethosechoicesaremadeatthesametime(andthe"minus"signisduetothefactthattheyearsofprisonareobviouslyconsideredasa"handicap"score).So,giventhiskindoftablerepresentation,it'stimetounderstand,throughsomemathematicalhelpofcourse,whichchoiceispotentiallythemoreconvenientforus.Morespecifically,thefirstthingtodowouldbetryingtounderstandwhetherwecouldsuccessfullyadoptsomethingcalled"apurestrategy".Inotherwords,wemustevaluatewhetherthereisachoicethat'sALWAYSmoreconvenientthantheotherone,nomatterwhatour"opponent"does.Buthowdowerecognizethechanceofsuccessfullyadoptingapurestrategy?Well,it'sverysimple:wecanadoptitincaseEACHvalueonaspecificrowisbiggerthanthecorrespondingvalueontheotherrow.In this case, for example, the pure logic indicates that confessing is always more convenient thandenying, considering that 0 > -1 and -5 > -10. So, confessing is universally the best choice tominimizeourrisks,nomatterwhattheotherprisonerisgoingtosay.Butlet'stakeanexampleof"game"withoutanychanceofpurestrategiesand,inordertounderstandwhat's the best thing to do in that case, let's examine a quite recurring situation during the pokermatches:callorfold?2000dollarsareinthepot,whose300dollarsareourlastbet.Wechangedtwocardsinourhandyet.Ourlastopponentingameraisesby1000dollarsandourhand,afterchanging,isgoodbutnothingextraordinary.So,dependingonwhetherouropponentisbluffingornot,wemustdecidewhethertocallhisbetornot.Let'sputthissituationinour"gametable":

Inthiscase,aswemayimmediatelynotice,therearenovaluesonanyrowthatareclearlybiggerthanthecorrespondingvalueontheotherrow(2000>0,but-1000<300).So,thebestwaytoactinthiscasewouldbe:

TakethehighervalueineachrowSubtractthesmallestvalueineachrowfromthehigheroneinthesamerow.CallthisvalueDV(DifferenceValue),andputitnexttotheOTHERrow.

So,inthiscaseyou'llhavetogotothe"call"row,calculate2000-(-1000)=2000+1000=3000,andputitnexttothe"fold"row.Atthesametimeyou'llhavetogotothe"foldrow",calculate300-0=

300andputitnexttothe"callrow".Thenyou'llhave:

Now,whatdothosevaluesmean?Simplythat,giventhisspecificsituation,foldingcanbeasituationstatistically ten times more convenient than calling.More in detail, if we write the two DifferenceValuesrespectivelyasanumeratorandasadenominatorof thesamefraction,andwesimplify thisfractionuntil the smallest value is1 (which is simply the same as dividingonevalueby theotherone),wegetanevaluationof "howmany timesonechoice isgoing tobemoreconvenient than theother one". Which, returning to our case, in which one choice is specifically 10 times moreconvenientthantheotherone,means:

If it's a recurring situation with exactly the same values, take the two choices in a 10-to-1proportion.Whichmeans, take in average the "10" choice 10 times, the "1" choice once andrepeatituntiltheothervariablesdon'tchange.

Ifit'saone-casesituation(whichislikelynearesttorealityinthepokercase),sumthetwoDVs(inthiscase,10+1=11)anddothe"10"choicewithaprobabilityof10outof11,anddothe"1"choicewith a probability of 1 out of 11.This could simplymean "just take the choicewith ahigher DV". Or, in case you want to do something mathematically more accurate (which isadvised in case the ratio between the two choices is not so clear), use any tool capable to"extract"a randomnumber inorder to let itbe inaccordancewithyourexactprobability.Oh,andyoudon'tnecessarilyneedacointofliporadietorollhere:forexample,youcouldsimplylookatthesecondhandofyourwatchandthencallonlyifit'spointingatthefirstsixsecondsofthecurrentminute.Infact,ifthesecondsinaminuteare60,theprobabilityweareinthefirst6secondsareexactly6outof60=1outof10,whichisquitenearto1outof11.

Ifthelastreasoningisstillnotverycleartoyoudon'tworryaboutit,becausewewillexplainmoreaboutprobabilitytheorylaterinthebook.Evenbecause,obviously,allofitstartswiththeassumptionthatyoucan'tdecipheryouropponent'sbehaviouratall(aswashappeningintheprisonerdilemma,the fundamental premise is that there is no chance of "information leak" between the twocounterparts).Butif,forexample,it'sknownthatyouropponentlikesbluffing,youcanrefineyourwayof thinking throughprobabilitystrategieswe'llsee later.Bynow,anyway, justbeawareof thefactthatthatthechoicewithahigherDVhasahigherprobabilitytobetherightchoice.

Ofcoursenowsomebodymaynoticethat,incaseofpokerpotsoryearsofprison,theprocedureissimplifiedbyhavingveryclearquantities involved in the situation,while the reality ismadeupofhues and unknown variables. Well, in case of not clearly definable or computable quantities, theanswer is in something we said in Chapter II: try to give a mark to what could happen as aconsequenceofspecificchoices.Ofcoursethiscouldnotbeveryaccurate,but it'sstillmuchmoreadvantageousthanacompletelyblinddecisionmaking.

For instance, let's supposeweare inasituation that'scompletelydifferent fromallof thepreviousones:wemustdealwithanimportantoralexamatouruniversity.Wehavereallyalittletimetostudy,andwecanchoosetostudyonlyonebetweentwoparticulararguments:averysimpleoneandavery

difficultone. Inaddition to this,we'requitesure thatourprofessor isgoing toaskusonebetweenthosetwoarguments.So,let'strytogiveamark,from1to10,toanypotentialsituations,like:

Westudythesimpleargumentandtheprofessorasksusaboutthedifficultone:1,becauseinthiscasewewilljust...makeaverypoorfigure!Westudythesimpleargumentandtheprofessorasksusaboutthesimpleone:9,becauseinthiscaseourexaminationwillquitelikelygoverywell.Westudy thedifficultargumentandtheprofessorasksusabout thesimpleone:4,becauseourpoorfigurecouldbesoftenedbythefactthatourprofessorcanappreciatethewaywededicatedourtimetosomethinghardertolearn.Westudythedifficultargumentandtheprofessorasksusaboutthedifficultone:8,becausewecould,despiteeverything,showsomedoubtduetothefactthatthatspecifictopicisquitehardtolearn,difficulttohandleandnotverysimpletotalkabout.

So,our"gametable"couldlooklikethis:

Herethereisnochancetoadopta"purestrategy"aswell.So,let'scalculateourDV:

9-1=8and8-4=4.4/8=1/2.So,the"difficultargument"choiceistwicemoreconvenientthantheotherone.Thatmeans,consideringthatthisisnotarecurringsituation,thatwecanjuststudythedifficultone.Otherwise,forsomethingmoreaccurate,wecansumthetwoDVs(4+8=12),andtakethe4choicewithaprobabilityof4outof12(4/12=1/3),andthe8choicewithaprobabilityof8outof12(8/12=2/3).Andthiscanbeeasilydonebytossingadie,anddecidingtostudythesimpleargumentif1or2comesout,andthedifficultargumentotherwise.I'llrepeatit:iftheprobabilityquestionisnot

verycleartoyounow,don'tworryaboutit.Itwillbewhenwe'llintroduceitintheChapterXIX.Fornowjustbeawareofthegiveninstrumentsandofthefactthatinanymomentofwork,study,leisureorrelax,you'lllikelyface"2x2games"circumstances.Andso,learninghowtotakethebestoutofsituationslikethesemeanslearninghowtotakethebestoutofanenormousamountofsituationsinyoureverydaylife.

Butbefore leaving thechapter, let'sput aside the "2x2"games inorder to takea look toawidertopic.Intheearly60s,amathematiciannamedJohnNash,youprobablyhaveheardaboutbecauseofthemovie "ABeautifulMind", formalisedoneof themost importantmathematical conceptsof thepast century. Nashwas living in a society thatwas grown andwas prospering according to thoseeconomicprinciplesexpressedbyAdamSmith,tellingthatthemaximumabundanceinagroupcanbeachievedonlywheneveryoneputshisbesteffortintopursuinghisownpersonalprofit.However,incontrastwiththat,afteranalysingandstudyingtheevolutionofcomplex"games"inwhichevenlotsof decision-makers take decisions from within a wide range of choices, Nash mathematicallydemonstratedthat"themaximumabundanceinagroupcanbeachievedonlywheneveryoneputshisbest effort intogiving thebiggest advantageas to himself as to theothers "players".Hemade thatsmall,butessentialadditiontowhatSmithsaidandfoundthemathematicalproofthatinanystrategicsituationthebiggestprofitforeveryonecanbegainedonlybyadoptingawider,group-orientedpointofview.Andthetruthfulnessofthisconcepthasbeenactuallyproveninthefurtheryearsbynumerousstudies,experimentsandsimulations.

The "golden rule" one can learn from all of this is actually very obvious and seemingly trivial:wheneveryoutakeanydecision,learntothinkandactintermsofwin-winresolutionforyou,allthepeopleinvolvedandthewholeenvironmentinwhichyouare.Ofcoursethisisnotalwayspossible,butifyouatleasttrytowidenyourperspectivethisway,you'lldefinitelyhaveagreatgainintermsofhappiness,successandpersonalsatisfaction.Dishonesty,ontheotherhand,enrichestheindividualbutpauperisestheenvironmentinwhichhelives.Anunbridledegotismcanboosttheoverallcapacitiesofasinglehumanbeing,butatthesametimeitlimitsthedevelopmentoftheotherpeoplehehastodealwith.Andtherightabundanceandastrongself-developmentaresterile,uselessandfragileforanybody if he's lacking of a proper human, social and structural environment that can adequatelysupports all of these things.Cooperationandwin-win-orientedprospective after all seem tobe theunique, realkey to the future for anyhumangroup, from the smallest familyuntil thatquite largecrowdlivingonthisplanet.It'smuchmorethanasetofbanal,orevenrhetoricalrulesofcommonsense:it'ssomethingasrealandundeniableasthelawsdefiningtheorbitsoftheplanetsaroundthesunortheaccelerationoffree-fallingobjects:it'smathematics!

IX-Smash,simplifyandsave

Inthischapterwe'llstartalongandcompletedissertationaboutfastmultiplicationandfastdivision.Asyou'llinfactnoticebyyourself,allthemathematiciansandthespeedmathstudentsgaveastrongimportancetothesetwooperations,developingaverybigamountofstrategiestocalculatetheminaveryfastandefficientway.And,aswesaidbefore,biggerthechoiceamongthestrategies,greaterthefreedomtoapproachtheproblemsolvingprocessinanentertaining,recreationalandcreativeway.

Inadditiontothis,multiplicationsureisacriticalandfundamentaloperationintotheeverydaymath:justthinktoitsimportanceinbasiceconomics,inpercentagecalculations,or(aswe'llseesoon),inanystrategicdecisioninvolvingprobabilitytheory!

Sonow,beforemoving towardour strategies, Iwant togiveyou justone, last advice:don't try tolearn all of them at the same time, but read them slowly, exercise yourself and try to repeat myexampleswithpenandpaper,tryingatthesametimetounderstandwhichkindofstrategiesaremorefittingforyou.

Now,let'spickupaprinciplewesaidinthepreviouspages:yourmindcanmuchmoreeasilyperformalongersetofeasiercalculationsthanasmallersequenceofmorecomplexcalculations.Infact,thepurpose of the first three techniques we'll introduce will be exactly about decomposing a hardermultiplicationintoasequencemadeupofverysimpleones.

Andthefirstofthesethreetechniquesisthestraightforward"decompositionintoaddends",whichcanbeusedonlyonmultiplicationsandcanbeappliedlikethis:

Decomposeonebetweenthemultiplicandsintoasumorsubtraction.Typicallytheeasiestthingto do is decomposing it into a sum between its units plus its tens plus its hundreds, etc. (forexample:456=400+50+6)

Multiplyseparatelythesepartsusingthedistributivepropertyofaddition.So,ifforexampleyouhave334x456,thiswilltransformyourmultiplicationinto334x(400+50+6)=(334x400)+(334x50)+(334x6).So, instead of performing a multiplication of three-digit numbers that, if made through theclassiccolumnmethod,wouldhaverequiredaconsistentamountoftime,you'llhavetoperform

asimplesumofsingle-digitmultiplications,having,asanonlyannoyance,toaddsometrailingzeroes(youdidn'tforgethowtoperformthemultiplicationby10,100,1000,right?).I clearly could also decompose the 334 into 300 + 30 + 4 and so transform the originalmultiplication into (456 x 300) + (456 x 30) + (456 x 4). The commutative property ofmultiplicationinfactsaysthatnomatterwhichnumberweapplythedecompositionto,becausethefinalresultwon'tchange.Andthat'swhyit'suptous,fromtimetotime,tochoosethekindofdecompositionthatcanleadustothemoreimmediatesetofmicro-operations.

Pleasenoticethatthedecompositionintounits,tensandhundredsisonlyoneoftheoptions: in fact,anynumbercanbedecomposedintoinfinite,differentways.Ifforexampleyouwanttoperform44x7,theninsteadofdoing44=40+4,youcouldapplythistechniquebydecomposing7into(10-3),andthus44x7=44x(10-3)=(44x10)-(44x3)=440-132=308.Anddecomposingintoasubtraction instead of a sum is a strategy that greatly simplifies your life whenever you have tomultiplybetweennumbersthatarecloseenoughtoamultipleof10.

Butlet'shaveanotherexampletoclarifythisconceptonceandforall:ifyouhavetomultiply400x59,insteadofdecomposingthe59into(50+9),youcandoitas(60-1),andso400x59=400x(60-1)=(400x60)-(400x1)=(400x60)-400=24000-400=23600,whichwasmucheasiertoperformthantheotheroption.

So,let'srapidlygotothesecondtypeofdecomposition:thefactorization,whichcanalsobeusedforthedivisionandrequiressomeextrapremises.

Firstofall,rememberthatdecomposinganumberintofactors(orfactorizing)meanstransformingitinto amultiplication among smaller integers,whose result equals the original number. In addition,decomposingitdownintoitsprimefactorsmeanstransformingitintoamultiplicationamongprimeintegers that returns the first number as a result. And, as you surely know, a prime integer is anintegerthat'sdivisibleonlybyoneandbyitself.

For example, you decompose 16 into factors when you say that 16 is equal to 4 x 4. But youdecompose16intoprimefactorsinsteadwhenyousaythat16isequalto2x2x2x2.Infact,2isaprimenumber,while4it'snot.

Now,justthinkabouttheadvantagesofbeingabletodecomposeanumberintofactors,evenbetterifprime.Infactif,forexample,youhavean88andyouhavetomultiplyitby16,youwillhavethat88x16=88x(2x2x2x2).Theassociativepropertyofmultiplicationtellsyouthatyoucanperformthesetasksinanyorder,andsomultiplyingby16willbeequivalenttothemuchmorerapidoperationofdoublingyouroriginalnumberfourtimes.

Andasimilarthinghappensifyouwanttodivideanumberby16,forexample256.Infact,256/16=256/(2x2x2x2).Thedivisiondoesnothaveassociativepropertyandthereforeweareobligedtoproceed by applying the simple properties of the fractions and transforming the above expressionintoanequivalent((((256/2)/2)/2)/2),thatsimplymeansthatyoumustdivide256by2fourtimes.Similarly,ifyouwanttodivideanumberby96,and96=2x2x2x2x2x3,youshouldfirsthalvethenumberfivetimesandthendivideitby3.

Inotherwords,dividinganumber "a"byanumber "b" is equal todividing "a"by the sequenceoffactorsofb.

Ofcoursethefactorizationisnotanimmediateprocedure,especiallyincaseoflargenumbers,andthatiswhyitbecomesnecessarytolearn,oratleastbrushupon,theso-calleddivisibilitycriteria.

Giventhatmultiplicationistheinverseoperationofdivisioninfactyoucansaythat,ifanumber"a"isdivisiblebyanothernumberb,thenyoucanfactorize"a"intoaproduct"bmultipliedsomething".Furthermore,it'sobviousthatthismysterious"something"isexactlyequaltoa/b.

Forinstance,supposeyouwanttofactorize256step-by-step:

Afterfindingoutthat256isdivisibleby4,youcanimmediatelysaythat256canbefactorizedintoa"4multipliedsomething"Youcanmorespecificallygetthat"something"byperforming256/4=64.So,4x64=256.Youknowthat4isdivisibleby2andthatit'sequalto2x2.Even64isdivisibleby2,andbeingthedoubleof32it'sobviouslyequalto32x2.Therefore,byreplacingthe4and64withtheirdecompositionsIgotthat256=4x64=2x2x2x32.And so on, stopping your factorization when you get all prime factors, or simply when thedecompositionwillallowyoutoworkinanenoughsimpleway.

Clearlyafterobservingthisprocedureitcouldbeimmediatelydeducedthatitnolongermakessensetoapplyitwhereafactorizationstartstobetoolongortoohardtoperform.Andthat'swhyknowingthedivisibilitycriteriawill letyouoperatewithgreatersimplicity,andevenhelpyoutounderstandwhenthemoreappropriatethingtodoisdiscardingthemethodatall:

Criterionofdivisibilityby2:Thisisprobablythesimplestandmostwellknown:anumberisdivisibleby2onlyifitsunitsdigitis0,2,4,6or8.

Criterionofdivisibilityby3:Anumberisdivisibleby3if,aftersummingitsdigits,yougetamultipleof3.Takeforexample476.4+7+6=19.If19isamultipleof3,sowillbe476,andtocheckthelastconditionyoucanapplythesamestrategyagain:1+9=10.10isnotamultipleof3andthenneitheris476.

Criterionofdivisibilityby4:Anumberisdivisibleby4ifitslasttwodigitsare00oranytwo-figurenumberdivisibleby4.Takeforexample56000932.Icantellimmediatelythatit'sdivisibleby4becauseitendsin"32"and32isdivisibleby4.

Criterionofdivisibilityby5:Anumberisdivisibleby5ifitsunitsdigitis0or5.

Criterionofdivisibilityby6:Anumberisdivisibleby6ifitmeetsthecriterionofdivisibilityby3ANDit'seven.

Criterionofdivisibilityby7:Anumberisdivisibleby7ifisdivisibleby7thenumbergivenby("numberwithoutitsunits"+"unitsdigityouremoved,multipliedby5").So,supposeyouwanttounderstandwhetherthenumber8422isdivisibleby7:you'llhavetoseeif isdivisibleby7 the sumof842 (whichofcourse is8422without itsunits)+ (2x5) (unitsremovedmultipliedby5).842+2x5=852.Tounderstandnowif852isdivisibleby7youcan

applyagainthesamerule:(85+2x5)=95.Then,again(9+5x5)=34,whichisnotdivisibleby7.Thenneitherwas8422.

Criterionofdivisibilityby8:Anumberisdivisibleby8ifthenumbermadeupofitslastthreedigits isdivisibleby8.Or, if this is toodifficult,youcanuse thisalternative strategy: take itsthirddigitstartingfromright,doubleit,addittothepenultimateone,doubletheresultandsumittothelastone.Ifthefinalresultisamultipleof8,sowillbetheoriginalnumber.So,ifforexampleyouhave865341,takethe3,doubleitbygettinga6andaddthelastresultto4,getting10asaresult.Afterdoubling10youwillgeta20andafteraddingittothelast1ontherightyouwillget21.21isnotdivisibleby8,then865341isnotdivisibleaswell.

Criterionofdivisibilityby9:Anumberisdivisibleby9ifthesumofitsdigitsisdivisibleby9.

Criterionofdivisibilityby10,100,1000:Anumberisdivisibleby10,100,1000oranyotherpowerof10ifithasanamountoftrailingzeroesequaltothoseofthepowerconsidered.

Criterionofdivisibilityby11:Startingfromrightandgoing left, sumtogether thedigits thattakeupanevenplaceinyournumber.Thendothesamewiththedigitsinanoddplace.Subtractthe smaller from the larger and if the result is 0, 11 or a multiple of 11, then the examinednumberisalsodivisibleby11.So,forexample,wehavethat8291778isdivisibleby11because:-Digitsinanevenplace:7,1,2and(7+1+2)=10.-Digitsinanoddplace:8,7,9,8and(8+7+9+8)=32.-Thelargerminusthesmaller:32-10=22.

Criterionofdivisibilityby12:Anumberisdivisibleby12ifit'sdivisiblebyboth3and4.

Criterionofdivisibilityby14:Anumberisdivisibleby14ifit'sdivisibleby7ANDit'seven.

Criterionofdivisibilityby18:Anumberisdivisibleby18ifit'sdivisibleby9ANDit'seven.

Criterionofdivisibilityby22:Anumberisdivisibleby22ifit'sdivisibleby11ANDit'seven.

In addition to knowingwhen a number is divisible by another one, it can also be useful to knowwhetheranumberisprime,andthereforenotdecomposableintofactors.Thesearchforaformulatohelptounderstandwhetheranumberisprimeornothasbeenthesubjectof study by mathematicians for several years but, despite this, still has not been found anythingsufficientlyeffectiveandefficient.Becauseofthis,beforeyoutrytofactorizeanumber,youshouldalways try to apply the most immediate criteria first and, if none of them gives its fruits, checkwhether thenumber iswritten in theprimenumbers table (and since I reallycareaboutyou,you'llfindaprimenumberstablerightattheendofthisbook).

The last type of decomposition we'll talk about is the so-called "decomposition into expressions",whichconsistsindecomposinganumber intoasetofmultiplicationsanddivisionsthatshouldmaketheoriginaloperationeasiertoperform.Forexample,50isequalto100/2.So,wecansaythatthemultiplicationofanumber"a"by50can

be just performed by "following the expression", and so bymultiplying our "a" by 100 and thenhalvingtheresult.Thedivisionby50insteadcanbeperformedby"invertingtheexpression"whichmeansbyswappinganymultiplicationwithadivisionandviceversa,andsobydividing"a"by100andthendoublingtheresult.

There is no specific criterion for decomposing a number into an equivalent expression useful forapplying this strategy, so let's just sayyoucando itafter trainingyoureye to recognizewhetheranumberisafactorof10(ormultiples)andit'snotconvenienttoapplyanyotherkindofcalculationstrategy.Forexample:

5=10/2.Andsomultiplyingby5isthesameasmultiplyingby10andthenhalvingtheresult.Dividing by 5 instead leads to reversing the expression, and so to dividing by 10 and thendoublingtheresult.

75=300/4.Andsomultiplyingby75is thesameasmultiplyingby100, thentriplicatingtheresult andhalving it twice. Insteaddividing it is the sameasdividingby100, thendoubling ittwiceandfinallydividingitby3.

250=1000/4.Andsomultiplyingby250isthesameasmultiplyingby1000andhalvingthesumtwice.Anddividingisequivalenttoapplyingtheinversestrategy.

Youcouldevencombinethedecompositionintoexpressionswiththedecompositionintoaddendsandsosimplysolve,forexample,amultiplicationx26.26isinfactequalto(100/4)+1.So,since"a"x((100/4)+1)=(ax100/4)+a,you justgot thatmultiplyinganumber"a"by26 is thesameasmultiplyingitby100,halvingtwicetheresultandthenadding"a"attheend.

I'llrepeatonceagainthatthisisaprocedurethatgreatlysimplifiestheoperations,butthat,atthesametime,doesnot implyaprecisepattern to followandso it just requiresaverywell trainedeyeandintuition to be properly performed. For this reason, whenever you realize that trying to use it iscomplicatingthingstoomuch,justforgetaboutitanduseonlythefirsttwostrategies.

So, after facing themore complicated things, here's a very useful fastmultiplication and divisiontable,preparedexactlybyapplyingalltheshowndecompositionstrategies.Ofcourseyoudon'thavetomemorizeit,butmysuggestionhereistounderstandhowwasitbuilt,inordertoeasilybecomeabletouseitscriteriatosolvealargeamountofoperations:

Multiplicationx4:Doubletwice.Division/4:Halvetwice.

Multiplicationx5:Multiplyby10andthenhalvetheresult.Division/5:Divideby10andthendoubletheresult.

Multiplicationx6:Doubleandthentriplicatetheresultorviceversa.Division/6:Halveandthendividebyonethird,orviceversa.

Multiplicationx7:Multiplyby10andsubtractthreetimestheoriginalnumber(noticeherethatyoucannotcreateanycriterionofdivisionby7,becausethedecompositionintoaddendsisnot

applicable to the division, factorization is not feasible since 7 is a prime number and adecompositionintoexpressionsisnotfeasibleaswell.)

Multiplicationx8:Doublethreetimestheoriginalnumber.Division/8:Halvethreetimestheoriginalnumber.

Multiplicationx9:Multiplyby10andthensubtracttheoriginalnumber.Division/9:Divideby3twice.

Multiplicationx11:Multiplyby10andaddtheoriginalnumber.

Multiplicationx12:Multiplyby10andaddtwicetheoriginalnumber.Division/12:Halvetwiceandthendivideby3.

Multiplicationx13:Multiplyby10andaddthreetimestheoriginalnumber.

Multiplicationx14:Multiplyby7andthendoubletheresult.Division/14:Halveandthendivideby7.

Multiplication x15: Multiply the original number by 10. Then add the original numbermultipliedby10first,andthenhalved.Division/15:Divideby3andthendividetheresultby5.

Multiplicationx16:Doublefourtimes.Division/16:Halvefourtimes.

Multiplicationx17:Multiplyby10.Thenaddthesamenumbermultipliedby10,butwhichyouhaveremoveditstriplefrom.

Multiplicationx18:Double,thenmultiplyby10andsubtractfromtheresulttwicetheoriginalnumberDivision/18:Halveandthendivideby3twice.

Multiplication x19: Double, then multiply by 10 and subtract from the result the originalnumber.

Multiplicationx20:Doubleandmultiplyby10.Division/20:Halveanddivideby10.

Nowyou canuse these strategies toperformamathematical trick thatwill come inveryhandy inyourpersonalfinancemanagementand,morespecifically,willletyouunderstandtheannualimpactofadailyexpense:

Takeyourdaily expense, for example those innocent5dollarsyoupay fordiningoutduringyourofficelunchtimeinsteadofbringingyourownfoodfromhome.Multiplythisamountbythedaysoftheweekyouactuallyspendthismoneyin.Inthiscase,let's

saythey'rethe5daysofanaverageworkweek=5x5=25dollars,thatwillbetheweeklycostofyourdailyexpense.Nowyouhave todonothingbutmultiplying theweeklycostby52andyou'llgetyourannualimpact!Buthowcanyoudothatinthefastestandmostefficientway?Well,asamatteroffactit'sverysimple:youcansplitthat52into50+2andthenget25x52=25x(50+2)=25x50+25x2.Multiplyingby50isaseasyastheproverbialpie:since50=100/2youjusthavetomultiplyitby100(whichmeansaddingtwotrailingzeroes)andthenhalvetheresult=25->2500->1250.Thislastestimatedexpenseisgoodenoughyettoletusunderstandhowbigtheannualimpactofthose"innocent"daily5dollarswillbe.Butifyoureallywanttobepreciseyoucanaddthoselasttwoweekstothatresultandgettingthenthat25x2=50,whichaddedto1250equals1300.Ouch!It'smuchmoreconvenienttoprepareyourfoodbyyourselfathome,isn'tit?

Ok,thischapterwasalittlebitmorechallenging,butfirstofalldoalwaysrememberyourself thatbiggertheeffort,bettertheresult.Andnowjustsitdownandrelax,becausenexttwochapterswillbemuchlighter,andwillintroduceyousomeofthemostcuriousandentertainingfactsintheworldofmathematics.

X-CuriositiesfromNumberworld

Coffeebreak!Ifyouhavebeenexhaustedbythepreviouschaptersthenit'stimetorelaxandtoenterinto the uniqueworld ofmathematical and numerical curiosities. So ... click on "play", start yourfavourite music playlist, sit down and read: some of these facts will show you the deep andextraordinaryharmonyofthemathematicalworld.Othersofthemwillprojectyouamongthemanyhistorical peculiarities of this discipline. While others again will be even useful for a betterunderstandingofsomesimplebutimportanteventsofyoureverydaylife.Let'sgo!

Theoddsofdyinginaplanecrasharethesameasflippingacoin21timesandalwaysgettinghead.Trytodothatandyouwillimmediatelyunderstandwhyit'ssaidthattheplaneisthesafestwaytotravel.

The Pythagoreans did not consider the number "1" to be odd nor even, but both. And that'sbecauseafteradding"1"toanaturaloddnumberyoualwaysgetanevennumber,andviceversa.Inthissense,itwasconsideredtobethegeneratorofallnumbers,evenandodd,andthereforeofallthings,finiteorinfinite.

There is an EEC regulation (specifically, the n.55 of 21/11/1994) that formalizes how themultiplesofathousandmustbecalled:inorder,thousands,millions,billions,trillions,billiardsandtrillions).Thereareinfact,apparently,no"official"namesforthelargernumbers.Andthestrangest thing of all, perhaps, is that this definition was written ... in a law about the safetransportationofdangerousgoods!

In theRussianroulettegame, ifonedoesNOTrotate thedrumaftereachattempt,who'sat theendoftheturnismuchmorelikelygoingtodiethantheothersbeforehim.Otherwiseeachonehasexactlythesameprobabilitytodieastheothers.Thisdoesnotmeanthatyoushouldrisklifeandlimbinagamelikethat.Butifitunfortunatelyhappensthatamobsterkidnapsyouandtakesyouintohislair,incompanyofshadyguyswhoforceyoutoplay,you'vegotastrategyforincreasingthechancesofsavingyourlife.Yeah,thisisnotgoingtobeveryprobable,too.

ThefirstnumbersystemhasbeenprobablyinventedbytheSumeriansin3000BC.Inthissystem,"one"alsomeant"man","two"meant"woman"and"three"alsomeant"many".

Somecicadasonlyhatchafterprimenumbersofyearsterms,inordertopreventpredatorswithvitalcyclesequaltodividendsoftheirhatchingperiodstoevolvespecializingintheircatch.Butthisisnottheonlyexampleofhowdeeplythelawsofnaturearewritteninthesamealphabetasmathematics: several animals in facthaveaquiteclear ideaof theconceptsofquantityandsize.Somebees,forexample,areabletocommunicateandunderstandtheexactdistancebetweenthehiveandapollenreserve.

Itissaidthattheinventorofthechessgame,asarewardforhisinvention,askedfor:"Putagrainofwheatonthefirstsquare,anddoublethequantityforeachsquarecomingnext".Arequestlikethat, ifaccepted,wouldhaverequiredmorethan18billionbilliongrainsofwheat.Anamountfargreaterthantheonecouldbeachievedafterusingtheentireplanetforcultivatingwheatandgatheringseveralmonthsofharvest.

It looks like that the ability of the great mathematicians to be easily distracted from theireveryday tasks iskindof legendary. It is said, forexample, that IsaacNewton,duringa lunch,wenttotakesomewineand,aftercompletelyforgettingwhathewasdoing,lockedhimselfinhisroomtowork,leavingthedinersatthetablewaitingforhiminvain.

1234567891, 12345678901234567891 and 1234567891234567891234567891 are all primenumbers.

All the numbers obtained by removing one digit at a time from "197933933 ", starting fromright, are prime numbers, except "1". In fact, in spite of the fact that "1" is clearly a numberdivisible only by 1 and itself, it's never considered as a prime one, because otherwise thepremisesofmanyimportantmathematicaltheoremswouldbecompletelylost.

Thereare7mathproblems,theso-called"millenniumproblems"that,ifsolved,wouldprovideaone million dollars prize because of their important implications in fields like physics andcryptography.Well,asamatteroffactoneofthemhasactuallybeensolvedyet,butthewinnerof the prize, a Russian mathematician, refused the reward. Furthermore, there is actually thepossibilitythatasolutionfortheother6problemsdoesn'tevenexistatall.

Ifitwasphysicallypossible,youcouldfoldapieceofpaperjust23timestoletitbeashighasMountEverest.And42timestoletitmatchthedistancebetweentheEarthandtheSun.

There is amathematical equation, called the "DrakeEquation"as formulatedby theAmericanscientistFrankDrake,thatshouldprovidethenumberofextra-terrestrialcivilizationsshouldbepossibletocommunicatewith.Infact,hemultipliedtogether:-Thenew-bornstarcreationrateinthegalaxy-Thefractionofstarshavingplanets-Theaveragenumberofhabitableplanetsperstarwithplanets-Thefractionofhabitableplanetslifecouldactuallybebornon

-Thefractionofsuchplanetsthislifecouldhavebecomeintelligenton- The fraction of planets where this intelligent life could have developed the ability (and thewillingness)tocommunicatewithotherlifeforms.And obviously after basing himself on nothing more than conjectures, Drake seems to havecometotheconclusionthattheresultofthisequationisequalto10.Anddespitetheutterlackofmathematical precision of the result, it's probably nothing but themathematical expression ofournaturalinstincttoraiseourheadsandrefusetobelievethatwearetheonlylivingbeingsinsuchimmensity.

The Soroban is a special type of abacus, introduced into Japan from China in the fifteenthcentury,whichletsanybodyefficientlyrepresentverylargenumbersandveryquicklyperformevencomplexcalculations.ManyAsianguysalsotrainthemselvessinceaveryyoungagetomentallyvisualizeaSorobanandperformoperationsonit, thusbecomingabletoachieveextraordinarymentalcalculationswithextremeefficiency.ThetechniqueaboutmentalcalculationsperformedthroughtheSorobaniscalled"Anzan"andcanbeproperlymanagedonlyaftertrainingalotwiththeabacusnumberrepresentation.Infact,while themajorityof the techniquesexplainedin thisbookcaneasilybemasteredinafewdaysandcangiveexcellentresultswhencombinedwithjustafewmemorytechniques,theAnzancouldevenrequireyearsoftrainingbeforegivingsatisfactoryresults.However,ifthisisatopicyouareinterestedinandyouwouldliketolaunchyourselfintothispath, you can find a lot of interesting tutorials simply by searching "Soroban calculation" or"Anzan"onYouTube.Also, ifyouhaveany touchscreendevice,youcanfindsome interestingappthatletyoueasilyemulateoperationsonarealSorobanabacus.

Throughaformula,youcandeterminethedayoftheweekofanydaybetweentheseventeenthandtwenty-secondcenturies.Asafirstthing,let'sintroduceatablethatassignsaspecificnumbertoeachmonth.You'llneedtomemorizeitifyouwanttousethisstrategywithoutanypenorpaper:

January=0February=3March=3April=6or-1May=1June=4July=6or-1August=2September=5October=0November=3December=5or-2.

Now,we'llneedasecondtablethatassignsaspecificnumbertoeachcenturyfromthe17thuntilthe22nd.Morespecifically,wehavethat:

Seventeenthcentury=6Eighteenthcentury=4Nineteenthcentury=2Twentiethcentury=0Twenty-firstcentury=6Twenty-secondcentury=4

So,takethelasttwodigitsoftheyearyouwanttoknowthedayoftheweekfrom.Let'scallit"a".Divideitby4withoutconsideringtheremainder.Wecallthisnumber"b".Wecall"c"thenumbertakenfromthe"monthstable"above".Wecall"d"thenumberofthedayinthemonth.Andwecall"e"thenumbertakenfromthe"centurytable".Nowcalculate(a+b+c+d+e)/7andconsidertheremainderofthedivision:0asaremaindermeansthattheconsidereddayisSunday,1meansMonday,2meansTuesday,andsoon.

ThesumofNconsecutiveoddnumbers,startingfrom1,issimplysquaredN.So,forinstance,youwanttocalculate1+3+5?Squared3=9.Wanttoinstantlycalculate1+3+5+7+9+11+13+15+17?Squared9=81!

Couldyou tellmehowlikely,givenanyclass fromanyschool, twoboysstudying therewerebornon thesameday?1%?2%?5%?Noneof this.According to the theoryofprobability, infact, a class exceeding23people is a sufficient condition for thatprobability tobemore than50%.

TheRomannumbersystem,which isperhaps themostwidelyknownamong theancientones,wasincrediblydifficulttomanageandnotveryintuitivewhenitwasnecessarytoperformanycalculations.And this is the reason inancientRome thecalculationswerealmostalwaysmadewiththehelpofpebblesorschedulesfor.Ontheotherhand,fromtheLatin,calculus=stone.

Thereareseveralformulasforcalculatingtheidealbodyweightofaperson,butonethat'satthesame time the most simple (requires no measurements apart from the height) and the mostreliable (though still theoretical, as of course it does not take into account anybody's specifichealthandbodystructurecase)istheso-called"KeysFormula",whichsaysthattheidealweightofanyperson(expressedinkilograms)isequalto:(Squaredheightinmeters)x22.1formen(Squaredheightinmeters)x20.6forwomenAndifthiscalculationnowseemscomplicatedtoyou,don'tworry...becauselateryouwilllearnallthenecessarytechniquesforquicklyandeasilyperformingitmentally!

Passionate about footballor anyother sports?Andyouwant toknowhowmanymatchesyoumustwatchonTVuntil theendofaspecific tournament inorder tonot tomissanyof them?Well, remember that in any tournament whosematches are a 1 vs. 1 with knockout rule, thenumberofgamesisequaltothenumberoffinallosers(so,ofcourse,it'sequaltocontenders-1).

The likelihood of having four of a kind right after the initial distribution of cards in aclassicpokergameisabout1in50000.

Amathematician namedCharlesBrenner has created a formula that can be used to find yoursoulmate.Morespecifically,hewas inspiredbyaspecificmathematicalproblem: there isadeckoffacedowncardsonatable.Oneachcardthereisanumberwecannotsee.Now,ouraimistograbthecardwiththehighestscore,respectingtherulethatwecanchooseitonlyafterhavingtakenit,andoncedecidedtotryanewone,thecardinourhandsmustbediscarded,ofcoursetakingtheriskthatinthedeckareremainingonlycardshavingalowerscore.Thiscanbringustoaskourselvesaspecificquestion:"Howmuchlargemustbethesampleofcardstodraw,inordertobesuretogetatleastoneofthehigherscores?"AndhereBrennermadehisanalogywiththerelationshipworld,comparingthedeckofcardstothe "golden period" in which anybody can actually try to achieve the highest number ofconquests,andthescoreoneachcardtothelevelofconnectionandempathyyoucanfeel inarelationshipwithsomebody.AlsoBrenneraddedthatthebesttimetogetanenoughlargesampletochoosetheidealpartnerisequalto1/e(e=Eulernumber=2.718...)multipliedbythe"goldenperiod".Assuming,then,forexample,thataman'sgoldenperiodgoesapproximatelyfromtheageof20to45,thebestsampleisreachedattheageofabout29and,oncereachedthatage,itshouldbecometypicallyconvenienttochooseasapartnerthefirstpersononefeelsastrongerconnectionwith.Well, clearly I think that love shouldn't be really chosen bymessingwith numbers, but apartfromthemoralofthestory("beawareofyourlimitsinordertogetthebestyoudeserve")itisalways interesting to seehow theapplicationsofmathematicscango farbeyondwhat anyonecould imagine.And then,who could ever believe that theEuler number (for thosewho knewhim)hadanyrelationwithloveconquests?Anextranote: theutilityof this formula,asamatterof fact,doesn'tendhere. Infact itcanbeapplied anytime we have to take one specific decision among a wide range of choices: justmultiply 1 / e (about 0.37) by the number of choices (or by the amount of time inwhich thechoicecanactuallybetaken)andassoonasyouseethatoneofthemisbetterthantheotheronesinsidethatspecificrange...that'syourchoice!

The so-called "perfect numbers" are those numbers that equal the sum of their divisors. Forexample,6isaperfectnumberbecauseit'sequaltoboth1+2+3and1x2x3.SaintAugustineevenwrote that"6 isaperfectnumberandnotbecauseGodhascreatedall things in sixdays.Indeedtheoppositeistrue:Godcreatedallthingsinsixdaysbecausethisisaperfectnumber".Even28isperfect(since28=1+2+4+7+14)andeven28hasalwayshadastrongsymbolicmeaning,beingthenumberofdaysinthelunarmonthintheJewishcalendarandthedurationofthewomenmenstrualcycle.

The "narcissist numbers" are those numbers that equal the sum of their cubed digits. Forexample,153isnarcissistsince153=1^3+5^3+3^3=1+125+27.

According to rumours, therewouldbenoNobelPrize formathematicsbecauseAlfredNobel

oncediscoveredherlovercheatinghimwithafamousSwedishmathematician,MagnusGustafMittag-Leffler.

The"twinprimenumbers"arethoseprimenumbersthatare"separated"inthenaturalnumberssequence just by a single number. For example, 5 and 7, or 11 and 13 are twin primenumbers.Alsotherearetheso-called"cousinsprimenumbers"thatdifferby4(like7and11),andthe"sexyprimenumbers"thatdifferby6(like5and11).

It has been reported thatmany prodigiousmental calculators ended up gradually losing theirextraordinaryabilitiesforapparentlynoreasons.EnrySafford,forexample,borninVermontin1836, at the age of 10was able to instantly performmultiplications like 365365365365365 x365365365365365365, but just after a few years its mathematical skills became absolutelyordinary. Even Zerah Colburn, born in 1804, at the age of 8 was able to calculate in a fewsecondsthefifteenthorsixteenthpowerofdifferentnumbers,butitlookslikethatattheageof20hadlostmostofitsability.Andwhatifthisisanotherproofthatsomeskillsaremoretheresultofasetofcircumstancesthansomethingthat's"divinely"givenfromnature?

TheheightofthePyramidofCheopsisexactlyequaltoonemillionthofthedistancebetweentheEarthandtheSun.PurecoincidenceortheancientEgyptianswereinpossessionofextraordinaryastronomicalknowledge?

The"0"symbolwasintroducedintheOccidentalWorldonlyinthe11thcentury.PriortothattheconceptwasquiteunknowninEurope,ifnotconsideredeven"taboo",becauseofitsrelationshipwith the concept of "nothing" or "emptiness". In other cultures instead it was not being usedsimply becausewere represented through pebbles and thereforeno stones corresponded to nonumber.

Thelargestnumberexpressiblethrough3digitsisnot999,but9rosetothe9thpowerrosetothe9thpower,whichisanumbermadeupof369693100digitsandthat, tobewritten,wouldrequire a 924 km long line of paper. If you wanted to write it in a simple Notepad text fileinstead,itwouldoccupyalmost2gigabytesonyourharddisk.

In2004itwascalculated that theasteroidApophiswouldhavehada2.7%chanceofcollidingwith Earth in 2029. Given the catastrophic event that would have followed, it's a rather highprobability(about1in30)andquitesimilartotheoddsoflaunchingtwiceadieandgettingbothtimes"1".Fortunately,subsequentcalculationshaveloweredtheprobabilityfrom1in30to1in230000,whichisthesameprobabilityasrollingseventimesadieandalwaysgetting"1"asafinalvalue.

NowIwanttotrytofrightenyoualittle:launchingseventimesadieandalwaysgettinga"1"isaneventwhichhasexactlythesameprobabilityaslaunchingitseventimesandgettinganyothercombinationofvalues.Yes, even if you roll anydie seven times just now, you'll get a combination of values that, ifcalculatedbefore,wouldhavehadexactlythesamechancetocomeoutastheannihilationforall

ofusin2029.So?Yourrollingdiehas justdemonstrated that theworld is likelygoing toendsoon?Not really.Attention in fact to the words "if calculated before", because they emphasize animportantconceptualpointintheprobabilitytheory.Thepointinfactisnotthecombinationyou'vegot.Thateventisnowhappened,itbelongstothepastandsoit'saneventwithaprobabilityof100%.Everythinghappenedinpasthasnowa100%probabilityofoccurrence.In fact itmakessense tospeakaboutprobabilityonly if related to futureevents nooneknowswithcertaintytheoutcomeof.Estimatingaprobability,afterall,meansmakingapredictionandthengivingameasureof theplausibilityof thatprediction.But this isprecisely thepointwe'retalkingabout:beforeyourolledthedice,youdidn'ttakeanypredictionabouttheresultofyourlaunches.Andsaying"theresultofmyrollingdicehadthesameprobabilitythattheasteroidhitstheearthin2029"hasthesamelogicasthecleverfortune-tellerwho,afterextractingtarotcardsoutofhisdeckssays,"Youknow,Iknewthatthisexactsequenceofcardswouldhavecomeout."So,itrathermakessensetosaythatifyourolladieseventimesandyouguessalltheresultsatyourfirstattempt,thenithappenssomethingthat'sexactlyasplausibleastheendoftheworld.Trythatandletmeknow!

Ifsomebodygivesyoutheopportunitytocombine(only)the7classicnotesofthemusicalscalein order to create a 12-notes long melody and you want to understand how many differentmelodiesyoucan freelycomposewithpreconditions like these,youcan simply raise7 to the12thpower.Andtheresultisthatyoucancompose13841287201possiblemelodies,allofthemdifferentbyatleastonenoteonefromeachother.Generalizing, if you want to know howmany are the possible n-long combinations of (alsorepeated)elementstakenfromasetof"m"choices,youhavetodonothingbutraisingmtothen-thpower.Anotherexample: takenthelettersoftheAnglo-Saxonalphabet(m=26), ifyouwant toknowhow many 3-characters words are obtainable by combining those letters (including even thenonsenseonesofcourse),you'llneed toperform26^3=17576combinations that, if sortedalphabetically,wouldgofrom"aaa"to"zzz".Ifthewordlengthwas5instead,thecombinationswouldhavebeen26^5=11881376.And this is precisely the reason forwhich anywebsite requires us to invent a password that'sboth composed by uppercase letters, lowercase letters and numbers, and that's as long aspossible.Infactapasswordcomposedbyalargesetofelementsincreasesm,whilelengtheningthe string increases n.So these choicesboth increase the combinations necessary to let somemalicioussoftwaretoguessyourpasswordandhackyourpersonalaccount.

If you have a poker deck and youwant to know howmany different combinations it can beshuffledinto,theproblemisslightlydifferentfromthepreviousone.Here, in fact, "n" (the combination length) is fixed and exactly equal to "m" (number ofelements),withtheextrarulethatthereisnopossibilityofelementsrepetitions(shufflingadeckofcoursedoesn'tchangethecards,butjusttheirorder).Inthiscasethentheoperationtobeperformedisdifferent,andit's"n!"(incaseyoudon'tknow,thatexclamationmarkmeansthatyouhavetomultiplytogetherallthepositiveintegersfrom1ton).

Butbeforeperformingthisoperationwithourcardsdeck,let'shaveaneasierexample.Let'simagineyouhavetwoslipsofpaperonatable,onewithan"a"writtenonitandonewitha"b".Nowyouwanttounderstandinhowmanydifferentordersyoucanarrangethemandhavingn=2you'lljusthavetocalculaten!=2!=1x2=2.Infact,thepossiblearrangementsoftwocardsareonly2:the"ab"orderandthe"ba"one.Otheranddifferentarrangementswiththesepremisesofcoursearenotpossible.Iftheslipsofpaperwerethree,andoneachofthemwasrespectivelywritten"a","b"and"c",thepossiblearrangementswouldhavebeenmore.Forn=3infactyouhavethatn!=3!=1x2x3=6andyoucanorderthemasabc,acb,bca,bac,cab,cba.So,let'sgobacktoourpokerdeck:inthiscaseyouhavethatn=52andso,inordertocalculateallthepossiblearrangements,you'dneedtoperform52!=1x2x3x4x5x6x...x50x51x52.Thisnumberisequalto80658175170943878571660636856403766975289505440883277824000000000000anditsimpressivevaluemeansthatit'salmostsurethatnodeckenoughandtrulyshuffledinhumanhistoryhaseverhadthesamecardorderasanotherdeck.More specifically, taken a hundred billion stars, each with a trillion planets, each with onemillion inhabitants,eachwithonemilliondecksofcards,andpresumedthateach inhabitant isabletoshuffleathousanddeckspersecondfor14billionyears(thesupposedageoftheuniversesincetheBigBang),onlytodaywouldstarttoappeartwodeckshavingthesamecardorder,anditwouldtake14billionmoreyearsofsuchanonsensetoletthishappenagain.Themathematicalcomplexityoriginatingfromsuchasimpleeverydayitemisquiteimpressive,isn'tit?

Thischapterendshere,butofcoursewewon't stophere talkingaboutmathematicalcuriosities! Infactinthenextchapterwe'llgobacktothemaintopicofthisbook,facingsomecuriositiesthataremuchmore closely linked to speedmath and examining the awesome harmony behind some very"special"numbers.

"The mathematical sciences particularly exhibit order, symmetry, and limitation; and these are thegreatestformsofthebeautiful."

(Aristotle)

XI-15goldennumbers

Nowlet'slaunchourselvesagaininthespeedmathworld,butstillremainingatthesametimewithintheorbitofthemathematicalcuriosities:let'sintroducetheso-called"goldennumbers!".The"goldennumbers"areallthosenumbershavingaparticularinner"harmony"thatmakesitridiculouslysimpletousethemtoperformsomespecifickindofcalculations.Ofcoursesomeofthemwillbefarlessuseful in theeveryday life thanothers,butanywayeven themostpointlessonecouldbehelpful inbuildingsomenicemagictricks.Enjoy!

3

Very often, in human history, to number 3 have been attributed various symbolic and esotericmeanings.Nottomentionthatithasoftenbeenconsideredtobethe"perfectnumber"parexcellence(althoughithasnotanythingtodowiththemathematicaldefinitionof"perfectnumbers"wesawinthepreviouschapter).Beyondthesymbolismandesotericism,however,whatwewanttoexamineinthischapterisaveryfastmethodtoinstantaneouslysquarenumbersonlymadeupofrecurrent"3"digits.Inparticular,toinstantlyperformthisoperationwemust:

Writeasmany"1"astheamountof"3"digitsinouroriginalnumber,exceptone.Writea0.Writeasmany"8"asthe"1"wewroteinthefirststep.Writea9.

So,forexample:Squared33=1089Squared333=110889Squared3333=11108889Squared33333=1111088889Andsoon.

9

Theproximity betweennumbersmade up only of "9" digits and powers of 10 has always given afascinatingharmonytotheresultsoftheoperationsperformedonthem.Forexample,inordertoaddtoanynumber"a"toanothernumbermadeuponlyofrecurrent"9"digits,youjusthavetosubtract1from"a"andthenwritethis"1"totheleftofthenumberitself:88+99=187543+999=15422342+9999=12341

Thisstrategyclearlyhasalotofinterestingpracticalapplications,sinceitletsyouquicklysumanynumber to another number that's very close to 99, 999, 9999, etc. For example, if you want tocalculate567+997,youcanapplythistechniquefirstandthen,consideringthat997islesserthan999by2,youcansubtract2togetthefinalresult.

Butit'salsoextremelyeasytomultiplybetweennumbersmadeuponlyofrecurrent"9"digitsandanynumbershavingthesameamountofdigits.Infact,toperformthisoperationyoucanjust:

Subtract1fromthenumberyouwanttomultiplyby9andwritethedifferenceonthe"leftside"oftheresult.Applytotheoriginalnumberthe"Allfromnineandthelastfrom10"SutrawediscussedinthechapteraboutVedicMathematics,andwriteitonthe"rightside"oftheresult(youdorememberthat the Sutra is about subtracting from 9 all but the last digit of the number, which must besubtractedfrom10instead,right?).

Now,let'shavesomeexamplesinordertoshowtheextremesimplicityofthistechnique:

89x99=write88ontheleftand11,whichistheresultoftheSutra,ontheright=8811768x999=write767ontheleftand232,whichistheresultoftheSutra,ontheright=7672323451x9999=write3450ontheleftand6549,whichistheresultoftheSutra,ontheright=34506549

Butanotherthingwecansayaboutthepropertiesof9isthatmultiplyingitbyanynumbermadeupofthesamerecurrentdigitisvery,veryeasy:

Multiply9bythe"recurrent"digit.Splitthetwodigitsoftheresultandputtheunitsontherightandthetensontheleft.Now,betweenthosetwodigits,putasmanynineasthedigitsoftheoriginalnumber,exceptone.

So,forexample:

9x666=6x9is54.Splitthetwodigitsandputtwo"9"betweenthem=59949x7777=699939x88888=799992

Another great property of this golden number? Well, it's even actually possible to almost

instantaneouslydivideanynumberbyit.Inthecase,forexample,ofa2-figurenumber(let'ssayyouwanttocalculate68/9),youcandolikethis:

Thefirstdigitof"a"istheresult(inthisexample68->6)Sumtogetherthefirstandseconddigitof"a".Thatwillbethedivisionremainder(forexample,6+8=14).If the resulting remainder is lesser than 9, it's all done. In the other case the result should beincreasedby1(becoming7inthiscase,sincetheremainderis14>9),whiletherealremaindercanbeobtainedbysubtracting9fromtheactualone(so,since14-9=5,inthiscasethefinalresultofourdivisionis7withremainder5).

Ifyournumberisa3-figurenumberinstead(forexample,327/9),youcandolikethis:

Takethefirsttwodigitsof"a"andaddthemtothehundredsdigit(inthiscase32+3=35).Thatwillbetheresult.Sumtogetherthethreedigitsof"a"(3+2+7=12).Thatwillbethedivisionremainder.Ifyourremainderislesserthan9,it'salldone.Ifthissumisgreaterorequalthan9instead(ashappensinthiscase),youmustsubtract9fromtheremainderuntilit'slesserthan9(andso,inthiscase,12-9=3)andthenyoumustadda"1"totheresultforeach9youhavesubtractedthisway(sointhisexample,sinceyouhavesubtractedonlya9,theresultbecomes36).

Now,let'sendthislongdissertationabout"9"byexplaininghowtoinstantlysquareanynumbermadeuponlyby"9"digits:

Writeasmany9asthedigitsofyournumber,exceptone.Writean8.Writeasmany0asthe9youwroteinthefirststep.Writea1.

So:

99x99=9801999x999=9980019999x9999=9998000199999x99999=9999800001

11

Eventheparticularharmonybehind11(andbehindallthenumbersmadeupofrecurring"1"digits)certainlycomesoutfromitsproximityto10.Butlet'sstartimmediatelytotalkaboutpracticalmath,andlet'sintroducethestrategyforrapidlyperformingamultiplicationby11.Atfirst,a technique thatcouldbeused toperformthisoperation is thedecomposition intoaddendsandwouldsimplyconsistinmultiplyinganumberby10andaddingthenumberitself.But,asamatteroffact,youcanevenusethisalternativeandprobablymoreeleganttechnique,whichisdirectlytakenfromtheVedicMathematicsandworkslikethis:

Takethefirstandlastdigitofthenumberyouwanttomultiplyby11andwritethemasfirstand

lastdigitoftheresult.Sum,startingfromleftandgoingright,twodigitsatatime,thedigitsofthenumberyouwanttomultiplyby11.Write,fromlefttoright,theresultsofthesesums.Andifanyofthemwasgreaterthan10,writeonlyitsunitsdigitandadd"1"tothedigitimmediatelyonitsleft.

Herearesomeexamples:

11x23=write2ontheleftand3ontheright,gettingsoapartial2_3.Nowsumtogethertwoata time, thedigits of23 (that of course areonly2, giving thenonlyone sum toperform) andyou'll get 5. Write it in the middle of your partial result and that's the final result of yourmultiplication:253.Thatwaseasy,wasn'tit?

11x387=write3ontheleftand7ontheright,gettingsoapartial3_7.Nowsumtogethertwoata time the digits of 387: start with 3 + 8 = 11.Write 1 and carry 1 to the next 3 on the left,transformingthepartialresultinto41_7.Nowcalculate8+7=15.Write5,carry1andgetthefinalresult:4257.

11x4709=write4ontheleftand9onright,gettingsoapartial4_9.Now4+7=11andthenthenewpartialresultis51_9.7+0=7sothepartialresultis517_9.Andfinally0+9=9,getting51799asafinalresult.

But let's levelupand let's try tounderstandhow toquicklymultiply111bya2-digitnumber.Asamatteroffacttheprocessisverysimilartothe11one,withjustsomelittledifferences:

Takethefirstandlastdigitofthenumberyouwanttomultiplyby111andsplitthemexactlyasyoudidforthemultiplicationby11.Sumthetwodigits,multiplytheresultofthesumby11andwritetheresultofthelastproductinthemiddleofyourpartialresult.Ifthelastoneexceedsthehundreds,putinthemiddleonlyitstensandunitsandcarrythehundredsdigittothefirstdigitofthepartialresult.

Anexample?Hereitis:let'strytocalculate111x76:

Write7ontheleftand6ontheright.Partialresult7_67+6=13.Multipliedby11itequals143.Thepartialresultwouldbe7_143_6,butwejustsaidwemustcarrythehundredsdigitiftheresultinthemiddlehasone.Soourfinalresultisnothingbut8436.

Now, let's end the "11" topicwith a very simple technique for squaring numbersmade only up ofrecurrent"1"digits:

Count.Yes,exactly,startingfrom1,countthepositiveintegersandstepforwarduntilyouhaveanamountofdigitsequaltothedigitsofthenumberyou'remanagingtosquare.Nowcontinuetocount,butsteppingbackwardsuntilyougobackto1again.That'syourresult.

Someexamples?Heretheyare:

Squared11=121Squared111=12321Squared1111=1234321Squared11111=123454321

37

37multipliedby3equals111.And,becauseofthat,37canbemultipliedbyanymultipleof3(<30)andtheresultwillalwaysbeanumberconsistinginasequenceofthismultipledividedbythree.So,forexample:37x12=444(infact,12/3=4)37x18=666(18/3=6)37x24=88837x27=999

Thisalsomeansthatany3-figurenumbermadeupofasequenceofthesamerecurrentdigit(suchas555or777),ifdividedbythesumofitsdigitsalwaysgives37asaresult.Trytobuildamagictrickbasedonthisprinciple!

143

143 has a very singular property: it can be multiplied by any three-digit number simply afterjuxtaposingtwocopiesofthisthree-digitnumberanddividingthelastby7.Let'ssayforexampleyouwanttoperform143x887.Inordertoquicklyperformthismultiplicationyou can just "build" thenumber887887 and thendivide it by7.Of course a one-digit division ismuchmoreimmediatethananythree-digitmultiplicationandthat'swhyeven143isreallyperfectforbuildingsomegoodspeedmathmagictricks.

666

Asamatteroffactitwouldbemoreappropriatetospeakaboutnumbersonlymadeupofrecurrent"6"digits,but the symbolismbehind thisnumberwas toostrong tonot to letmedecide towrite itdown exactly like this.More specifically, numbersmade up of recurrent "6" digits have a curiousproperty:justashappenswithnumbersmadeupofrecurrent"1"or"9"digits,youcanactuallysquarethemveryrapidly.Just:

Writeasmany4asthe6are,exceptone.Writea3.Writeasmany5asthe4are.Writea6.

So,forinstance:Squared66=4356

Squared666=443556Squared6666=44435556Squared66666=4444355556

1089

Takeany3-figurenumber,with theonlycondition that thehundredsdigitmustbegreater than theunitsoneby2atleast.Forexample542.Now subtract from this number the one you obtain by swapping the units digit with the hundredsdigits(inthiscasethen542-245=297).Thentaketheresultandsumittothenumberyougetbyswappingunitswithhundredsagain.So297+792.You'vejustgot1089,right?Theresultofthisprocedureisinfactalways1089,nomatterwhat'sthe3-figurenumbersyoustartedfrom.And that'swhysometimes1089 isconsidereda"magic"numberand it's theperfectbase forsome good number-basedmagic tricks.What if, for example, you justask to a friend of yours torepeatthepreviousstepsandthenyoushowhimyoucanguesstheresult?Yeah,dothatonlyonce!

2025,3025,9801

Thesethreenumberscanbecountedamongthe"goldennumbers"becauseit'spossible tocalculatetheirsquarerootinanalmostimmediateway.The"secret"hereinfactisjusttosplitthemintotwogroupsmadeupoftwodigitsandtosumthem.Infact:√2025=45=20+25√3025=55=30+25√9801=99=98+1Itwouldbereallygreatifcalculatingsquarerootswasalwayssoeasy,wouldn'tit?

3367

3367hasverysimilarpropertiesto143.Morespecificallyyoucanknowtheresultof3367multipliedbyanytwo-digitnumbersimplybyjuxtaposingthreecopiesofthelastnumberandthendividingitby3.So,forexample:3367x98=989898/3=3299663367x55=555555/3=185185Now,aquestiontostimulateyourmind:whytheresultofsuchadivisionwillalwaysbean integernumber?Ahint:youcanfindtheanswerinChapterIX.

37037

37037isverysimilarto37.Infact,ifmultipliedby3equals111111andbecauseofthat,ifmultipliedbyanymultipleof3(<30),theresultwillalwaysbeasequenceofthatmultipledividedby3.So:37037x6=22222237037x15=55555537037x24=88888837037x27=999999

142857

Ifyoumultiply142857byanynumberbetween2and6,theresultwillalwayskeepexactlythesamedigits,withtheonlydifferencethattheywillbe"shifted".Infact:142857x2=285714142857x3=428571142857x4=571428142857x5=714285142857x6=857142

Inadditiontothis,ifyoumultiplythisnumberby7,theresultwillbe999999.Thisstillcouldnotbevery helpful in the everydaymath, but again could be useful for building somenicemathematicalmagictricks.Justuseyourimagination!

12345679

Thisnumberthat'snothingbutthenaturalnumberssequencefrom1to9without8,hasverysimilarpropertiesto37and37037.Infact,ifmultipliedbyanumber"a"multipleof9andlesserthan90,itgivesbackasaresultanumbermadeupofasequenceof"a/9"repeated9times.So,forinstance:

12345679x9=11111111112345679x36=44444444412345679x54=66666666612345679x63=77777777712345679x81=999999999

1016949152542372881355932203389830508474576271186440677966

Iseriouslydoubtyou'llevercomeacrossthenumber1016949152542372881355932203389830508474576271186440677966inyourlife.Nevertheless,ifonedayakingpromiseshiskingdomtoallthosewhoareabletomultiplyitby6withinthreeseconds,simplydeletethelast6ontherightandputitontheleft.Yes,whilethe9and11wereprobablythemostusefulgoldennumbers,thisisdefinitelythemostuselessofall.

Thesectionrelatedtothemathematicalcuriositiesendshere...withoutendinghereatall!Inthenext

chapter, in fact, we will talk about fast multiplication by introducing a multiplication table that'scompletelyalternative to theone introducedat theendofChapter IXandwaswrittenbyaRussianmathematicianduringtheirstayinaNaziconcentrationcamp.

XII-Multiplicationtablesfromhell

Was the SecondWorldWarwhile JakowTrachtenberg, JewRussianmathematician always hardlycriticalagainst theNazism,wasimprisonedinaGermanconcentrationcamp.So, inordertotrytokeep hismind away from the horrors of that hell, he tried tomentally "escape" into theworld ofnumbersandmanagedtobuilddaybydayaninnovativeanduniquefastcalculationmethod.Andthemostextraordinarythingwasthathedevelopedhissetofstrategieswithoutwritinganythingbutsomelines on the occasional scraps of paper he could fortunately put his hands on. Fortunately then, in1944,Trachtenbergwasable toescapefromtheprisonwherehehad justbeen transferredwith thehelpofhiswifewhopledgedallherjewelleryandcorruptedsomeguards.SohefledtoSwitzerland,where he ended up teaching his mathematical method that won several awards and became reallypopular,especiallyamongthosewhohadproblemswiththetraditionalcalculationmethods.

Inparticular,inthischapterwewillanalyseTrachtenberg'smethodofrapidmultiplicationby5,6,7,9, 11 and 12, while later we'll face his more general "graphic multiplication" that has the strongadvantagetoreducethenumberofdigitstokeepinmindifcomparedtothe"classic"columnmethod.

As amatter of fact,Trachtenberg alsowrotemethods for rapidmultiplicationby8, 3 and4, but Ideliberatelydecidedtonottotalkaboutthembecausethey'requitecumbersomeandcomplex,andifthegoalofourentirebookistomakemathfaster,easierandmoreefficient,Ifirmlybelievethatthenatureofthesemethodswouldcompletelybeincontradictionwithourfundamentalintentions.

So,whichexactlyarethemainadvantagesofthismethod?Well:

Exceptforsometrivialoperationslike"halve"or"double",everymultiplicationistransformedinto a sequence of simple subtractions and additions. And apart from the obviously reducedoverall complexity, this will be very helpful if you still have some little problems with the"classic"multiplicationtables.

Itsspeedcomparedtoclassicmultiplicationenormouslyincreasesasfarasthemultiplicand(thenumberthatyouwanttomultiplyby5,6,7,9,11or12)islarger.

Itonlyoperatesonthemultiplicanddigits.Infactifforexampleyouwanttomultiply567345by12,allyouhavetodoisperformingsumsontheinternaldigitsof567345.Clearlythemethodthesesumsmustbeperformedthroughwillvarydependingonthemultiplier.But the importantthinghere is that, inanycase,onceyou'veseenwhich themultiplier isandyou'veunderstood

what's the right strategy to use, you can also "completely forget" about it.And this of coursereducesthenumberofdigitstokeepinmindandso,asweshouldwellknow,itfurthermakesthewholementalcalculationprocedureseasiertoperform.

Apartfromeverythingelse,thisisanextrainstrumentforyourmentaltoolboxand,asweshouldwell know, having a choice among various tools for solving the same problem represents atremendouscreativeandstrategicresource.

So,let'sstarttoenterintotheheartofthematter,andlet'sdosomefundamentalpremises:

Asalreadymentionedyoumustonlylookatthemultiplicandandonlyworkonthat.

Youmuststartfromtheunitsdigitofthemultiplicandandproceeddoingtheoperationsonedigitata time,going left.Wewillsee indetailHOWtodo these thingssoon.Fornowjustkeep inmindtherightmodeandordertoproceedwith.

The next digit to the right of another digit is called "its neighbour". And since everyTrachtenbergstrategyisbasedonperformingcalculationsbetweenadigitanditsneighbour,thisisafundamentalconcepttokeepinmind.An example?Here it is: taken3792, the neighbour of 3 is 7, the neighbour of 7 is 9, and theneighbourof9is2.

The"neighbour"oftheunitsdigitisalwaystobeconsidered0.

The leftmost digit of any number should be handled as if it was the neighbour of a "ghost"leading zero. And this "ghost leading zero", mind you, unlike the one we mentioned in thepreviouspoint,shouldbeconsideredasifitwasarealdigitofthemultiplicand.So,onlyafteryou"handled"itproperly,thewholeTrachtenbergprocedureends.

Aneasywaytoremindyourselfaboutthetwozeroeswetalkedaboutinthelasttwopoints?Atleastinafirstmoment,beforeyouoperate,youmayactuallywritethemintothemultiplicand,takingcareofwriting the trailingoneafteradecimalpoint.Youcan rewrite so, forexample,1234as01234.0or452as0452.0.Twozeroeswrittenlikethat, infact,won'talter theoriginalvalueof thenumberandat thesametimewillhelpyoutorememberexactlywhenthemethodstarts(totheleftofthedecimalpoint)andwheredoesitend(attheleadingzero).

Anytime is required tohalveadigit, and thatdigit isodd,youshouldalways round the resultdown to thenearest integer.Sowhenyouare asked tohalve1write0,write1whenasked tohalve3,2when5,3when7,andsoon.

Identicallyashappened in thecaseof some techniquesdiscussed in thepreviouschapters, anysumrepresentsadigit in the final resultandso, ifoneof themgivesa resultgreater than10,onlytheunitsdigitmustbeputintheresult,whilethetensmustbecarriedtothenextdigitontheleft.

Butnowlet'sgototherealtable:

Multiplicationx11

AfterthemethodbydecompositionandtheVediconeseeninthechapteraboutGoldenNumbers,here'sathirdwaytomultiplyby11:startingfromtheunitsdigitandgoingleft,"addeachdigittoitsneighbour".That'sit!

Well,let'smakeanexampletobetterexplainwhatdoesthatmeanandlet'strytocalculate56782x11:

Atfirst, ifyouwant,write56782as056782.0.Aswesaid, thevaluedoesnotchangeifwrittenlikethat, and at the same time it will help you to remember how to act. Attention: we'll use this"simplified"notationonlyinthisfirstexampleandwe'lltrytoworkwiththeoriginalnumberslaterinordertogetusedtothem.

056782.0Startfromtheunitsdigit.Itsneighbour,asmentionedbefore,is0.Sothefirstthingtodoissumming0+2.Andhere,verysimply,yougetthattheunitsdigitofyourresultis2

056782.0Nowgoleftandadd8toitsneighbour.So8+2=10.Thesecond(tens)digitoftheresultis0,witha1tocarrytothenextsumyou'llperform.

056782.07+8=15+the1wecarriedfromtheprevioussum=16.Thethird(hundreds)digitintheresultis6.1mustbecarriedagain.

056782.06+7=13+thecarried1=14.Thefourthdigitoftheresultis4.1mustbecarried.

056782.0Thenext5+6=11+the1wecarried=12.Thefifthdigitoftheresultis2.1mustbecarried.

056782.0Wesaidwemustconsider,astheleftmostdigitinthemultiplicand,a"ghostleadingzero".So,0+itsneighbour5=5,plusthe1carriedfromtheprevioussum=6.Sothesixthandlastdigitoftheresultis6.

FinalResult=624602

Asyoucanseethismethod,exceptforsomedetails,isalmostthesameastheVedicone.AndsincethetimeTrachtenberglived,VedicMathematicswasn'tabsolutelyknownintheOccidentalworld,it'svery fascinating to notice how people who lived in completely different times and places camebasicallytobuildexactlythesametheories.

Multiplicationx12

If you need tomultiply by 12 youmust "Double each number and then add the neighbour". Thismeans,moresimply:justoperatelikeyoudidwiththe11,butdoubleanynumberbeforeperformingthenecessarysums.

A curiosity: like happened with the multiplication by 11, there is a specific Vedic Sutra saying"Ultimate and twice the penultimate" and reveals, except for some little details, exactly the samemethod.Butlet'sjumptothepracticeandlet'susethistechniquetomultiply829x12:

829Doubled9equals18.Itsneighbouris0+18=18.Write8intheresultandcarry1.

829Doubled2equals4.Afteraddingtheneighbour9itequals13andafteraddingthecarried1itequals14.Write4andcarry1.

829Doubled8equals16.Plustheneighbour2andthecarried1equals19.Write9andcarry1.

0829Youarenowatthe"ghostleadingzero".Afterdoublingit,itstillequalszero,plustheneighbour8=8andplustothecarried1=9.

FinalResult=9948.

Multiplicationx6

Inordertomultiplyby6youhaveto"addtoeachdigithalfofitsneighbour,andadd5ifthenumberisodd".Sotheprocedureisbasicallythesameasthe11oneagain,butthistimeyoumusthalveyourneighbourbeforeaddingittoyourcurrentdigit(rememberingtorounditdowntothenearestintegerifodd,asIsaidintheintroduction).Inadditiontothat,ifyourcurrentdigitisodd,youhavetoadd5tothesumbeforemovingtothenextone.Butlet'shaveademonstrationofthemethodbytryingtomultiply6821x6:

6821Theneighbouris0here,butyoumustadd5tothefinalsum,too,because1isanoddnumber.Sothefirstdigitintheresultis1+5=6.

6821Theneighbourof2is1,whichhalvedequals0.5.0.5roundeddowntothenearestintegerequals0,whichdoesnotaffectthesum.Inaddition,2isanevennumberandsoyoudon'thavetoaddanythingelse.Theseconddigitoftheresultissimply2.

6821Youmustaddto8thehalfofitsneighbour2:8+2/2=8+1=9.8alsoisanevennumber,thereforeyoudon'tneedtoaddanythingelse.Sothethirddigitintheresultissimply9.

Attention:manypeopleheregetconfusedandsay"oh,ok,9isanoddnumber,soImustadd5attheend".Try toavoid thiserrorandalways rememberyoumustcheckwhether isoddonly the initialdigityoustartedoperatingwith.

6821Afteradding6tothehalfofitsneighbouryouget6+8/2=6+4=10.Write0andcarry1.Don'taddanythingelsebecause6isanevennumber.So,thefourthdigitoftheresultisnothingbut0.

06821Sowearrivedatthe"ghostleadingzero".Addhalfofitsneighbour6,whichis3,plusthe1carriedfromtheprevioussum=thefifthandlastdigitoftheresultis4.

Finalresult=40926

Multiplicationx9

Hereistheproceduretomultiplyby9:

For each digit (except the ghost leading zero), subtract it from 10 if it's the units digit andsubtract it from9 if it's oneof theotherdigits.Don't subtract anything if you're at the "ghostleadingzero"andperformthissubtractionseparatelyoneachdigitandbeforeeverystepinsteadofdoingitonthewholenumberatthebeginning.Thisprecautionwillhelpyoutoavoidsomebadmistakes.Addeachdigittoitsneighbour.Attention:theneighbourwillalwaysbethedigitthatoriginallywasinthemultiplierbeforedoingthesubtractionsexplainedinthepreviouspoint.Whenyouarriveattheghostleadingzero,insteadofaddingtheneighbour,addtheneighbour-1.

Inthistechnique,asyoumayhavenoticed,thereissomelittle"shadow"ofaVedicMathematicsSutraexplainedinthelastpages.Purecoincidenceorinevitablyrecurrentpatterns?

Inaddition,thismethodmayseemabitmorecomplicatedthantheothers,butasamatteroffactyou'llnoticeit'smuchmoreimmediatethananyclassicalmultiplicationby9.So,let'scalculate9x3825:

3825(10-5)+0=5

3825(9-2)+theneighbour5=12.Write2intheresultandcarry1.

3825(9-8)+theneighbour2=3.Plusthepreviouslycarried1=4.

3825(9-3)+theneighbour8=14.Write2intheresultandcarry1.

038250+theneighbour3-1=2+thecarried1=3.

Finalresult:34425

Multiplicationx7

Therulehereis:"doubleeachdigit,addhalfofitsneighbour,andifthedigityoustartedoperatingfromwasodd,add5"

So,let'simmediatelyshowhowthestrategyworks,bymultiplying2894x7:

2894Doubled4equals8,plushalved0thatisstill0=8,whichistheunitsdigitintheresult.

28949x2+4/2=18+2=20.Add5since9isanoddnumber=25.Write5intheresultandcarry2.

28948x2+9/2=16+4=20.Plusthecarried2=22.Write2intheresultandcarry2again.

28942x2+8/2=4+4=8.Afteraddingthecarried2equals10.0soisthefourthdigitintheresultandI'llhavetocarry1tothenextsum.

028940x2+2/2=0+1=1.Afteraddingthecarried1equals2.2,so,isthelastdigitintheresult.20258infactisthefinalresult.

Multiplicationx5

Let'sgobacktoaprocessthat'sslightlyeasiertoperformthanthepreviousones.Asweshouldknow,afirstmethodtorapidlymultiplyby5consistsinhalvinganumberandthenmultiplyingitby10,orviceversa.TheTrachtenbergmethodactuallyisnotthatmuchdifferent,despitebeingmucheasiertokeepinmind.Infact,tomultiplyby5byusingthistechniqueyouhaveto:

Halvetheneighbour.Ifthedigityoubegantooperatewithisodd,add5.

Acuriosity: foreachextrakilogram(=2.20pounds)of fat,ourbodybuildsanaverageof5extrakilometres (=3.11miles) of bloodvessels.So if you've recently gainedweight, you canhalve theamount of extra kilograms, apply this technique for seeing howmanymore kilometres of bloodvesselsyourheartisforcedtosupport...andmotivateyourselftoimmediatelystarttoloseweight!

Butlet'shaveademonstrationofhowthismethodworksandlet'strytocalculate24568x5:

24568Theneighbourof8is0.Halved0stillequals0,andso0isthefirstdigitintheresult.

24568Theneighbourof6is8.Halved8equals4.Thentheseconddigitoftheresultissimply4.

24568Theneighbour5is6.Halved6equals3.5isanoddnumberhowever,soyoumustadd5.3+5=8,whichisnothingbutthethirddigitoftheresult.

24568Theneighbourof4is5.Justafterhalvingandroundingityougetthefourthdigit,whichis2.

24568Theneighbourof2is4.Afterhalvingityougetthefifthdigitoftheresultthat's2again.

024568Theneighbouroftheleading0is2.Afterhalvingityougetthesixthandlastdigitintheresult,whichis1.

Finalresult:122840

Doyouthinkthischapterwaschallenging?Hereisalittlemnemonictable,usefulformemorizingalloftheexplainedmethods:

The5and11strategiesaretheeasiest:thefirstrequireshalvingtheneighbourandadding5ifthedigitisodd,whilethesecondjustrequiresaddingeachdigittoitsneighbour.The12multiplicationstrategyisthesameasthe11one,withthedifferencethatyoumustdoublethedigitbeforeaddingittoitsneighbour.The9multiplicationstrategyisthesameasthe11one,withthedifferencethatyoumustapplytheVedicSutrafirst,andaddtheneighbour-1tothe"ghostleadingzero".The6multiplicationstrategyisthesameasthe5one,withthedifferencethatyoumustaddthedigititselftohalfofitsneighbour.The7multiplicationstrategyisthesameasthe6one,withthedifferencethatthedigitmustbedoubledbeforeperforminganysum.

Butrelaxnowandtakeabreath,becauseinthenextchapterwewillintroducetwofastmultiplicationstrategies so immediate to learn and to perform that you'll soon start wondering how it can bepossiblethatnobodyevertaughtthematschool.

XIII-Move,invertandamaze

Themultiplication method we'll introduce, and that's extremely faster andmore efficient than theclassic"column"method,directlycomes fromthe thirdVedicMathematicsSutra, the"Paravartya -Yojayet"thatinourlanguagemeans"Move,invertandapply".Asamatteroffactitalsohasaprecisedemonstration,whichcouldbeperformedusingourpolynomialalgebra,butsincewelovepracticemuchmore than theorywewillput itasideandwewilldirectlyshowhowthiscalculationstrategyworks. So, let's explain the two-figure numbers version of this method right by showing how tomultiplytwotwo-figurenumbers,forinstance28x45:

Write theproductof the twotensdigits in the leftpartof thepartial result.So, in thiscasewehave2x4=8____

Writetheproductofthetwounitsdigitsintherightpartofthepartialresult.So,inthiscaseis5x8=40andthepartialresultbecomes8___40

Nowmultiplythetensdigitsofeachnumberbytheunitsdigitoftheotherone,sumtheresultsofthetwoproductsandwritethelastinthemiddlepartofthepartialresult.Sointhiscaseitwillbe(2x5)+(4x8)=10+32=42.Sothepartialresultwillbe8__42__40

Now you just have to carry the "extra digits": as seen inmost of themultiplicationmethodsdiscussed so far,only theunitsdigitsof eachnumbermustbekept,while the tensdigitsmustalwaysbecarriedtothenextnumberontheleft.Soherewehavethat:The4inthe42mustbecarriedtothe8ontheleft=12_2_40.Westillhaveatwo-figurenumberintherightpartofthepartialresult(ofcoursethe12mustn'tbeconsideredsinceit'stheleftmostnumberandsoitcan'tbecarriedanywhere),sonowthe4inthe40mustbecarriedinturntothe2onitsleft=12_6_0

Whenthelasttwonumbersareonlymadeupofasingledigityou'redoneandhere'syourresult:1260!

Noticeyoujustobtained1260inaquarterofthetimeitwouldhavetakentousethecolumnorthe

decompositionmethod,andwiththeneedtostoreinmemorymanylessdigits.Infact,thismethodisperfectlysuitableforapurely"mental"calculation,performedwithoutwritinganypartialresult.Well,asamatteroffact,ifyouwanttoperformatwo-figuremultiplicationlikethatwithoutanypenorpaper,thereisanextratrickyoucandotooptimizetheusageofyourmemory:insteadofstartingfrom the first step, start from the last one (the one consisting in summing the results of the twoproducts).Infact,ifyouperformthemostcomplexoperationfirst,thenthebrainwillbecompletelyfreetorapidlyandefficientlyperformtheeasiestones.Butlet'smakeanexampleof"pure"mentalcalculationperformedexactlybyfollowingthisprocedure:

Problem:multiply78x44

Firststep:Multiply the tens digits of a number by the units digits of the other one and viceversa.Or,ifit'smoreimmediateforyou,multiplythe"internaldigits"anddothesameforthe"externaldigits".Ofcoursethelastisjustadifferent,moreintuitivewaytosayexactlythesamething.Infact:78x44:theunitsdigitofthefirstnumberandthetensdigitofthesecondoneareonthe"innerside"oftheoperation.78x44:thetensdigitofthefirstnumberandtheunitsdigitofthesecondoneareonthe"outerside"oftheoperation.Oh,andofcourse,attheendsumtheresultsofthetwoproducts.Inmindthencalculate7x4=28and4x8=32.Afteradding28and32weget60andwe'llkeep it inmind.Thatwaseasy,wasn'tit?Secondstep:multiplythetwotensdigits:7x4=28.Putitontheleftandkeepinmind"28_60".

Third step:multiply the two units digits: 8 x 4 = 32. Put it on the right and keep inmind "28_60_32".

Nowyouhavetodonothingbutthecarryingtheextradigits,so:Carrythe3from32tothe60ontheleft,"transforming"yoursequenceinto"28_63_2"Carrythe6from63tothe28,finallyobtaining"34_3_2".Hereisyourresult=3432!

Hereyoucannoticebyyourselfthatifyoustartedfromthesecondorthethirdstep,thendoingthefirstonewouldhavebeenmuchharderduetothebiggersequenceofnumberstokeepinmind.Justtrythatandyou'llimmediatelyunderstandwhatI'mtalkingabout.

Itmayalsobeworthobservingthat,incaseofoperandswithidenticaldigits,youcantakesomelittle,extrashortcut.Infact:

If the numbers have the tens digits in common you can replace the step involving themultiplicationbetweenthetensdigitswithasimplesquare,andtheoneaboutthemultiplicationbetween"internalandexternaldigits"witha"sumoftheunitsdigits,thenmultipliedbythetensone".So,forexample,ifyouwanttomultiply76x72youcandolikethis:-Squared7=49.Writeitontheleft.-Sumofunitsmultipliedbythetens:6+2=8,whichmultipliedby7gives56asaresult.Writeitinthemiddle.

-Unitsbyunits:6x2=12.Writeitontheright.-Aftercarryingtensfrom49_56_12youget5472,that'syourfinalresult.

If thenumbershave theunitsdigits incommon youcan replace the step that consists in themultiplicationbetween"internalandexternaldigits"with"sumofthetensdigits,thenmultipliedby the units" and the units multiplication with a simple square. For example, if you want tomultiply28x38,youcandolikethis:-2x3=6.Writeitontheleft.-Sumofthetensmultipliedbytheunitsdigit:(2+3)x8=5x8=40.Writeitinthemiddle.-Squaredunits:64.Writeitontheright.-Aftercarryingthenecessarydigitsinthe6_40_64sequenceyou'llgetthefinal1064.

Nowlet'sextendthismethodtothemultiplicationbetweenthree-figurenumbers,whichcanbedonelikethis:

Multiplythefirsttwodigitsfromonenumberbythefirsttwodigitsfromtheotheroneandwritetheresultontheleft.Soifforexampleyouhavetomultiply143x388,multiply14x38=532_

Multiply the first twodigits of onenumberby theunits digit of theotherone andviceversa.Then sum the results of the twoproducts andput it to the right of the number yougot in thepreviousstepThus,incaseof143x388,you'llhavetoperform:38x3=11414x8=112114+112=226Sonowthepartialresultis532_226

Multiply the twounitsdigits andwrite the result of theproduct in the right sideof thepartialresult.So,hereyouhavetodo3x8=24andthepartialis532_226_24.

Aftercarryingtheextradigitsasusual,youhave:-532_228_4(aftercarryingthe2from24to226)-554_8_4(aftercarryingthe22from228to532)Therearenomoredigitstocarry,so55484isexactlytheresultofthemultiplicationwewerelookingfor.

Andhereonecouldrightlyarguethatthelastprocedureismoredifficultandcompletelyunsuitablefora"pure"mentalcalculation.Butwemustanywayconsiderthat:

Wecansimplifythementalprocedure,ashappenedinthefirstcase,simplybystartingwiththemore complex steps (so, in this case, it's convenient to operate exactly in the order as shownbefore).

Morewe'lltrainourmemoryandourconfidencewithtechniqueslikethe"phoneticconversion"andmoreeasilywe'llbeabletousethistechniquewithoutanywritingsupports.

Thequickerwelearntomultiplytwo-digitnumbersandthebetterwe'llbeabletomanagethistechniqueatitsbest.Its"harderchallenge"infactjustintothefirststep.So,oncewelearntodo

thatrapidlyandefficiently,the80%ofthehardworkisdone.

Hey,afterallyoudon'tnecessarilyhavetouseitwithoutwritinganything,andamongthewrittenmultiplicationtechniquesthisisstillthefastestandmostefficientatall!

Lastthingbeforemovingtothenextchapter:asamatteroffact,thisfastmultiplicationstrategycanbeextendedinordertomultiplyeven4or5figurenumbers,butwewilldiscussthatlaterbecausethebasic concept of this extended version of the technique is slightly different. In fact what we'll betalkingaboutisbasicallyagraphicaltechnique,thatcanbeperformedaccordingtoaprecisevisualpatternandthatwillletyoucompleteevenverycomplexcalculationsintheblinkofaneye.

XIV-Theprodigiousconnection

Afterhavingintroducedamethodtoperformanykindofmultiplicationbetweentwo-figurenumbers,let'sgobacktosome"nonuniversal"techniques.Inthischapter,infact,we'lltalkaboutstrategiesthatwillletyourapidlyandeasilymultiplytwonumberstogether,butonlyattheconditionthattheyarelinkedbysome"special"mathematicalconnection.Andevenif thisconnectionmaysometimesbeabit'restrictive',thesestrategiesareasamatteroffactsoimmediatetolearnandexecutethatanytimeyou'reluckyenoughtofindthem,you'llbecomeabletoperformextremelycomplexcalculationsinabunchof seconds. Furthermore, as youwill notice, from these specific cases itwill be possible todeducemoregeneralonesandsothepreconditionswillneveractuallybetoo"strict"foryou!Well,enoughsaid.It'stimetostart!

Two-figurenumbershaving the tens (orunits)digit in commonandwhoseotherdigits,whensummedtogether,equal10.

Thistechniqueisreallysimpletoapply,evenifit'snecessarytodistinguishbetweentwocases:

Case A: The tens are the same and the units, if summed together, give 10 as a result. Thishappens,forexample,ifyouwanttocalculate58x52(2+8=10).Case B: The units are the same and the tens, if summed together, give 10 as a result. Thishappens,forexample,ifyouwanttocalculate23x83.

Inparticularyouhavethat,inordertorapidlymultiplynumbersasdescribedin"CaseA"(58x52),youmust:

Multiplythetensbythemselves+1(Inthiscase5x(5+1)=5x6=30)Multiplytheunitsdigitsandwritetheresulttotherightofthenumberyougotinthepreviousstep.(Inthiscase,2x8=16.Finalresult:3016,obtainedinfewsecondsandthroughextremelysimplecalculations)

Inthe"CaseB"(forexample23x83),youhaveinsteadto:

Multiplythetenstogetherandaddtheunitsdigittotheproduct(Inthiscase2x8=16.Addtheunitsdigitsandit's16+3=19)Asincase"A",multiplytheunitstogetherandwritetheresulttotherightofthenumberyougotinthepreviousstep(inthiscase,3x3=9,andsothefinalResultis1909)

Becareful:inbothcases,followthisruletoavoiderrors:ifanyofthepreviousstepsgivesasingle-digitresult,writea0asitstensdigitbeforewritingitintothepartialresult.Otherwise,forexample,in the last example you could have written 199 instead of 1909, getting so a completely wrongnumberasaresult.

Asyou can see thesemethods are extremely simple and immediate to perform.But how couldwereducetheeffectsoftherestrictionsimposedbytheirpreconditionsinordertousethemtoperformevenmoregeneralmultiplications?

TheanswerhereisrightinChapterIX,whenwetalkedaboutdecompositions.Infact,allwehavetodo is to use the decompositionswe know in order to transform our generalmultiplication into amultiplication respecting one of our previous preconditions. For example, suppose you want tocalculate58x54:

54 is very close to52, the number having the same tens digit as 58 andwhose units digit, ifsummedto8,gives10asaresult.Buthowcanweapplythetechniqueifwehave54andnot52?Wemayusethe"decompositionintoaddends"!Infactherewecandecompose54into(52+2),andso:58x54=58x(52+2)==(58x52)+(58x2).58x2isextremelysimpletocalculate,while58x52canbeinstantlyresolvedusingourmethod.

Thesecret,then,istorecognizewhenit'spossibletodecomposethemultiplicationinsuchawaythatoneofthetwofactorshasthesametens(orthesameunits)oftheotherone,whiletheotherdigits,whensummedtogether,give10asaresult.Andthiscangenerallybedonewhen:

ThenumbershavethesametensdigitorareoneveryclosetoeachotherThenumbershavethesameunitsdigitThenumbersaresuchthattheirtensdigit,ifsummedtogether,give10asatotalThenumbersaresuchthattheirunitsdigit,ifsummedtogether,give10asatotal

Note:thismethodcanbeextendedtothree,fourorhigher-figurenumbers:

Extensionof"CaseA":Thismethodcanbeapplied if the twonumbershave thesame leftmostdigit,whilethenumbersmadeupoftheremainingdigits,ifaddedtogether,giveapowerof10asaresult.Forexample,itcouldbeappliedto567x533,since67+33=100.Inthiscase,theproceduretoapplyisexactlythesame:youmustmultiplytheleftmostdigitbyitself+1,andtheproductgoesontheleftinthepartialresult:(5x(5+1)=30_).Thenyouhavetodonothingbutmultiplyingtheremainingnumbersandwritetheresultofthisproducttotherightofthepreviousnumber:(67x33=2211).Finalresult:302211

Extensionof"CaseB":Youcanapplythismethodif thetwonumbershavethesamerightmost

digit,whilethenumbersmadeupoftheremainingones,ifsummedtogether,giveapowerof10asaresult(forexampleyoucanapplyitto441x561.Infact,44+56=100).Heretheprocedureisverysimilartotheoneadoptedinthebasiccaseaswell.Ashappenedthereinfact,thefirststepconsistsinmultiplyingthenumberswithouttheirunitsdigit(i.e.44x56=2464)andwritingtheproductontheleft.Nowbecareful:thereisanew,"extra"steptoperform:takethenextlowerpowerof10totheonegivenbythesumofthefirstdigits(inthiscasethesumofthefirstdigitsequals100andthenitsnext lowerpowerof10 is simply10)andadd it to thepreviousproduct. In thiscase,10+2464=2474.Finally,multiplytheunitstogetherandtriviallywritetheresultofthisproducttotherightoftheprevious one, as happened in the basic method (then in this case it's 1 x 1 = 1). Even here,however,becarefulandifyouseethattheproductoftheunitsismadeupofasingledigitonly,writea0asitstensdigit.Infact,inthiscase,theresultis247401andnot24741!

Ofcourse,whatever thepowerof10considered, thediscussionaboutusing thedecompositions inorder to further extend themethods still remains valid, but since the procedure is a little bitmorecomplicated,theuseofthelasttrickshouldbeevaluatedwiththepropercaution.

Numbersbetween11and19

Thismethodisverysimpletoapplyandlearn.Let'simagineforexampleyouwanttomultiply17x18:hereyousimplyneedto:

Sumanumbertotheunitsdigitoftheotherone.So17+8=18+7=25.

Multiplytheprevioustotalby10.So25x10=250.

Addtothisresulttheproductoftheunitsdigits.So,inthiscase,7x8=56,whichaddedto250equals306,that'sthefinalresultofourmultiplication.

Unlikethepreviousmethod,thisisquitelimitedanddoesnothaveanyspecialvariants,exceptfortheuniversalpossibilityofgettingamultiplicationofthistypethroughthedecompositionintoaddends.Forexample,ifyouwanttocalculate12x21,youcoulddecompose21into(19+2).Then12x(19+2)=(12x19)+(12x2).Thefirstmultiplicationcanthenbecalculatedrightbyapplyingthismethod,whiletheotheroneisquiteimmediate.

Butanotherextensioncanbeobtainedbyusing the factorization,whichcanbe appliedquiteoften,since thismethod is based on themultiplication between essentially low numbers,which can veryeasilybeseenasthefactorsofhighertwo-digitnumbers.

Forexampleifyouwanttocalculate26x28youcanfactorise26into13x2and28into14x2.Thenyouwouldhave26x28=13x2x14x2.Atthispoint,ifyouusethecommutativeandassociativepropertiesofmultiplicationitwillbecome=(14x13)x2x2.Sonowyouhavetodonothingbutapplying this multiplication method to 14 x 13 and then doubling the result twice. Of coursetransforming a general multiplication into a multiplication between numbers between 11 and 19requiresawell-trainedmathematicaleye,butwithsome timeandexperienceyou'll immediatelybeabletounderstandwhenit'stherightmomenttodoit.

Numbersequidistantfromaninteger

Thismethodbasicallyconsistsin"replacing"amultiplicationwithasquarecalculation.Needlesstosay,therefore,thatinordertoletitbeconcretelyuseful,you'llhavetomemorizeoratleasttolearnhowtoveryquicklysquaresometwo-digitnumbers.Butwe'llseethatlater,solet'sdoonethingatatime.

Asfirst,let'sspecifythattwonumbersareconsideredas"equidistant"(=havethesamedistance)fromanintegerif they'rebothevenorbothodd.And this integerwe're talkingabout isnothingbut theirmathematicalmean.Infact,ifweimagineallpositiveintegersarrangedinsequenceonastraightline,wehavethatthemathematicalmeanofanycoupleofnumbers"a"and"b"correspondstothepointthat'sexactly"inthemiddle"ofthem.

Forexample,34and36areequidistantfromanintegerbecausethey'rebotheven.And,rememberingthatthearithmeticmeanoftwonumbersisobtainedbysummingthenumbersanddividingthenby2,wehavethattheirmeanisnothingbut35.Or,equivalently,wecansaythattheyareequidistant from35andthattheirdistancefrom35is1,since35=36-1=34+1.

Yes,thedistanceisnothingbuttheabsolutedifferencebetweenanumberanditsmean.Oh,andincaseyoudon'tknowit,"absolute"inshortmeansthatifthevalueisnegative,youjustmustignorethatandalwayswrite isasapositivevalue.A"negative"distance, in fact,wouldmakenosense,even in itsmorecommonlyusedmeaning.

67and89arealsoequidistantfromanintegerbecausethey'rebothodd.Infact,aftercalculatingthemeanIhave that67+89=156,whichdividedby twoequals78.And theirdistancefrom78 is11,since78=67+11=89-11.

55and34ontheotherhandwouldnotfitthepreconditions.Ofcoursethey'restillequidistantfrom"something" and of course theirmean could be calculated aswell, but it would not be an integernumber.Infact55+34=89,whichdividedbytwoequals44.5.

Butnow,howcanweuseallthisinformationtoperformourmultiplication?Well,themultiplicationoftwonumbers"equidistantfromaninteger"isequivalenttotheirsquaredmeanminusthesquareddistancefromtheirmean.

Confused?Let'sgetanexampleandlet'stake34and36again.Wesaidthattheirmeanis35andtheirdistancefromthemeanisnothingbut1.Thesquaredmeanis1225,whilethesquareddistanceisstill1,sotheresultofthemultiplicationis1225-1=1224!

Anotherexample:67x89.Thattheirmeanis78andtheirdistancefrom78,asmentioned,is11.

Theirsquaredmeanequalsthen6084,whiletheirsquareddistancefromthemeanequals121.

6084-121=5963,whichistheresultof67x89.

Beforeconcludingthispoint,we'llexplainwhytwonumbersequidistant fromanon-integernumberdon'tfit.Asfirst:theformula"thesquareofthemean-squareofthedistance"worksfinewiththemaswell.Whythenhaveweexcludedthemfromthebeginning?Theanswerisactuallysimple:becauseiftheyarebothevenorbothodd, theirmean(aswellas theirdistance) turnsout tobeanon-integer

numberandthismakesourcalculationsmorecomplicated.

Forexample,given55and34,theirmeanis44.5,whilethedistancebetweenthemis10.5.Tryingtosquarethreedigitnumberscanbedefinitelycumbersome,andsothewholemethodwouldofcourseloseintermsofspeedorefficiency.

Numbersthatarecloseenoughtoapowerof10

ThismultiplicationtechniqueborrowsitspowerfromtheVedicSutras.Inparticularitwouldbegoodto brush up on the one introduced in Chapter V, the "All by 9, the last from 10" that consists insubtractingallthedigitsofanumberfrom,9exceptforthelastonethatmustbesubtractedfrom10,inordertoletuscalculatethedifferencebetweenthatnumberanditsnexthigherpowerof10.So,ifapplied to 87 it gives the result of 100 - 87, if applied to 343 it gives the result of 1000 - 343, ifappliedto7384itgivestheresultof10000-7384,andsoon.

Furthermore it's important to specify that this technique can only be applied to multiplicationsbetweennumbershavingthesamenumberofdigits.Inaddition,ithastheclearadvantageofhavinganon-restrictivelimitation:namely,thefactthatthetwonumbersshouldbecloseenoughtoapowerof10isnotamandatoryprerequisite.However,aswe'llseemoreindetaillater,ifthenumbersarenotcloseenoughtoapowerof10,theconsequentcalculationscanbereallycomplicate,andthewholeprocedurecanbecomecompletelyinefficient.

Atfirst,let'sdistinguishbetweenthethreecasesinwhichthistechniquecanbeapplied:

Case1: The two numbers are both slightly smaller than a power of 10.For example, 9948 x9975.Case2:Thetwonumbersarebothalittlelargerthanapowerof10.Forexample,1018x1023.Case3:Thetwonumbersaresuchthatoneisalittlesmallerandtheotheralittlelargerthanapowerof10.Forexample,1007x883.

So,let'sstartexplaininghowtobehaveincase1,becauseit'stheeasiestandbecausetheotheronesarenothingbutitslittlevariants:

Asfirstkeepinmindthatinthiscasetheresultshouldhaveanumberofdigitsequaltothesumofthenumberofdigitsoftheoriginaloperands.Calculatethedifferencesofthetwooperandsfromthenexthigherpowerof10.Ifnotimmediate,applytheSutra"Allfrom9,thelastfrom10".Calculatetheproductofthesedifferences.Subtractanybetweenthetwonumbersfromthedifferenceoftheotherone(fromitsnexthigherpowerof10)andwritetheresulttotheleftoftheproductyougotinthepreviousstep.If, after juxtaposing the two results, the totalnumberofdigitsyougot is smaller than theonecalculatedinthefirststep,thenyouneedtodonothingbutwritingasmany"zeroes"betweenthetwonumbersasit'snecessarytomatchthatquantity.

Butsinceitcouldlooklikeaquitetwistedprocedure,let'sseeanexampleofhowitcanbeappliedbymultiplying9948x9975:

Thetwonumbersaremadeupof4digitsandsotheresultmustbemadeupof8digits.Thedifferencebetween10000and9948is52.Thedifferencebetween10000and9975is25.Multiply52x25=1300.Calculate9975-52or9948-25(it'sthesamething),getting9923,whichmustbeputtotheleftofthepreviousnumber.Hereisthefinalresult:99231300,obtainedinafewsecondsandwithoutanyeffort!

Note that the extraordinary efficiencyof thismethod is due to the fact that it transformsa4-digitsmultiplicationintoa2-digitsone.Andthemorethetwofactorsareclosetotheirnexthigherpowerof10,and themore themethod is simplified.Forexample,a9997x9998multiplicationwouldbereduced to a simple2x3.And, in the sameway, if thenumbers start tobedistant from theirnexthigherpowerof10,theactualmultiplicationwillbetransformedintoa3or4digitsone,losingthenthewholeadvantagethetechniquewasgiving.

Butlet'simmediatelymovetocase2,verysimilartothepreviousone:

Keepinmindthatheretheresultshouldhaveanumberofdigitsequaltothesumoftheamountofdigitsofthetwooperands,minusone.So,forexample,theresultof1002x1011mustbemadeupofsevendigits.Calculatethedifferencebetweenthenumbersandtheirnextlowerpowerof10.Hereofcoursethecalculationisstraightforward.Calculatetheproductofthesetwodifferences.Sum to a number the difference of the other one from its next lower power of 10 (nomatterwhichoneyouchoose,theresultwillbethesame)andwritetheresulttotheleftofthenumberyougotinthepreviousstep.If the number of digits is larger than the one calculated in the first step, you must take theleftmost digit from the right section, and carry it to the left one. For example, if you have1345_4393,andyoucalculatedthattheresultshouldhavehad7digits,youmusttakethat4from4393andsumittothe1345,gettingso1349393asaresult.

Nowlet'shaveanexampleofhowthisprocedureworksandlet'scalculate1072x1048:

Eachmultiplicandhas4digits.4+4-1=theresultmustbemadeupofsevendigits.Thedifferencesfromtheirnextlowerpowerof10areobviously72and48.Multiplythem,getting3456asaresult.1072+48,or1048+72=1120.Writesothepartialresultas1120_3456.Attention,thetotalnumberofdigitsexceedstheoneitshouldbe.So,carrythat3from3456to1120,obtaining1123_456.1123456,infact,isourfinalresult.

Finally,case3,inwhichanumberisslightlyhigherthanapowerof10andtheotheroneisslightlysmallerthanthesamepower:

Sumtothesmallestonethedifferenceofthehighestfromtheconsideredpowerof10,orsubtractfrom the largest one the difference of the smallest from the considered power of 10. Theseoperationswillgiveboththesamenumberasaresult,sojustdotheeasiestone.Write,intheprevioustotal,asmanytrailingzeroesastheconsideredpowerof10has.Multiplythetwodifferencestogetherandsubtracttheresultofthisproductfromthenumberyou

gotinthepreviousstep.

Andsincethisprocedurecouldlookalittlebitharder,let'sshowhowitworksbymultiplying989x1024:

Thetwodifferencesare11(989+11=1000)and24.Soherewecanadd24to989orsubtract11from1024,theresultwillalwaysbe1013.Theconsideredpowerof10hereis1000,sowritethreetrailingzeroes:1013000.Multiply11x24,getting264.Doyourememberhowtorapidlymultiplyby11,right?Subtract 264 from 1 013 000.This could seem hard, but can be made very easily through asimpletrickwesawinChapterV:if1013000=1012000+1000,youcancalculate1000-264firstbyapplyingtheproperSutra,andthenadd1012000.Sowe'llhave:1000-264=736.736+1012000=1012736,whichisalsotheresultof989x1024.

The chapter about the "prodigious connections" in the multiplication ends here. In the next twochapterswewillstartexploringtwowaystographicallyperformanykindofmultiplication,andevenif theycannot literallybeconsideredas"rapidmentalcalculation" techniques, they'reanyhowveryinterestingbecauseoftheiroriginalstructureandgreatsimplicityofuse.

"Tothosewhodonotknowmathematics,itisdifficulttogetacrossarealfeelingastothebeauty,thedeepestbeauty,ofnature. If youwant to learnaboutnature, toappreciatenature, it isnecessary tounderstandthelanguagethatshespeaksin."

(RFeynman)

XV-Chinesemultiplication

Thefirstgraphicaltechniqueformultiplicationsolvingwe'llintroduceinthisbookisthatso-called"ChineseMultiplication"thatalwayshasacertaincharmforthosewhodiscoveritforthefirsttime.And this fascination is probably due as to its extraordinary simplicity, as to its great speed whencompared to the classic multiplication methods, as to the fact that it lets one calculate anymultiplicationwithout knowinga singlemultiplication table.Also, if you do not have any pen andpaper,youcaneveneasilyperformitbyproperlydisposinglittletoothpicks,matchesor...spaghettisticksonatable!

Andhere'showthiscurioustechniqueofcalculationworks:

Foreachdigitofthemultiplicanddraw(orarrange),fromthebottomgoingupwards,asmanygroups of oblique lines oriented as the symbol "\" (therefore, for example, two digits = twogroupsoflines,threedigits=threegroupsoflines,andsoon).Eachgroupalsoshouldhaveanumberoflinesequaltothevalueofthedigititself.Forexampleyoucanrepresenta"12"thiswayjustbydrawingalineandthentwomorelinesarrangedthisway:

For each digit of the multiplier draw (or arrange), now starting from the top and goingdownwards, as many groups of lines, but now oriented as the symbol "/". And even here ofcourse anygroupmustbemadeupof anumberof lines equal to thevalueof the considered

digit.For example, if your multiplier is "34" then you have to draw or dispose seven lines,overlappingthe12likethis:

Youwillimmediatelynoticethatinsomesectorsofthepicturethelinesarecrossingeachotherandthat,indeed,youcangroupthenearestintersections.Circlethesegroupsandcounthowmanyintersectionsthereareineachgroup.

(Here you can actually choose whether to merely count the intersections or to multiply thenumberoflinescomingintothatgrouphorizontallybythelinescomingvertically.Ofcoursedowhat'stheeasiestandmostimmediatetoyou.)

Youwillnotice that someof the resultinggroupsof intersectionsareverticallyaligned.Thenhereyouhavetodonothingbutsummingtheamountsofthesoalignedgroupsandwritingthemdownyourpicture.Here,forexample,thetwocentralgroupsareverticallyalignedandtheirsumis10.

Now, again, it's time to carry the extra digits.Wherever a product or a sum gave a numberlargerthan10asaresult,carrythetensdigittotheleft.Inthiscase,forexample,youwillneedtocarrythe1from10totheleftmost3,gettingso4asaresult.

Justputtogetherthedigitsobtainedthiswayandhereisyourfinalresult:408!

But since the best way to understand a graphical procedure at best is through more visualdemonstrations,let'shaveanadditionalexample.Forinstance,ifyouwanttomultiply32x14,here'swhatdoyouhavetodraw:

Hereforthe"32"wedrewontheleftthreelinesandthentwomoreobliquelinesfrombelowgoingupwards, while for the 14 we drew a single line and then 4 more lines, from the top goingdownwards.

So let's lookat thegroupsof intersections, let'scircle themand let'scounthowmany intersectionsthereareineachgroup:

Now we must sum the numbers from the vertically aligned groups. In this case there are only 2alignedgroupsagainand12+2=14.So:

The last sumgave us a number larger than 10. So, after carrying the 1 from14we'll obtain 448,whichisthefinalresult.

Let's have now a last demonstration of the method, but multiplying three-figure numbers. Forinstance,123x311:

Asyoucansee,the123isgivenbythegroupsofone,twoandthenthreelinesontheleft,whilethe311isgivenbythegroupsofthree,oneandonelastlineontheright.Nowhighlightthegroupsmadeupoftheclosestintersectionsandcounttheintersectionsforeachgroup.

It'stimetoperformthesums:sumallthenumberscorrespondingtotheverticallyalignedgroupsandwritethetotalsdown:

Aftercarryingthe1from12,weget38253,that'sourfinalresult.

Themethodcanbeusedevenwith4,5ormore-figurenumbers.Clearly increasing theamountofdigits will increase in turn the group of intersections to count and the consequent sums, thusinevitablyincreasingtheoverallcomplexityandthelengthofthewholemethod.Anyway,evenincaseof larger numbers, thismethod can still be considered as a very efficient, simple and even funnycalculation strategy, especially for all those who want to make experiments with somethingcompletelydifferentfromwhat'smorecommonlyusedineverydaylife.

XVI-Theslidingcross

Let's return to a more "traditional" multiplication method and let's introduce theTrachtenberg'scrossingmultiplicationsystemwementionedafewchaptersago.Thissystemisverysimpleandeffective to learnandapply,even in thepresenceofvery longoperands,andunlike the"multiplicationtable"builtbythemathematician,doesnotneedyoutomemorizelongorcomplicatedprocedures,butonlytomakesomelittle"visualtraining",necessarytounderstandthe"cross-shaped"graphicalpatterntofollow.

Inorder to simplify things, also,we'llonlyexplainhow tomultiply twonumbershaving the samequantityofdigits. In fact ifyouneed tomultiplynumbershavingdifferentsamountofdigitsyou'llstill be able to use this technique, but in addition you'll have towrite, in the smallest between theoperands,asmanyleadingzeroesasthey'reneededtomatchthedigitsamount.

Enoughsaid,solet'sstarttoshowthe2digitsnumberx2digitsnumbermultiplicationtechnique.

Step1:Alignthenumbersincolumn,multiplytheirunitsandwritetheresultofthisproductontherightin

thepartialresult.

Step2:

Multiplytheunitsdigitsofanumberbythetensoftheotheroneandviceversa.Then,sumthetwoproductsandwritethetotaltotheleftoftheresultyougotinthepreviousstep.

Step3:Multiplythetensdigitsandwriteagainthelastresulttotheleftoftheresultobtainedtheprevious

step.

Step4:Ifanyamongtheresultsobtainedinthepreviousstepsisbiggerthan10,thencarrythetensdigitto

thenumberontheleft.

Yes,thisisnothingbutthemultiplicationmethodintroducedinChapterXIIIandIwroteitagainherebecause, asalreadymentioned, itsbasicpattern is the sameas theonewe'll see formultiplicationsbetween larger numbers, and therefore trainingyour eye toworkwith this visualmethod since itsbasic case is definitely the best thing to do to become completely confident with it. So, let's seeanotherexampleofthisprocedureandlet'smultiply54x78:

54x

78=

Step1:Multiply4by8=32andwriteitontherightinthepartialresult.

Step2:Multiply5x8andaddtheresulttothe4x7one:

(5x8)+(4x7)=40+28=68.Writeitontheleft,obtaining68_32.

Step3:Multiply7x5andwriteitontheleftagain.You'llgetso35_68_32

Step4:Carrytheextradigits:

Carrythe3fromthe32tothe68:35_71_2.Carrythe7fromthe71tothe35:42_1_2.

Yourfinalresultis4212!

Butsincethiswasawell-knownissue, let'sgoimmediatelytothenext level,andlet's introducethegraphicaltechniqueforthe3-digitsnumberx3-digitsnumbermultiplication:

Step1:Multiplytheunitstogetherandwritetheproductintherightpartofyourpartialresult.

Step2:Multiplytheunitsdigitsbythetensoneandviceversa.Then,sumthetwoproductstogetherandwrite

thisresulttotheleftoftheoneyougotinthepreviousstep.

Step3:Andhereisastepweneverfacedbefore,evenifthepatterntofollowisalwaysthesame:multiplythe

symmetricallyoppositenumbersandsumalltheresultsoftheproducts.Inthiscase,then,calculate(UnitsofaxHundredsofb)+(TensofaxTensofb)+(Hundredsofax

Unitsofb).Thenwriteagaintheresultofthissteptotheleftofthepreviousone.

Step4:Nowyou'removingyour"cross"totheleft,andtheprocedureisalmostfinished.Multiplythe

hundredsbythetensandtensbythehundreds.Thensumthetwoproductsandwritetheresulttotheleftofthepreviousone.

Step5:Multiplythehundredsdigitstogetherandwritetheresulttotheleftofthepreviousone.

Step6:Carrytheextradigits.

So,whatvisuallyhappens,irrespectiveofthenumberofdigitsisthat:

Youalwaysstartontherightsidebymultiplying"unitsbyunits".

One digit at a time, going left, you go on by considering an always larger piece of the twonumbers.

Foreach"piece"consideredlikethat,youhavetomultiplythesymmetricallyoppositedigits,sumthe results of these products and then write the total to the left of the results you got in theprevioussteps.

When the "piece" you're actually considering exactly corresponds to the whole number, youmust start taking inconsiderationsmallerandsmallerpiecesof the twonumbers,obtainedby"cutting" thenumberson their right sideonedigit at a time.Then,you justhave to repeat thesameprocedureasinthepreviousstep.

Youhavetostopwhenthe"cross"hasbecomea"line"andthenyouarrivedtothemultiplicationbetweentheleftmostdigitofanumberandtheleftmostdigitoftheotherone.

Carrytheextradigitsanytimeastepgaveanumberlargerthan10asaresultand...it'sdone!

Thisofcoursedoesnotpretendtobearigorousorformaldescriptionofthemethod,butonlyabriefdescriptionofthevisualschemetoadopt.Onceyoulearnthat,youcaneasilyapplyittoanykindofnumber!

Butlet'shaveapracticalexampleofthe3x3procedurebytryingtomultiply673x231:

673x

231=

Step1:Multiplytheunits:3x1=3.Partialresult_3

Step2:Unitsbytensandtensbyunits:

(7x1)+(3x3)=7+9=16.Partialresult16_3

Step3:Unitsbyhundreds,tensbytensandhundredsbyunits:

(6x1)+(7x3)+(2x3)=6+21+6=33.Partialresult33_16_3.

Step4:Hundredsbytensandtensbyhundreds:

(2x7)+(6x3)=14+18=32.Partialresult32_33_16_3.

Step5:6x2=12.Partialresult12_32_33_16_3.

Step6:Carry-overs:

Carrythe1fromthe16tothe33:12_32_34_6_3.Carrythe3fromthe34tothe32:12_35_4_6_3Carrythe3fromthe35tothe12:15_5_4_6_3

Finalresult:155463!

NowI'llconcludethischapterbyshowingyouhowtoperforma4x4anda5x5digitsmultiplicationthroughthismethod.SincetheschemeisessentiallyalwaysthesameandsinceIthinkit'smuchmoreusefultogetitvisuallyratherthanendlesslyreadingthesamethings,thistimeI'llleaveyoujustwiththeimagesshowingthepatterntofollow,withoutdescribingitstep-by-stepagain:

Multiplya4-digitnumberxa4-digitnumber

Step1

Step2

Step3

Step4

Step5

Step6

Step7

Step8:Carry-overs

Multiplya5-digitnumberxa5-digitnumber

Step1

Step2

Step3

Step4

Step5

Step6

Step7

Step8

Step9

Step10:Carry-overs

(Hereyoumayhavenoticedthatthenumberofstepstoperformsimplyequalsthenumberofdigitsofthenumbersyou'reoperatingwith,multipliedby2)

Atipforlearningtomentallyapplythistechniqueconsistsinusingsomemnemonictechniqueasthe"phoneticconversion"seeninChapterIII,inordertoeasilykeepinmindthepartialresultsandthecarriedamounts.Withalittlemnemonictraininginfactyouwon'tonlybeabletomentallyperformverylongmultiplications,butyou'llalsolearnhowtocalculatethemfasterandfaster.

The "multiplication techniques" part of this book ends with this chapter. In the following chapter,however,we'llpracticallyapplythem,addressingtwoissuesthatareasapainforlotsofpeople,asabsolutely indispensable in daily practice: the percentages and the calculations involving numberswithadecimalpoint.

"Evenmathematics isa sciencecreatedbyhumanbeings.And then, ineachage,aswellas ineachcommunity,hasadifferentspirit."

(HermannHankel)

XVII-Discounts,badpoliticiansandinternationalmeasurements

The percentages are a notation of the "parts" of something, represented as an answer to thequestion "if this thing was made up of one hundred parts, how many of them we'd beconsidering?"andsince it's a verywidelyused concept,knowinghow toworkwith them is reallycrucial in the everyday life: just think about the opinion polls, the simple store discounts, or tothatprobabilitytheorywe'lljustfaceinacoupleofchapters!

Anddespiteit'sabroadlyintuitiveconcept(mostofpeopleareabletoquitesimplyvisualizethe50%or 25% of a pie, for example), the situation starts to be much less intuitive when it comes torequiremorestrictcalculationsofpercentages.Anexamplecanbefoundinthisproblem:"Ifthepriceofalitreofpetrolis100,afteraweekisreducedby10%andafteranotherweekincreasedby10%,howmuchwilleventuallycostalitreofpetrol?"

Themajorityofpeopleinfacthereanswers"100",forgettingthatthepercentageofsomethingshouldalwaysberecalculatedonthebasisofitscurrentvalue.Infact,100-10%=90%.Butthen90+10%=99%,sincenowthe10%mustnowbecalculatedon90andnomoreontheinitial100!

In the sameway, very often themedia take advantage of this difficulty people have in concretelyreasoning with percentages.And an actual example of this can be found in the words of a U.S.reporterwho,asa resultof thepresidentialelectionof2008,declared:"In2004,37%ofvotershaddeclaredhimselfaRepublican,37%aDemocratandtherestanindependent".Headdedthenthat,fouryears later,39%ofvotershaddeclaredhimselfDemocratand32%Republican.So, theconclusionwasthat"Infouryearsthe5%oftheRepublicanshadbecomedemocratic".

ClearlytohavebecomeaDemocratwasthe5%ofthetotalvoters,notthe5%oftheRepublicans,nottomention that thenumberofvotershaddefinitelychangedover time, and therefore that statementwaslackingofanyconsistency.

Fromallofthiswecandeducetwofundamentaltruths:

Amathematicalone,whichrecommendstoalwaysbecarefultocalculatethepercentagesontheactualandcurrentvalueofsomething.

Alessmathematicalone,whichrecommendstoneverbelievetothepercentagesproclaimedbyanypolitician.

Giventherightpremises,it'stimetointroducesomemorepracticaltoolsforcalculatingpercentages.First,asalreadysaid,theX%ofaquantitymeansdividingitby100andthenmultiplyingitby"X".This, of course, can lead to simplifications that, after applying some principles from the rapidmultiplication table we introduced in Chapter IX, can let you quickly calculate a lot of specificpercentages.Andhere isahandypercentage tablebuilt rightusing theseprinciples.Clearlyyoudonot have to memorize it, but rather try to find out which specific criteria were used to get theseresults:

Tocalculate:

The2%ofanynumber=Dividethatnumberby100andthendoubleit(orviceversa,theorderisnotimportant).

5%=Dividethenumberby10andthenhalveit(orviceversa).

10%=Dividethenumberby10.

15%=Triplicatethenumber,thenhalveitanddivideitby10(inanyorder).

Anothermathematicallifehack:theselasttwopercentagesarequiteusefulwhenyouhavetocalculatetipsforrestaurantbills.Forexample,whenyouhavetocalculatea15%tipona50dollarsbill,youcanjust:-Triplicateit:150-Halveit:75-Divideitby10:7.5dollarsIt'sveryeasy,isn'tit?

20%=Calculateits10%andthendoubleit(orviceversa).

25%=Halveittwice.

30%=Triplicateitandthendivideitby10.Ifyoudon'tcareabouthavingatoomuchaccurateresult,youmay also approximate 30% to 33.3%, and thenget the desired percentage just by dividing theoriginalnumberby3.

40%=Calculatethe10%andthendoubleittwice.

50%=Justhalveit.

60%=Doubleit,triplicateitandthenfinallydivideitby10(inanyorder)

70%=Therearenotricksherebutmultiplyingitby7andthendividingby10.Butifyoudon'treallycareabouthavingatoomuchaccurateresult,youmayalsoapproximate70%to66.6%,andsogetthedesiredpercentagejustbydividingtheoriginalnumberby3andmultiplyingitby2.

75%=Triplicateitandthenhalveittwice(orviceversa)

80%=Doubleit3timesandthendivideitby10(orviceversa)

90%=Triplicateit2timesandthendivideitby10(orviceversa)

So,forexample,doyouwanttoknowhowmuchyouwillpayifthereisa60%discountonthat2500dollarscar?Doublethepriceto5000,triplicateittoget15000andfinallydivideby10togetyour1500discount!Isitworththecostnow?

Andwonderinghowmuchdoesitcostthatbathtubhavinga1955dollarspriceanda15%discount?Triplicate the price to get 5865, halve it to get 2932.5 and finally divide it by 10 by moving thedecimalpointtotheleft,gettingyour293.25dollarsdiscount.Nowthatwegotaresultlike"293.25",andsinceit'ssomethingdeeplylinkedtothepercentagestopic,let'saddsomeimportantinformationaboutspeedmathinvolvingnumbershavingadecimalpoint.

Roundingandaligningincolumn

Thefirstthingyoucandoifyouwanttomorerapidlycalculatewithnumbershavingadecimalpointisroundingthem:justremoveoneormoretrailingdigitsafterthedecimalpointandiftheleftmostdigitamongtheonesyoudeletedwasgreaterthan5,thenincreaseby1therightmostdigityou"left"inthenumber.It'seasierdonethansaid:forexample,ifyouwanttoround5.489byremovingthelasttwodigits(89),you'lltransformitinto5.5sincetheremovedleftmostdigit(8)wasgreaterthan5.

Infact,asfaraswedeletedigitsafterthedecimalpointtheresultwillbelessaccuratebutatthesametimewill letyouoperatemuch faster.Andofcourse it'sup toyou todecidehowmuchprecision isneededinyourcalculationsandwhethertheprecisionlossisworththeeasegain.

If,forexample,youaremeasuringawallinordertounderstandwhetheryouhaveenoughspacetoputyournewwardrobethere,andyouhaveanultra-preciselasermetretellingyouthatyourwallis1.784532metreswide,eliminatingthedigitsafterthe4won'tcreateyouanyproblem.Foraproblemlikethis,infact,it'snecessarytobeaccuratebythecentimetre,bythemillimetreatmost,butthetenthorhundredthofamillimetreareofcoursecompletelyirrelevantparameters.

Nothingspecialtosayincasewemustaddorsubtractnumbershavingadecimalpoint:youmustjustalignthemincolumnproperlybeforeperformingthecalculation,handleanypossiblenumberswithoutdecimalpointasiftheyhada".0"nexttotheirrightmostdigit,andthenyoucanuseallthenormaltechniquesofadditionandsubtractionyousawinthepreviouschapters.

Formultiplicationanddivisionbetweennumberswithdecimalpoint,ontheotherhand,wehavetomakesomeextraclarifications.

Multiplicationanddivision-Decompositionintoexpressions

Afirst,"classic"waytomultiplybetweennumbershavingadecimalpointconsistsin:

Removing thepoint (leaving thedigitsunchangedandsowithoutconfusing thiswith roundingthem)fromalloftheoperandsCounting howmany digits there were after the decimal points in each of the operands beforeremovingit

Summingthesequantities(let'scallthistotal"b")PerformingthemultiplicationRe-writingthedecimalpointintotheresultinsuchawaythatafteritthereareexactly"b"digits.

Soifforexampleyouhavetomultiply7.76x9.31,youcanmultiply776x931first(andhereyoucould use the technique for "numbers close to a power of 10" seen inChapterXVI) to obtain 722456.Then,sinceinbothnumberstherewere2digitsafterthedecimalpoint,and2+2=4,you'llneedtohavefourdigitsafterthedecimalpointandyourfinalresultisnothingbut72.2456.

With the division instead you should simply remove the point from the operation by using itsinvariantiveproperty(let'srepeatit:multiplyingordividingdividendanddivisorbythesameamountdoes not change the final result of the division). So, you should just multiply both operands by apowerof10suchthatitletsyouremovethepointbybothofthem.So,forexample,ifyouhave1.24/2.33youcanmultiplybothofthemby100inordertoget124/133.Samethingifforinstanceyouhave133/0.002:inthiscaseyoucaninfactmultiplybothnumbersby1000andobtainthattheresultequalsthe133000/2one.

But a technique that's much more efficient than the classic one is the "decomposition intoexpressions", very similar to the one we saw in Chapter IX. In particular, in order to apply thistechniqueyoucan:

Decomposethemintofactors...liketheydidn'thavethedecimalpointatall!Forexampleyoucanhandlea4.5likeitwasa45that,decomposedintofactors,equals3x3x5.Ok,youjustbehavedlikethedecimalpointdidn'texist,butofcoursemultiplyinganumberby3,3and5insequenceisthesameasmultiplyingitby45,butnotthesameasmultiplyingitbythe4.5weneed.So,howcanwehavebackour4.5?Well, it'sverysimpleactually:you justhave toadd to thepreviousdecompositionadivisionbyapowerof10suchthatthedecimalpointreappearswhereit'ssupposedtobe.Forexample, inorder to transformbackthe45intoa4.5youwill justneedtodivideitby10(divisionbyapowerof10,doyouremember?)andso,thefullexpressionequivalentto4.5isnothingbut3x3x5/10.

Soyoujustfoundanexpressionthat'sequivalenttoyourfractionalnumberandthat,ashappenedinChapterIX,describesnothingbutasimplesequenceoftaskstoperform.Infact,let'srememberthatmultiplyingbyanumberdecomposedintoanexpressionisequivalentto"followingthatexpression".Thenmultiplying a number by 4.5 is equivalent to triplicating it twice, then quintuplicating it andfinallydividingitby10.

Dividinganumberdecomposedintoanexpressionisequivalentto"reversingtheexpression"instead,which means replacing any multiplication with a division and vice versa. So, dividing by 4.5 isequivalenttomultiplyingby10,thendividingby3,thendividingagainby3andatlastdividingby5.

Here isstillvalidwhatwesaid inChapter IX:check thedivisibilitycriteria first (using thenumberwithoutitsdecimalpoint,ofcourse),andifatleastthefastestonesdonotgiveanyusefulresult,checkwhetheryournumberisprimeornot.

Decompositionsintoaddends

AnothercalculationmethodwecanapplytonumbershavingadecimalpointisthedecompositionintoaddendswesawinChapterIXaswell.Thismethodontheonehandpresentstheusuallimitationofnotbeingapplicable to thedivision,but theotheronehas theenormousadvantageofbeingusableindependentlyof theactualpossibility to factorizeanumber.But let's imaginewewant to use it tocalculate4x1.25:1.25herecouldbedecomposed into(1+0.2+0.05)andat thispointwecouldseparatelymultiply 4 by 1, by 0.5 and then by 0.05 and at the end sum together the results of theproducts.Butofcoursethisstrategywouldnotbereallyefficientifcomparedtothe"classic"one.

Thereisinfactatricktomakethistechniqueinfinitelymoreusefulandeffective.Infact0.25=1/4andmultiplyingby1/4meansnothingbutdividingby4.So:4x1.25=4x(1+0.25)==(4x1)+(4x0.25)==4+(4/4)==4+1=5

Thismeansthatifyoucandecomposeanumberwithdecimalpointintoasequenceofkeyaddends,themultiplications coming from the decomposition can be transformed into quite straightforwarddivisionsandthewholeprocessbecomesridiculouslysimple.Butwhichexactlythese"keynumbers"are?Let'sseeit:

0.1-0.01-0.001-etc.Infactmultiplyingbythesenumbersisthesameasdividingby10-100-1000etc.0.2-0.02-0.002-etc.Infactmultiplyingbythesenumbersisthesameasdividingby5-50-500etc.0.25-0.025-0.0025-etc.Multiplyingbythesenumbersisthesameasdividingby4-40-400etc.0.33-0.033-0.0033-etc.Multiplyingbythesenumbersis(almost)thesameasdividingby3-30-300,etc.0.5-0.05-0.005-etc.Multiplyingbythesenumbersisthesameashalvingthenumber-halvingandthendividingitby10-halvingandthendividingitby100,etc.0.66-0.066-0.0066-etc.Multiplyingbythesenumbersis(almost)thesameassubtracting1/3fromtheoriginalnumber1/3-subtracting1/3andthendividingby10-subtracting1/3andthendividingby100,etc.0.75 -0.075 -0.0075 - etc.Multiplying by these numbers is the same as subtracting from thenumber1/4-subtracting1/4andthendividingby10-subtracting1/4andthendividingby100,etc.0.8-0.08-0.008-etc.Multiplyingbythesenumbersisthesameassubtractingfromnumber1/5-subtracting1/5andthendividingby10-subtracting1/5andthendividingby100,etc.0.9-0.09-0.009-etc.Multiplyingbythesenumbersisthesameassubtractingfromthenumber1/10-subtracting1/10andthendividingby10-subtracting1/10andthendividingby100,etc.

Ofcourseyoucan'talwaysdecomposeanumberintoasequenceofkeyaddends,butusingsomelittleapproximationcanalmostalwayshelpyoutogetwhatyou'relookingfor.Forexample,let'simagineyouwanttomultiply8by1.77.Inthatcaseapproximatethat1.77to1.75andthencalculate:8x1.77="almost"8x1.75==8x(1+0.75)==8x1+(8x0.75)==8+(8minus1/4)="almost"14

Oh,andofcourseeveninthiscaseit'suptoyoutodecidewhetherthelossintermsofprecisionisworththegainintermsofspeedandefficiency.

Lastthingbeforemovingtothenexttechnique:ofcoursenotallthemultiplicandsare"compatible"withthekeyaddends.Forexampleifinthepreviousoperationwehad7insteadof8,performinga"7minus1/4"ofcoursewouldnothavebeensosimple.

So,hereyoudon'thavetodoanythingbutrememberingtoapplythemethodwhenevertheresultingaddendsgivebackaneasydivision.25x0.8,forinstance,canbequitesimplebecauseremoving1/5from25 is a straightforward calculation.On theother hand, 25x0.66 canbenot so simple, sinceremoving1/3fromthesamenumberisnotsoimmediate.Butapartfromthespecificcases,somegood training and experience are the best way to let your eye immediately recognize when thistechniquecanworkatitsbest.

Decompositionintopercentages

And here is where the percentages topic completely embraces the decimal point numbers one. Soimagine, for example, you want to multiply a number by 3.6. By using the previous system youshouldhave triplicated thenumber first and thenyou shouldhaveadded3 times1 / 5of the samenumber,whichcoulddefinitelysoundacumbersometasktoperform.Butthereis,asamatteroffact,amucheasierway,whichconsistsinmultiplyingitby4andthensubtractingthe10%fromtheresult.

Infact3.6isnothingbut4wesubtractedits10%from,andsowecansayasageneralprinciplethatifwe want tomultiply by a number having a decimal point (like 3.6)and the last equals an integernumberweaddedorremovedaspecificpercentagefrom(like4,minusits10%),thenwecansimplifythemultiplicationbymultiplyingbythatintegerfirstandthenbyapplyingthatspecificpercentagetothefinalresult.

Thisprinciplecouldstillsoundcomplex,andthat'swhyit'sbettertowriteacoupleofexamples:

Multiplicationby1.8:Doubleandeliminatethe10%fromtheresult.Infact1.8=2-its10%By2.5:Doubleandthenadd25%totheresult.Infact2.5=2+its25%By3.9:Triplicateandthenaddthe33%.Infact3.9=3+its33%.By4.4:Quadruplicateandthenaddthe10%By7.2:Multiplyby6andthenaddthe20%By8.9:Multiplyby9andthenremovethe10%

Thismethodiscertainlymuchlessintuitive,can'tbeappliedtodivisionandislessflexiblethanthe

previousonebuthasitsenormousutilityinthefactthatanytimeyoubecomeableto"grab"therightpercentage it's much faster than any other strategies seen so far. There is no general rule tounderstandwhenit'spossibletoapplyit:itwilljustcomeautomaticallytoyourmindwhenyou'llbeconfidentenoughwithcalculatingpercentagesandrecognizingthem.

Integersmultiplicationthroughdecimalpointtechniques

Anyamong the techniques for rapidlymultiplyingnumbershavingadecimalpointcanbeused toinstantaneouslymultiplyintegersaswell!

Imagine,forexample,youhavetocalculate758x36.Youcould:

Transformthat36intoa3.6,whichisequivalenttodividingitby10.Applyingthepercentagemultiplicationtechnique!Somultiply758by4(=3032)andsubtractthe10%.Theresultis2728.8.Youmultipliedby3.6insteadof36,sothat'snottheresultofyouroriginalmultiplication,right?Andhowcanweget the right one? It's simple!Sinceyoudivided themultiplier by10beforecalculatingeverything,youjusthadaresultthat'stentimessmallerthantheoneyouneed.Soyoudon'thavetodoanythingbutmultiplyingitby10againinordertogettheresultof758x36=27288.

So,asageneralprinciple,wecanperformanymultiplicationbetweenintegersby:

Checkingwhetheroneoftheintegers,ifdividedbyapoweroften,canbecomeanumberwithdecimalpointyoucaneasilyapplyafasttechniqueon.Ifso,performthatdivisionbythenecessarypowerof10.Performthemultiplicationbyusingthattechnique."Restore" the right result bymultiplying the previous product by the same power of 10 youdividedbybefore.

Let'shavealastexampleofthisinordertomakeitclearer.Incaseof4844x12575,whichnormallycouldbeaveryhard-to-performmultiplication:

Let'sdivide12575/10000,obtaining1.25751.2575=1+0.25+0.0075,whichareall"keynumbers".Let'sapplytheirproperties!4844x(1+0.25+0.0075)=4844+(aquarterof4844)+(4844-aquarter,dividedby100)=4844+1211+3633/100=6055+36.33=6091.33.Wedividedby10000?Nowwehavetodonothingbutmultiplyingby10000.Ourfinalresult,then,is60913300.

Numbersfortravellers

Inches,centimetres,pounds,gallons,litres...what'sthebestthingtodowhenyoustarttobeconfused

betweenquantitiesexpressedinmetricunitsandquantitiesexpressedintheimperialones?Thefirsttrickhere,ofcourse,couldbetoGooglethem,butwhatdoyoudowhenyoursmartphonebatteryisdead,youhavenosignaloryoujustwanttoquicklydoitbyyourself?Themostefficientsolutionafterall,due to itsgreaterspeed, reliabilityandpotential, isalways touse the toolyouhave insideyourhead.Buthow?Simplybyapplyingallthetechniquesseensofar!Let'sseemoreindetailhow(takingtheBritishimperialmeasurementsasareference):

From inches to centimetres:One inch is about 2.5 centimetres long. So, in order to convert yourmeasurefrominchestocentimetresyoucaneasilytakeyourinches,doublethemandaddtheirhalf.Done!

Fromcentimetrestoinches:Hereyoucanusethedecompositionintoexpressions:if2.5=25/10=10 / 4, then you can easily divide the centimetres by 2.5 by "reversing" the 10 / 4, which meansdoublingthemtwicefirst,andthendividingthemby10.

Fromfeet tometres:One foot isabout0.3metres long.Sohereyoucansimplyapproximate it to0.33anddivideby3.

Frommetrestofeet:Dotheoppositeandmultiplyby3.

Frommilestokilometres:Onemileequalsabout1.61kilometres.Ifyouapproximateitto1.60,youcandoubleyourmilesandthenremovethe20%fromtheproduct.Infact,2-20%=1.60.

Fromkilometrestomiles:If1.6=16/10,youcouldgetyourmilesbymultiplyingyourkilometresby10andthenbyhalvingthem4times.Alternatively,ifyouwantsomethinglessprecisebutalittlebit faster,youmightconsider theoppositeequivalence: ifakilometreequalsabout0.62miles,youcanmultiplyby0.62.0.62caninturnbeapproximatedto0.66,andthenyoucansimplyremoveone-thirdfromthemeasureinkilometresinordertogetitexpressedinmiles.

Frompoundstokilograms:Onepoundequalsabout0.45kg.Sogivenaweightexpressedinpoundsyoucanconvert it intokilogramsbyhalving it (multiplicationby0.5)and then removing its10%(0.50-10%=0.50-0.05=0.45).Thatwassimple,wasn'tit?

From kilograms to pounds: One kilogram equals about 2.2 pounds. Then in order to make yourconversion,youcansimplydoublethekilogramsandaddthe10%.Note:herethesequenceofoperationsistheinverseoftheoneseenfortheoppositeconversionbut,as a general rule, if you use the percentage technique for multiplications, there are no fastmathematical rules to get the "inverse" expression for dividing by the same number! Only thefactorisationanddecompositionintoexpressionsareinvertibleoperations!

Fromgallonstolitres:Onegallonequalsabout3.785 litres.Hereyoucaneasilyapproximate it to3.80andsomultiplyby4andthenremovethe5%(4-5%=4-0.20=3.80).

Fromlitrestogallons:Onelitreequalsabout0.26gallons.Ifyouapproximateitto0.25youcanjustdivideitby4andyourproblemissolved!

The techniques seen in this chapter can also be very useful for converting between differentcurrencies. Obviously we can't write a conversion table for that since the conversion rates areconstantly changing, but if you apply any of the techniques listed above you can immediatelyunderstandcostsandexpensesevenifyouareinanyforeigncountry,without...riskingtobetakenin

fromanybody!

Thesectionaboutpercentagesandnumbershavingadecimalpointendshere.Let'smovetothenextchapter and let's try to understand how we can make incredibly simple even the more complexdivisions.

XVIII-Howtomakethedivisionlessdreadful...andtriplicateyourcapital!

Division is probably the most problematic among basic arithmetic operations, as for the expertmentalcalculators,asforthosewhorecentlystartedstudyingmentalcalculation.Andthisisdefinitelyduetothefactthatit'samuchlessintuitiveoperation than, forexample,additionormultiplication,has fewer properties that can let us simplify it, and it depends from a more complex calculationprocedure.

In thischapter,anyway,we'll findaway toapproach thisarithmeticoperationfromanewpointofview,makingitmuchmore"intuitive"and"familiar".Ofcoursewehavealreadyseensomeusefultechniques for this purpose in Chapters IX and XVII, but now we'll see probably the easiest one,alwaysrememberingthatitseffectivenesswillincreaseanytimeyou'llbeabletocombineitwiththeothertechniquesseensofar.

This technique is called "The recursive subtractions strategy", isbasedon the fact that thedivisionconceptually indicates how many times a number can "contain" another number and consists inreplacingthecalculationofthedivisionitselfwithamucheasiersetofsubtractions.

Thismethodhasthedisadvantageofbeingnotreallyso"fast",butitsurecanmakelifeeasiertoallthose who know how to perform subtractions very quickly and / or are not very confident withdivisioncalculationsyet.

Andhereishowyoucanperformit:

Asithappensintheclassicaldivision,ifthedividend(note:thenumber beforethedivisionsign)is smaller than thedivisor (remember: thenumber after the division sign), youmustmultiplyyourdividendby10andwritea"0."inthepartialresult.So,forexample,iftheinitialdivisionis7/15,youhavetowrite70/15=0.andwritethenextdigitsyoucalculaterightafterthatdecimalpoint.

Ifdespitethepreviousstep,thedividendisstillsmallerthanthedivisor,writeanother0afterthedecimalpointinthepartialresultandmultiplyagainthedividendby10,repeatingthisstepasfarasthenumberbeforethedivisionsignisstilllargerthanthedivisor.

If any between the operands has a decimal points, use the invariantive property of division inordertosimplifythedivisionandremovethem.Forexample,ifyouhave56/0.23,multiplybothnumbersby100andtransformtheoperationintoa5600/23.Andthelastoneisexactlythedivisionwe'llapplythefulltechniqueto,inordertoshowhowitworks.Inaddition,we'llreferinthenextstepsto23astheoriginaldivisor,sinceit'sthebasicdivisorbeforestartingtoperformtheactualtechnique(thesethreestepswerejust"preliminaradjustments"toletthedivisionbeeasiertocalculate).

Andhere iswhere the techniqueactually starts: let thedividendand thedivisorhave the samenumberofdigits,bymultiplying thedivisorbyapowerof10 largeenough tomatch them. Forexample,inthiscase,inordertomakesurethat23hasthesamenumberofdigitsas5600,justmultiplyitby100again,havingso5600and2300.Warning: Ifaftermatching thedigits thisway thedividend is lesser than thedivisor, removeazerofromthelastbeforemovingtothenextstep.

Now subtract the dividend from the divisor, and do it as long as the result is still a positivenumber.So,inthe5600/23caseyouhave:5600-2300=3300.Subtractagain:3300-2300=1000.Youcan'tperformsubtractionsanymore,ortheresultwouldbecomenegative.Thencounthowmanytimesyou'vesubtractedthisway: inthiscaseit's2timesandthen2isthefirstnumbertowriteintheresult(totherightofthedecimalpoint,ifyouwroteitinoneoftheprevioussteps).

Now take your last difference (in this case, 1000) and start subtracting from it, but insteadofworkingwith thedivisorusedsofar,subtract thesamedivisordividedby10 (in thiscase thenyou'llhavetouse230insteadof2300),andcontinuedoingthatuntilthenumberisstillpositive.Inthiscase,then,youhave:1000-230=770.Still,770-230=540.Then540-230=310.and310-230=80.Stopherebecauseotherwiseyou'dgetanegativenumber.Youhavesubtracted4timesandthenyoucanwrite4inthepartialresultthatnowis24.

Iftheresultofyourlastsubtraction(inthiscase,80)isstillgreaterthantheoriginaldivisor(inthis case,23), repeat theprevious stepagainusing thedivisor /10 (insteadof230, thenworkwith23).Sointhiscaseyouhave:80-23=5757-23=3434-23=1111-23wouldbenegativeandyousubtractedjust3times.So,write3intheresult,whichnowis243.Attention, here: you just got a number, 11,which is even lesser than theoriginal divisor(23).Thenthepartialresult,243,isalsothefinalresultandthenumberyoustoppedsubtractingatisthefinalremainderofthedivision.

Ifinsteadofstoppingattheremainderyouwanttocontinueandcalculatethedigitsafterthedecimalpoint, you have to do nothing but dividing the divisor by 10 again and continuing with thesubtractionsasintheprevioussteps.Forexample,23/10=2.3andthen:11-2.3=8.78.7-2.3=6.46.4-2.3=4.14.1-2.3=1.8

Youhavesubtractedfourtimesandinfact4isthefirstdigitafterthedecimalpoint.Andyoucanstillcontinuethiswayasfarasyouwant.Ofcoursesometimestheresultscanhaveaninfiniteamountofdigitsandthenitbecomesnecessarytostopjustwhenyouhaveenough"significant"digits.

Nowlet'sapplythistechniquetoaformulathatdefinitelywillcomeinfinanciallyhandy:doyouwantto (approximately)knowafterhow longyour capital investedat a certain rateof interestdoubles?Thenyoujusthavetodivide70bytherate.Ifyouwanttoknowafterhowlongittriplicatesinstead,youhavetodivide115bythesamenumber.

Let's assume, then, that youwere lucky enough to deposit an amount ofmoney in a bank accounthaving an interest rate of 2.75%, that youwant to triplicate that amount and that now youwant tounderstandafterhowmanyyears this"automatically"happens.Soyou'llneedtodivide115by2.75andthiscanbedoneexactlybyapplyingthistechnique:

Remove the decimal points: do not calculate 115 / 2.75. Apply the invariantive property ofdivisioninstead,multiplybothnumbersby100andcalculate11500/275.

Matchthenumberofdigitsbywritingenoughtrailingzeroesinthedivisor.You'llget11500/27500,but...payattentiontotheresultingoperation!Hereinfact,aftermatchingtheamountofdigits, the divisor is greater than the dividend, so you must remove one trailing 0 from thedivisorbeforestarting,gettingthen11500/2750.

Startperformingtheusualsubtractions:11500-2750=8750-2750=6000-2750=3250-2750=500.You'vesubtracted4timesandyoucan'tcontinuewithoutgettinganegativenumber.Thenwrite4intheresult.

500-275=225.You'vesubtractedonceandhavetostop.Write1intheresult.

225 is lesser than theoriginaldivisor275.Thismeans that, ifyouwant tocontinue,youmustwriteadecimalpointintotheresultandwritethenewdigitsyougetrightafterthatpoint.Withoutwritingthewholeprocedureagain,wehavethat27.5canbesubtractedfrom225eighttimesbeforegettinganegativenumber.So thepartial resultwillbe41.8.Theprocedurecouldcontinueagain,butofcourseit'snotnecessary:wejustgotthat,inordertotriplicateacapitalatthereasonablerateof2.75%,weneedmorethan40years.Andthismeansthattheonlywayto

effectivelytriplicateanamountofmoneythroughaninvestmentistohavealotofpatience...ortolookforhigherratesinvestments,acceptingalltheinevitablerisksthatwillcomewithit!

But speaking about risks and randomness: in the next chapter we will talk about one of themostinteresting,excitingandevenusefulmathematicaltopics:theprobabilitytheory.Let'smoveon!

XIX-Gambling,guessingandgaining

SaidArthurBenjamin,afamousmathematicianduringaTEDconference,thatinsteadofconvergingto calculus, every mathematics programme in schools should converge towards a study of theprobability theory instead, since it's probably the most charming, helpful and funny branch ofmathematics.Headded, also, that a greater commonknowledge of statistics andprobabilitywouldalmostcertainlyhaveavoidedtherecentfinancialcrashandhelpedeveryonetotakewisereconomicandfinancialdecisions,withbetterconsequencesforthewholemodernsociety.ButapartfromBenjamin'spointofview,it'sundeniablethatunderstandingstatisticsandprobabilitytheorymeansacquiringpowerful tools toput togetherdata,conqueringagreaterunderstandingofthefuture...andevenincreasingourchancestowinsomemoremoneyingamblinggames!So, after bothering you with division, percentages and multiplication between numbers havingdecimalpoint, in this chapterwewilldiscuss their application inprobability theory, andwe'll startrightbyintroducingoneofthemostusefulandimportantprobabilityindexes:theso-called"indexofexpectedvalue",or"expectedvalue"wealreadymentionedinChapterVI.Theexpectedvaluecanbequitesimplydefinedastheaveragegainwegetbydoinganaction,incasewekeepdoingitinthelongterm.Itcouldbe,forexampletheexactanswerto:"HowmuchmoneyIgain(orlose)eachtimeontheaverageifIcontinuetobetthreeeuroonredattheroulette?"Theformulatocalculatethisindexisverysimple:(ProbabilityofgainxGain)-(ProbabilityoflossxLoss)anddecidingwhat todooncewecalculated it is reallystraightforwardaswell: in fact, if thecalculatedexpectedvalueispositive,thencontinuingtodothatactionwillletusgainanadvantageinthelongterm.If it'snegative,will letuswasteresourcesinstead.Inshort:do ifpositive,don'tdo ifnegative!Nowitcouldbelegittoaskaquestion:howcanIactuallycalculatetheprobabilityofoccurrenceofanevent?Well,theprobabilitytheoryexplainsthatthiscanbeessentiallydoneintwodifferentways,eachofthemhavingitsadvantagesanddisadvantages:

Oneconsists in lookingat theresultsofsimilareventshappened inpastand incalculating theprobabilityasthenumberoftimestheeventhasoccurred,dividedbythenumberoftimesthatit"couldhavehappened."Forexample,itcouldbesaidthattheprobabilitythatanaircrafthasanaccidentduringitsflightisequalto"Airplaneaccidentsoccurred"divided"Flightslandedwithoutanyaccident."Thismethodhas the disadvantage of being clearly impractical if youdonot have access to a

sufficientamountofconsistentdata,anditmayleadyouintomanydilemmas,reallynotworthdealingwithhere,aboutwhatcanactuallybeconsideredas"consistent" (for example, does itmakesensetousetheflightdatafromairplanesbuiltin1920tounderstandmychancestotravelsafelytoday?Andthosefromplanesfromdifferentairlines?Andthoseofcompletelydifferentmodels?).Clearly, on the other hand, this is a method that has the undoubted advantage of being veryprecisetocalculatetheprobabilityofoccurrenceofmorecomplexphenomena.

Anothermethodtocalculateprobabilitysimplyconsistsinsaying:"Ifaneventcanendupinxpossibleways,theprobabilitythatanyofthosehappensis1/x."Letmeclarify it throughanexample: if you flip a coin, andheadsor tails cancomeout (so,we'retalkingaboutaneventthatcanendupin2possibleways),theprobabilitythatoneofthemhappens,forexampleheads,is1/2,oroneoutoftwo,or0.5.Ofcoursethereisnoneedtohavethedataofallthecoinstossedintheworldfromtheancienttimestilltodaytogetthat.Samethingifyourollasix-sideddie:attheenditcancomeoutofanumberfrom1to6,soit'sanevent thatcanendupin6possiblewaysandtheprobability thatONEof themhappens(theprobabilitythata4comesout,forexample)is1/6.It'ssimpleafterall,isn'tit?

Let'saddsomethingmore:ifwehaveaneventthatcanendupinxpossiblewaysandwewanttoknow the probability that any among two, three, four, etc. of those happen, in that case theprobabilitywillbesimply2/x,3/x,4/x,etc.So if for example you have your die to roll and youwant to know the probability that "oneamong2,3or4comesout"(probabilitythataneventwith6possibleevolutionsendsupinoneif3possibleways),theprobabilityis3/6=1/2.Yeah,Ijustdividedthenumberof"interesting"eventsbythenumberoftotalevents.Samethingif,forinstance, thereisalotterywhosewinnerisdrawnoutof90total ticketsandyouboughtoneticket:thereyourchancesofwinningwouldbe1/90.Butifyoubought20ticketinstead,yourchancesthenwoulddrasticallyincreaseto20/90.

This second probability calculationmethod has the clear disadvantage of being further awayfromrealityandthereforemuchlessabletodescribecomplexphenomena.Forexample, ifwego back to the above-mentioned airplane case, itwould be absurd to say that a flight has 0.5probabilitytoendinanaccidentjustbecausewehavethetwo"Accident/noaccident"choices.However,thismethodismucheasiertouse,andiswellsuitableforrepresentingwithsufficientaccuracy phenomena that are simple enough to let each case have the same probability ofoccurrenceastheotherones.

A mathematical note before moving on: in these strategies we talked about, the probability isexpressedintermsoffractions,whosevalueisbetween0(impossibleevent)and1(sureevent)insteadof the more widely used percentages.And we decided to do like this because this notation willsimplifyourcalculations.However,theconversionbetweenthetworepresentations(thatarenothingmore than two different ways to express the same value) is quite trivial: in order to switch fromfractions to percentages youmustmultiply the probability expressed in fractions by 100, while inordertohavethefractionsbackyoumustdividetheoneexpressedinpercentagesby100.Andso,forexample:

"100%probability"infractionswillbenothingbut"1",tobeinterpretedas"anytime"."50%probability":aftercalculating50 /100 itwillbe0.5.Andsince0.5=1 /2, it's the samethingassayingthat"theeventhappensonceintwotimes."Conversely,ifyouhaveaneventthatcouldhappen"Fivetimesaweek"(fivedaysoutofseven),itsprobabilitywillbe5/7=0.71,whichmultipliedby100=71%probability.Ifyouhaveaneventthathappens"Sixteentimesoutofseventy",theprobabilitywillbe16/70=0.23,whichmultipliedby100equals23%.

Lasttwothingsworthknowing:

"AND" probability: the probability that, given two (or more) independent events (where"independent"meansthattheoccurrenceofoneofthemcannotinfluencetheotherone),bothofthem happen, can be calculated by multiplying the occurrence probability of an event by theoccurrenceprobabilityoftheotherone.Forexample,iftheprobabilityofhavingheadsafterflippingacoinis1/2,theprobabilityofhavingheadstwiceafterflippingtwocoins(orflippingthesamecointwice)is1/2x1/2=1/4.

"OR"probability: theprobability that,given twoormore independentevents,at leastoneofthemhappens,isgivenbythesumoftheprobabilitiesthattheeventsindividuallyhappen,fromwhichwemustsubtracttheprobabilitythattheyallhappen(andthelattercanbecalculatedlikewejustsaid).So if for example youwant to flip two coins and calculatewhat is the probability that headscomesoutatleastonce,youhavetocalculate:1/2(theprobabilityforheadstocomeoutfromthefirstcoinflip)+1/2(theprobabilityforheadstocomeoutfromthesecondcoinflip)-1/4(theprobabilitythatheadscomesoutfrombothflips)=1-1/4=3/4=0.75=75%.

Same thing if you roll a dice first, then you flip a coin and you want to know what's theprobabilitythatheadscomesoutfromthecoinflipor6comesoutfromthedieroll.Inthatcase,infact,you'llhavetocalculate:

1/2(coinflip,probabilityofoccurrenceofanyresult)+1/6(dieroll,probabilityofoccurrenceofanyresult)-1/12(theproductofthepreviousones,whichistheprobabilitythattheybothhappen)=7/12=0.58=58%.

Now you have all the necessary tools to read the curiosities seen in Chapter X from a widerperspectiveandtounderstandallofthemmoreindetail!So, let'smakeanexampleof thepracticalutilityof theexpectedvalueindexbyexamininghowtheAmericanrouletteworks.Asmostofpeopleknow,theAmericanRouletteisaspinningdiscdividedinto38numberedslots(18reds,18blacks,azeroandadoublezero).Aballislaunchedonthewheelandanyplayercanbetanamountofmoneyonthenumber(orcolour,group,etc.)theballwillstopat.Nowwe'llcalculatetheexpectedvalueofrepeatedlybetting1dollaronthered.Let'srememberagain

thattheexpectedvalueindexformulais(ProbabilityofgainxGain)-(ProbabilityoflossxLoss)andlet'strytounderstandmoreindetailwhichexactlyarethecomponentsoftheformula:

Gain=1euro,becausethisistheamountofmoneywe'llwiniftheballstopsonaredslot.Loss=1euro,becausethisistheamountofmoneywe'llloseiftheballstopsonablackslot,onthezerooronthedoublezero.Probability of gain: being 18 red slots, a bet on redwins 18 times out of 38.We can use thesecondmethodofprobabilitycalculationandwehavethatit's18/38=0.47(or47%).Probabilityofloss:weloseiftheballstopsononeamongthe18blackslotsorthe2greenones,soabetonredloses20timesoutof38andtheprobabilityoflossis20/38=0.53(or53%).Andhereyoucanstartnoticingthe"suspicious"factthattheprobabilityoflossishigherofthegainone.

Theexpectedvaluethenis(0.47x1)-(0.53x1)=0.47-0.53=-0.06.So,whatdoesthisresultmean?Thatforeacheurowebetonred,we'llgetanaveragesixcentslossifwe keepplaying.And you can verify by yourself that the valuewill not change if you calculate itconsideringblackinsteadofred.Therefore,inevitably,ifwecontinueplaying,nomatterwhatresultwebeton,thecontentofourwalletismeanttoshrinkmoreandmoreinthelongterm.Andthis isnothingbutthemathematicaltricktheCasinosgetricherandricherwith.Butthisprecioustoolcanbeactuallyusedinvarioussituationsandnotjustingambling.Forexample,wecanuseittoimproveourperformancesinmultipleansweringtests.Forinstance,let'simagineyouhaveatestinwhichthequestionscanbeansweredbychoosingamongfiveanswers:A,B,C,D,E.Letusassumethateachcorrectanswerwillletusearn3pointsandthatthere isnopenalty forwronganswers.Moreover let's imaginewedon't knowsomeof the correctanswersandthatsowejustwanttotryguessingthem.Solet'staketheexpectedvalueformulaandlet'sseehowcouldweguessinordertomaximiseourfinalresult:

Gain=3pointsLoss=0,becausetherearenopenaltiesforwronganswersProbability of gain = since the choice is among five possible answers, the probability ofguessingtherightoneissimply1outof5=1/5=0.2=20%Probabilityofloss=theprobabilityofguessingthewrongansweris4outof5=0.8=80%,butsincethelossis0thatdoesn'tmatter.Infact:

(1/5x3)-(4/5x0)=(1/5x3)-0=3/5

Theexpectedvalueindexhereispositive,whichmeanswe'llgetanaveragegainofsomepoints(3/5ofapoint,whichisstillbetterthannothing)foreachguessedanswer.Sotheruleyoucandeducefrom this reasoning is that if there is no penalty for wrong answers in a multiple answering test,guessingincasewedon'tknowananswerisalwaysmoreconvenientthanleavingablankbox.Butwhathappensif, inthesametest,weintroduceapenaltyof2pointsincaseofwronganswers?Well,we'llhavethattheformulabecomes:(1/5x3)-(4/5x2)=3/5-8/5=-5/5=-1

So,now,withsuchapenalty,theexpectedvalueindexbecomesnegative,andforeachguessedanswerourriskistoloseonepoint.Thereforeinthiscaseleavingablackboxisthebestthingtodo.Itwouldbedifferent,however,ifwecouldeliminatethreeanswersfromtherangeofpossiblechoices,andstartguessingjustbetweentheremainingtwo.Infact,inthatcase,theprobabilityofguessingtherightonewouldrisefromit1/5to1/2,whiletheprobabilitytobewrongwoulddropfrom4/5to1/2.Thenwewouldhave:(1/2x3)-(1/2x2)=3/2-2/2=1/2,whichisapositiveexpectedvalueagain.We just noticed then that the presence of a penalty is not enough to let us clearly take ourdecision. How to understand, then, more specifically, whether guessing is more convenient thanleavingablankbox?Justafterapplyingsomemathematicalequivalencetotheexpectedvalueformula(throughstepswewon'treporthere),wehavethatguessingismoreconvenientwhen:

Pointsforacorrectanswer>PenaltyPointsforawronganswerx(Possibleanswers-1)

This justmeans youmustmultiply your penalty points by (Possible answers - 1), and then checkwhether theresultingnumber isgreateror lesser thanthepointsforasingle,correctanswer. Ifso,guessingisconvenientwhenyoudonotknowananswer.Otherwiseyoujustshouldn'tansweratall.In addition, from this formulawecandeduce than ifwecan safely exclude some choices from thepossible ones, our expected value increases, the quantity after the "greater than" sign will besmaller,andsowecouldmeetthecaseinwhich,evenif"Pointsforacorrectanswer"islesserthantheothernumber,itwon'tbelikethatanymoreafterperformingourexclusions.

Beforeending thischapter, let's extend the last formula to thegenericcase, and let's transform theoriginalquestion itwasanswering to, intoa"When should I just try todo somethingandcontinuedoingitinthelongtermdespitetherandomcircumstances?"

Afterapplyingtothegeneralexpectedvalueformulasomestraightforwardmathematicalequivalenceagain,wehavethatweshouldcontinuedoingsomethingwhen:

ProbabilityofGainxGain>ProbabilityofLossxLoss

Despiteitssimplicity,thisformularepresentsanamazingtreasurebecause,ifusedintherightway,can help you to save thousands dollars and tomaximize your results inmany sectors, from yourbusiness toyour interpersonal relationships.For example,when somebody saysyou that youhave"nothing to lose" inasking thatgirlout, thatmeans that itdoesn'tmatterwhat'syourprobabilityofbeingactuallyrefused:yourcouragewillalwaysletyougainaconsistentadvantageinthelongterm.

"I like to look at mathematics almost more as an art than as a science; for the activity of themathematician,constantlycreatingasheis,guidedthoughnotcontrolledbytheexternalworldofthesenses,bearsaresemblance,notfancifulIbelievebutreal,totheactivityofanartist,ofapainterletussay."

(Bocher)

XX-Agameoftimeandprecision

Inthischapterwe'lltalkaboutestimationandproofby9,twoincrediblyvaluabletoolsforanymentalcalculator, since they respectively let anybodyperformmoreprecise calculations in exchange foraloss of speed or, vice versa, to calculate more rapidly in exchange for a loss of precision. Morespecifically,wehavethat:

Theestimationmaybeperformedbeforeanycalculationandletsyouknow"moreorless"andalmostimmediatelyitsresult.

Theproofbyninecanbeperformedafteranycalculationandcanalmostcertainlytellusifwemadeamistakeduringtheprocedure.Curiosity:thereisevidenceofusageofthistechniquesincethethirdcenturyAD,knownbythenameofabjectionovenaria.However,sincetheRomanswereperformingcalculationsthroughpebbles,theyquiteunlikelywereevenabletodemonstratewhyitworked.Thefirstdocumentedexplanation of its process in fact is much older, more precisely dated 1202, and was foundamong the pages of the Liber Abaci by Fibonacci, the same book that brought the ArabicnumeralstoEurope.

So,let'sstarttalkingrightaboutestimationandlet'sstartdistinguishingbetweenthetwobasictypesofestimationsyoucanapplytoyourcalculationsinordertomakethemextraordinarilyfaster:

Estimationbyrounding

Roundingconsists innothingbutreplacingwithzeroessometrailingdigits fromthenumbersyou'recalculatingwith.This,anyway,isnottheonlysteptoperform.Infact,inordertomaketheroundingalittlebitmore"consistent",youhavetocheckwhethertheleftmostremoveddigitwasgreaterthan5and,ifso,thenyouhavetoincreaseby1therightmostnon-nulldigitinyournumber.Sinceit'seasierdonethansaidlet'sexplainitthroughanexample,andlet'sapplythisstrategytoamultiplicationlike7845x9871:

Youcanreplacethelastthreedigitswithzeroes,getting7000and9000.

Theleftmostremoveddigitin7845wasan8,whichofcourseisgreaterthan5.Sowe'llhavetoincreasetherightmostnon-nulldigitin7000by1,gettingso8000.Eventheleftmostdigitweremovedfrom9871wasan8,sowemustincreasethesecondroundednumberaswell,getting10000.7845x9871=approximately8000x10000=80000000

Hereyoucannoticethattheactualresultof7845x9871was77437995,that'sverycloseto80000000.Obviously,inorder togetamorepreciseresultwecouldhaveappliedroundingbyreplacingjust the two trailing digits instead of the last three, calculating then amuch "closer" 7800 x 9900insteadof8000x10000.Theresultofthelastoperationisinfact77220000,whichisinfactevenmuchmore"similar"tothe"real"result.Sobasically,morearethedigitsyoureplacewithzeroes,lesstheestimateisclosetotherealresult,andfaster/easieristhewholeprocedure.Andofcourseit'suptoyoutodecidetimetotimehowmanydigitsit'sconvenienttoreplace,dependingonwhetheryouneedafasterresultoramorepreciseone.

Here you may notice that a similar technique, referring to numbers having a decimal point,was introduced in Chapter XVII. And it may be worth noticing that, in that case, the roundingtechniquewasmuchmore"painless"toperform: infact, ifwetakeanynumberandwejust removesomedigitsafteritsdecimalpoint(that'sthesameasreplacingthosedigitswithzeroes),weusuallychangeitjustbyaverysmallquantity,andconsequentlytheapproximationofacalculationinvolvingthatnumberusestobemuchmore"acceptable"thanwhatjusthappenedwithreplacingzeroes"totheleft"ofthedecimalpoint.

I'llexplainmyselfbetterabout the lastobservation: ifwewant tocalculate7.8793x6.4339andweroundbothnumbers to7.9x6.4,weget that theestimatedresult is50.56.Therealonewouldhavebeen 50.694628, which differs from the estimation just by 1 / 10, and this quantity ismuch moreirrelevant than the3millionsdifferencewehad inour first "7845x9871"example.So,always becarefultothe"orderofmagnitude"oftheapproximationyougetafterroundingyournumbers.

Estimationbyreplacementwithexpressions

Thistechniqueconsistsinnothingbutreplacinganumberwithafractionofamultipleof10and, ifcomparedtothepreviousone,givesonethechancetogetmuchmoreaccurateresultsinexchangeforamuchfasterandmoresimplecalculation.

However,itpresentsthesameproblemasthe"decompositionintoexpressions"wesawinChapterIX:it's not easy to immediately understand which expression you should replace your numberswith.However, onceyou trainedyourself toworkwith largenumbers, getting thatwill bekindofautomatic foryou.Also, trying tounderstandandmemorise thefollowingexampleswilldefinitelyhelpyoutogetthehangofthiskindofsubstitutions:

142=about1000/7.Soifforexampleyouhavetomultiply148x344,youcouldapproximate148to142and142to1000/7.Thenyoucangetanacceptablyestimatedresultofthemultiplicationjustbyaddingthreezeroesto344andthendividingitby7.Theresultofthelastoperationisabout49142,

while theoriginalresultwas50912.Asyoucansee, thedistancebetweenthe twonumbers ismuchsmallerthantheonewe'dhavehadafterusingtheroundingtechnique:inthatcase,infact,wecouldhaverounded148to100and344to300,obtainingthequiteimpreciseestimationof100x300=30000.

333=about1000/3.Thenifforinstanceyouwanttomultiply734x330,youcouldapproximate330to333,thenperformthemultiplicationbyaddingthreezeroesto734andatlastdivideitby3=about244666,veryclosetothe"real"242220.

428=about3000/7.

666=about2000/3.

999=about3000/3.

1666=about10000/6.

Andsoon.

Nothingelsetoaddabouttheestimationbut...exercise,exercise,exercise!Solet'sgotheotherwayroundandlet'strytounderstandhowcanweusetheproofby9inordertoincreasetheprecisionandtheaccuracyof the calculations involvingall four elementary operations, in exchange for ... somelittlemoretimetoperformitafterourbasicmathematicalwork!

Proofby9foradditionandsubtraction

First,here'showyoucanperformtheproofby9foraddition.Theproofforsubtraction,infact,willrequiretoexecutethesameprocedurewithjustsomelittlevariants:

Foreachsummand,calculatethesumofitsdigits.Thensumtogetherthedigitsofthelasttotaluntilyougetasingledigit.Wewillcallthisoperationthe"Mod9"ofthenumbersinceit'sequaltotheremainderyouwouldobtainafterdividingthatnumberby9.So,let'ssupposeyoujustcalculated237+344=581.InordertogettheMod9of237,you'llhavetosum2+3+7=12first,andthen1+2=3.Samefor344:3+4+4=11and1+1=2.

Nowsumthemodsyougotinthepreviousstepand,ifthetotalismadeupoftwoormoredigits,calculateitsMod9.Inthiscaseyoujusthavethat3+2=5.

NowcalculatetheMod9oftheresultofyouraddition.Inthiscase,5+8+1=14and1+4=5,again.

Did you notice that? The sum of themodules of the addends equals the module of the total.Andthisbringsus infrontof thebasicconceptbehind theproofby9,whichnamely is: if the two finalmodulesdon'tmatch,youcanbesurethattheresultofyourcalculationiswrong.However,ifthetwofinalmodulesarethesame,youcan'tbesurethatyourresultisright.

Inshort,theproofby9canclearlyletyouknowifyouwerewrongbutcan'tpositivelytellyouwhether

you've performed a calculation properly. Anyway,we have that the resultswillmatch in case of awrongcalculation1timeoutof9.So,asfarasthiscanconcretelycomeinhandy,wecansaythatit'squite"rare"toseetwomodsmatchingafterperformingawrongcalculation.

So, let's immediately go to the proof by nine for subtraction. As we said, it's very similar to theadditionone,exceptforjustsomelittledifferences.Let'sseethem:

CalculateagaintheMod9ofeachofyouroperands.Soifforexampleyoujustcalculated788-215=573,thenyouhavetosum7+8+8=23and2+3=5.Afterthat,justperformthesamecalculationontheothernumber,getting2+1+5=8.

Now,insteadofsummingthemodulesyouobtainedinthepreviousstep,justsubtractonefromeachother,andifyougetanegativenumber,simplyadd9.Theninthiscaseyouhave5-8=-3,andafteradding9youget6.

Nowtaketheresultofyoursubtraction,andcalculateitsMod9.Inthiscaseyouhave5+7+3=15and1+5=6,whichmatchesthepreviousresultagain.

Butwhat if the result of the subtractionwasnegative?Well, in that caseyou shouldhavedone theMod9asalways,butattheendoftheprocedureyou'dhavewrittenanegativesignnexttotheresultofitsmodule,andthenadd9.Butlet'shaveanotherexampleinordertoclarifythelastconcept.

Let'simagineyouwanttochecktheresultof45-388=-343.TheMod9of45is9,whiletheoneof388is1.Thefirstdigitweneedtoperformourproof,then,is9-1=8.

Nowlet'scalculatetheMod9oftheresult:3+4+3=10and1+0=1.Let'swriteanegativesignnextto1,getting-1andlet'sthenadd9,getting8.Done!

Proofby9formultiplication

Thisisperhapsthemostwidelyusedkindofproofby9,andhere'showyoucanperformit.You'llnoticethattheprocedureisverysimilartotheadditiononeaswell:

CalculatetheMod9ofyourmultiplicand.Forexample,ifyoujustcalculated45x62=2790,add4+5=9CalculatetheMod9ofthemultiplier.Inthiscase,6+2=8.Multiplytheresultsofthepreviousmodulestogetherandiftheresultingproductismadeupoftwoormoredigits,calculateitsMod9.Inthiscase,9x8=72and7+2=9.NowcalculatetheMod9ofyourresult.Inthiscase2+7+9+0=9+9=19,and1+8=9.Theverification tobemadehere isalways thesame: the firstmodulemustbe identical to thesecondone.Asusual,ifso,wehaveprobablyperformedourcalculationproperly,otherwisewedefinitelymadeamistake.

Proofby9fordivision

Theproofby9forthedivisionisexactlythesameastheoneformultiplication,withthenecessarypremisetoperformitonlyincaseofintegerquotients.

Morespecifically,sincethedivisionistheinverseoperationofmultiplication,inthiscaseyouhavetodonothingbut:

CalculatingtheMod9ofthequotient.CalculatingtheMod9ofthedivider.Multiplyingthepreviousresultsand,again,iftheproductisatwo-or-more-figurenumber,thencalculateitsMod9.ComparingthisresultwiththeMod9ofthedividendanddrawingyourusualconclusions.

An useful but not necessary notion before ending this topic: sometimes the proof by 9 can beperformedinunionwithamorecomplexbutmoreaccuratetest,whichisnamed"proofby11"andconsistsinexecutingexactlythesamestepsastheproofby9,butreplacingthe"Mod9"calculationwiththe"Mod11"one.Andwhat'sthe"Mod11"about?Well,exactlyashappensintheMod9case,it'snothingbuttheremainderyougetifyoudividethatnumberby11andcanbecalculatedthisway:

Startingfromright,sumtogetherthedigitsinanoddplace(sothefirstontherightplusthethirdontherightplusthefifth...andsoon).Nowsumthedigitsinanevenplace(sothesecondontherightplusthefourthplusthesixth...andsoon)Subtractthelasttotalfromthefirstone.So,forexample,theMod11of341is(1+3)-4=0.

SincetheMod11isnotasimmediatetocalculateastheMod9,the"Proofby11"isclearlyabitmorecomplexandslower toperformbutwhereveragreateraccuracy in results isneededandyouhaveenoughtimetoperformalltheadditionalcalculationsinvolved,itcanbeusedaloneorinunionwiththeproofby9inordertogetamuchmoreaccurateresultcheck.Infact,ifyouapplybothteststoacalculation andyou findout that the finalmodules are the same forboth, you canbe sure that theresultwillthenbecorrectin99%ofcases,whichisstillnot100%,butcanbeanywaymuchmorethanwhatweactuallyneed.

XXI-10strategiestosquarenumbers...andastrategytosaveyourlife!

It'ssaidthatwhenEuler,mathematicianwholivedintheageofEnlightenment,endedupworkingonperfectsquares,wassofascinatedbytheirinnerharmonythathestartedbelievingthattheyweretheunquestionableevidenceoftheexistenceofGod.Clearlythestudyofthephilosophicalbeliefsthatonecanaccrue fromhispersonal approach tomathematics is not a topic thisbook is supposed to talkabout,soIwilljustcarefullyobservethat,asEulernoticed,squaringintegersisoftenaverysimplecalculation to perform and, as seen in the "golden numbers" chapter, it often produces, as results,incrediblysymmetricalandharmoniousnumbers.

But let'smove immediately topracticeand let'sgo throughall thestrategiesuseful forperformingthesecalculationsinthefastestandmoreefficientway.Maybe,unlikeEuler,youwon'tseethehandofsome almighty God there, but I'm sure they will extremely simplify your life on more than oneoccasion.

1-Memorisethebasicsquares

Let's start this chapter right by introducing the basic squares table. Acquiring the ability toimmediatelyrecallfrommemorythesquaresofthe20smallestpositiveintegerscaninfactbeveryhelpfulinspeedingupthecalculationofthesquaresforlargernumbers:

Squared1=1

Squared2=4

Squared3=9

Squared4=16

Squared5=25

Squared6=36

Squared7=49

Squared8=64

Squared9=81

Squared10=100

Squared11=121

Squared12=144

Squared13=169

Squared14=196

Squared15=225

Squared16=256

Squared17=289

Squared18=324

Squared19=361

Squared20=400

Acuriosityforgourmetlovers:didyoujustnoticethatsquared9isalmosttwicebiggerthansquared7andthatsquared12isalmosttriplethanthesamequantity?WhyamIsayingthat?Well,sincetheareaofacircleexactlyincreaseswiththesquareofitsradius,youjustgotthata7''pizzaisbigalmostthehalfofa9''oneandonethirdofa12''one.Andsinceyou'llneverpaya12''pizzathreetimesthepriceofa7''one,thismeansthatbuyingalargerpizzaisalwaysthemostconvenientchoice,becauseitwillletyouhavemuchmorepizzainexchangeforamuchsmallerprice.Inthiscase,again,mathisyourmostpreciousally!

2-Usingfastmultiplicationtechniquefornumbersbetween11and19

Yeah,sometimesmemorisingthesquaresofthefirst20positiveintegerscanbejustaveryhardtaskto accomplish.Usually, in fact, most of people easily remember only the squares of the first 10integers(essentiallybecausethey'reinthemultiplicationtablestheylearntatschool)andso,inordertorapidlygettheremainingsquareswhenourmemorydoesn'treallyhelpus,wecansimplyusethefast multiplication strategy for numbers between 11 and 19 we introduced in Chapter XIV. Afteradaptingittotwoidenticalnumbers,infact,itbecomes:

Sumthenumbertoitsunits.

Multiplytheresultby10.

Squaretheunitsofthenumbertosquareandaddittotheproductyougotinthepreviousstep.

Sotheprocedureforrapidlysquaring17,forexample,wouldbe:

17+7=24

24x10=240

240+7x7=240+49=289

Thistechniquecanalsobeusedtoeasilycalculatethesquaresofnumberslike110,120,130,140,etc.Infact,inthiscase,youhavetodonothingbutrepeatingthesameprocedureastheywere11,12,13,14...withtheonlydifferencethatyoumustwritetwotrailingzeroesintheresultattheend.

3-Strategyfornumbersendingin"1"

Squaringnumbersendingin"1"isactuallyaveryquickandsimplecalculationtoperform:

Takethenumbertosquare,removeitsunitsdigitandsquarewhatremains.Doublethesamenumberwithoutitsunitsandwriteittotherightoftheresultyougotinthefirststep.Ifthelastnumberismadeupoftwoormoredigits,carrytheextradigittothesquareyougotinthefirststep.Writea"1"attheendofthepartialresult.

Let'simagineforexampleyouwanttocalculatethesquareof41.Youwillneedto:

Removetheunitsdigit,get4andsquareit=16.Double4=8.Nowwriteittotherightofyourpreviousresult=16_8.Write1attheend=1681.

Samethingcanbedonewithnumbersmadeupofthree,fourormoredigits.Theonlyprobleminthatcasewillbethatobviouslysquaringanddoublingthenumberwithoutitsunitswon'tbeasimmediateas in the two digits case. Anyway using some of the fast multiplication techniques we saw in thepreviouschapters, like theTrachtenbergone, shoulddefinitelyhelpus to simplifyourcalculationsevenincaseoflongernumbers.

4-Strategyfornumbersendingin"5"

Thestrategyforsquaringnumbersendingin"5"isprobablyeveneasiertoapplythanthepreviousone.Infact,incasesliketheseyoujusthaveto:

Takethenumber,removeitsunitsandmultiplywhatremainsbyitself+1.Write25attheend.

So,ifforinstanceyouwanttorapidlysquare75,youcan:

Removetheunitsfrom75andmultiply7x(7+1)=7x8=56Write25attheend:5625

Eveninthiscasetherulecanbeappliedtothree,fourormorefigurenumbersaswell,withtheonly,usualinconvenientoflongerandmorecomplexcalculations.Butevenherethetechniquesseeninthe

previouschapterscancometoourhelp.Infactifforexampleyouwanttosquare335,youcan:

Multiply33x34.HereyoucanusethetechniqueseeninChapterXIIIandeasilygetthattheresultis1122Writea25attheend:112225.

5-Strategyfornumbersendingin"25"

Evenrapidlysquaringnumbersendingin"25"isquiteanelementarytask.Morespecifically in thiscaseyoucan:

Takethenumberyouwanttosquare,removethe"25"andsquarewhatremains.Nowaddtotheprevioussquarehalfofthesame,originalnumberwithoutitsfinal"25".Multiplytheprevioustotalby10.Write625attheend.

Supposeyouwanttocalculatethesquareof825.Inthiscaseyou'lljusthavetoexecutethesesimplesteps:

8x8=64.64+8/2=64+4=68.68x10=680.Write625attheend=680625.

6-Strategyfornumbersbeginningwith"5"

Thisisanothercalculationstrategyverysimpletolearnandapply.Let'sstartlookingattheproceduretorapidlysquareanumberbetween51and59first,thenwe'llextendthetopictothreeormore-figurenumbers:

Squaretheunitsdigit.Subtract 25 from the number you want to square and write the difference to the left of theprevioussquare....done!

Soifforinstanceyouwanttorapidlysquare56,youcan:

Square6=36.Calculate56-25=31.Thefinalresultissimply3136.

Asyoucansee,it'saquitestraightforwardprocedure.So,whichexactlyarethedifferencesincaseofthree,fourormore-figurenumbers?Well,let'sseethem:

Squarethenumberwithoutitsleading5.Take25andwritenexttoitasmanytrailingzeroesasyouneedtomatchthenumberofdigitsof

thenumbertosquare(soyou'llhavetotake250ifyouwanttosquare512,2500ifyouwanttosquare5887,25000ifyouwanttosquare54321,andsoon).Subtractthemultipleof25yougotinthepreviousstepfromthenumberyouwanttosquareandwritetheresulttotheleftofthesquareyoucalculatedinthefirststep.

Thisstrategy,appliedforinstanceto531,wouldthereforeconsistinthefollowingsteps:

Squared31=961 (andhere you canuse the strategy for rapidly squaringnumbers ending in"1").Calculate531-250=281andwriteittotheleftofthepreviousresult.Thefinalresultisexactly281961.

7-Usingthe"equidistantnumbers"multiplicationtechnique

InChapterXIVweexplainedthatit'spossibletomultiplytwonumbersbycalculating theirsquaredmeanminustheirsquareddistancefromthatmean.

Sonowwecanjustreversethatformulaandthereforesaythatit'spossibletosquareanynumberjustbymultiplyingtwonumberswhosemathematicalmeanisthenumbertosquare,andthenbyaddingthesquareof theirdistance from thatmean.Transforming in fact apotentiallyhard-to-performsquareintoasequenceofeasiertaskscanseriouslysimplifyourlifeifwedoitproperly.

Well,thefirstquestiononecouldaskhereis:"howcanIfindtwonumberswhosemathematicalmeanisthenumberIwanttosquare"?

It'squitestraightforwardactually,sinceyoucanjustchoosethepreviousandthenextinteger,orthosenumbersyouobtainbyaddingtothenumber+2and-2,-3and+3,+4and-4,andsoon.Forexample,ifyouwanttofindtwonumberswhosemathematicalmeanis38youcanjustchoose37and39,36and40,35and41,andsoon.

More specifically since, taken an integer, there are infinite numbers whose mean is that specificinteger, the best rule tomake thismethod as easier as possible is: take two numbers that are easyenoughtomultiplytogether.Andthiscanbeachievedforexamplebychoosingacoupleinwhichatleastonenumberisamultipleof10(forexample,bychoosing36and40inthe38case).

Soifyouwanttorapidlycalculatethesquareof38youcan:

Multiply37x39andadd1(distance=1thenthesquareofthedistance=1).Multiply36x40andadd4(distance=2thenthesquareofthedistance=4).Multiply35x41andadd9(distance=3thenthesquareofthedistance=9)....andsoon.

Inthiscase,likewejustsaid,theeasieststrategyisdefinitelytocalculate36x40=1440andthenadd4,resultingin1444,whichisjustthesquarewewerelookingfor.

8-Takingadvantageoftheproximitytoanothersquare

Simplybyusingavariantof the formulawesaw in thepreviouspoint,wecaneasilycalculate the

squareofanumbernincaseweknow (orwecaneasilycalculate) thesquareofanother"referencenumber"rthat'scloseenoughtoourn.

Moreindetail,inordertosquareourn,wecan:

SumnandrMultiplythistotalbythedistancebetweennandr(rememberingthatthedistanceisobtainedbysubtractingthelessernumberfromthegreaterone).Takethesquareofryoualreadyknow(oryoucouldeasilycalculate)andsumittotheresultofthepreviousproduct ifyour"reference" r is smaller thann. If r isgreater thann instead,youhavetosubtractthatresultfromtheknownsquare.

Thestrongpointofthistechniqueisinthefactthat,asareference,youcantakeanyincredibly-easy-to-squarenumber,liketheclosestmultipleof10,orthenearestnumberendingin1or5.Butlet'shaveanexampleofthistechniqueandlet'suseittocalculatethesquareof58:

Let's take 55 as a reference. Its square in fact, which is immediately calculable through thestrategyforthenumbersendingin"5",is3025.55+58=113.Let'smultiply113by3,whichisthedistancebetween55and58,obtaining339.55issmallerthan58,sowehavetoaddthelastproducttothefirstsquare:3025+339=3364,whichisnothingbutoursquared58.

9-Usingthecrossingmultiplicationtechniquefortwoandthree-digitnumbers

The "crossing" multiplication technique we saw in Chapter XIII can come to our aid during oursquarecalculationsaswell.Ashappenedinfactinthemultiplicationtechniquefornumbersbetween11and19case,wedon'thave todoanythingbutadapting themultiplicationprocedure to identicalnumbers!

Itcanalsobeusefultonoticethatthistechnique,sinceit's"universal"anddoesn'thaveanyspecificprerequisitetobeperformed,representsanexcellentalternativetothetwopreviousones,andsoit'sup to you, time to time, to choose the best technique to calculatewith, depending onwhat'smorecongenialtoyou.

Let'sstartwiththetwo-digitsversion:

Squarethetensdigit.Multiplytheunitsbythetens,doubletheresultandwriteittotherightofthesquareyougotinthepreviousstep.Squaretheunits.Carryall theextradigits to the left incaseanyof thepreviousresultswasmadeupof twoormoredigits.

Soifforexamplewewanttocalculatethesquareof87,wecan:

Square8=64.Multiply7x8=56anddoubleit=112.Sothepartialresultisnow64_112.

Square7=49.Partialresult=64_112_49.Carrytheextradigitstotheleft:64_116_9(addedthe4from49to112)75_6_9(addedthe11from116to64)Nomoredigitstocarry,andinfact7569isexactlythesquarewewerelookingfor!

Whenyoucomeacrossthree-digitnumbersyoucangroupthetensandthehundredsdigitstogether,handle the resulting number like itwas the tens digit in the previous version, and repeat the sameprocedure,whichof coursebecomesa littlemore complicated, but sure is still adefinitelydoablechallenge.Infactinthiscaseyoucan:

Squarethenumbermadeupofthefirsttwodigitsofyournumber.So,forexample,inthecaseofthesquareof458,youmustsimplysquare45=2025.Nowmultiply the same number you started from in the previous step by the units digits anddoubletheresult.Sointhisexampleyoumustcalculate45x8x2=720.Thepartialresultisthen2025_720.Thenjustsquaretheunitsandwritetheresulttotherightofthenumberyougotinthepreviousstep.Inthiscaseyouhave8x8=64,andtheresultsofaris2025_720_64.Carrytheextradigits:2025_726_42097_6_4=209764,whichisthenumberwewerelookingfor.

10-Usingthemultiplicationtechniquefornumbersclosetoapowerof10

Likeanyothermultiplicationtechnique,eventhisonecanbeextremelyusefulforrapidlysquaringevenverylargenumbers,attheconditionthattheyarecloseenoughtoapowerof10.Andsincewhenweweretalkingaboutthattechniquewehadtodistinguishbetweentwospecificcases,wemustdothesamehere:

Howtosquarenumbersslightlylowerthanapowerof10:

Keepinmindthatthefinalsquareshouldhaveanumberofdigitsequaltotwicethedigitsoftheoriginalnumber.Calculatethedifferencebetweenthenumbertosquareandthenexthigherpowerof10.Ifit'snotimmediatetoperform,apply thesutra"All from9, the last from10".Thensquare theresultofyoursubtraction.Nowsubtractfromthenumbertosquareitsdistancefromthenexthigherpowerof10andwritethedifferencetotherightoftheresultyougotinthepreviousstep.Ifthetotalamountofdigitsinyourpartialresultislesserthantheoneyoucalculatedinthefirststep,youjustneedtowriteasmanyzeroesbetweenthetwonumbersyougot,asit'snecessarytomatchthatamount.Youjustgotyourresultotherwise.

So,let'simmediatelyhaveanexample,andlet'scalculatethesquareof9987:

Theresultwillbemadeupof8digits,since4x2=8.

Itsdifferencefrom10000,thenexthigherpowerof10,is13.Squared13=169.Subtract13from9987,getting9974asaresult.Writeittotheleftofthepreviousresult.Youjustgot9974_169.Thetotaldigitshereare7,andsincewefoundoutthattheresultshouldbemadeupof8digitsinstead,wemustwriteazerobetweenthetwonumbers:finalresult99740169.

Asnoticed incaseofmultiplication, this technique letsussquareevenhugenumberswithamazingspeed, but the more we move away from a power of 10 and the more the things start to getcomplicated.Andthisofcoursehappensbecausethedifferencescalculatedinthefirststepstarttobelargernumbers,andtheconsequentsquarestartsinturntobeamuchlongercalculationtoperform.

Howtosquarenumbersslightlylargerthanapowerof10:

Keepinmindthatthefinalsquareshouldhaveanumberofdigitsequaltotwicethedigitsoftheoriginalnumber,minusone.Calculatethedifferencebetweenthenumbertosquareandthenextlowerpowerof10.Andthisshouldbestraightforward.Sumtothenumbertosquareitsdifferencefromthenextlowerpowerof10andwritethetotaltotheleftofthenumberyougotinthepreviousstep.Ifthetotalamountofdigitsinyourpartialresultisgreaterthanwhatyoucalculatedinthefirststep,youjustneedtotaketheleftmostdigitinthenumberyougotinthepreviousstepandsumittothenumberyougotinthefirststep.

Butlet'sclarifythisprocedureonceandforallthroughanexample,andlet'ssquare1094:

Theresultwillbemadeupof7digits(4x2-1=7).The difference from the next lower power of 10 is of course 94 and its square can be easilycalculatedthroughoneoftheprevioustechniques=8836.Nowsum1094+94,getting1188,andwriteittotheleftofthepreviousresult.Nowyourhave1188_8836.Wesaidinthefirststepthattherealresultmustbemadeupof7digits,sowemusttaketheleft-mostdigitfromtherightsidenumber(thelast"8"from8836)andthesumitto1188.Soweget1196_836=1196836,whichisoursquared1094.

Nowthatyou'refullyawareofhowtorapidlysquareanumber,here'sanothercuriosity:wesawthatifyousquare55youget3025asa result.Now let's take anumber slightlyhigher than55:75, forexample. Its square equals5625,which is almostdouble than thepreviousone. Likewe saw in thepizzaexample,alittleincreaseinthenumbertosquareproducesabigincreaseinthecorrespondingsquarednumber.

Now let'sdigup somephysics:wehave that thekinetic energy (the energyassociated toamovingbody) of any object increaseswith the square of its speed.And you just calculated that the energyassociatedtoacarrunningat75mphisalmostdoublethanacarrunningat55mph.Littleincreaseinspeed,doubletheenergy,anddouble,ifnotinfinitelyheavier,theimpactdamageincaseofaccident.Andthat'swhyyoushouldneverrunfull-speedwhiledrivingyourcar:evensmallincreasesinspeed

enormously exacerbate the potential damage in case accident, as well as keeping the speed lowerreducesbyseveralordersofmagnitudeanyriskduringyourjourneys.

Meanwhile,thisspecificjourneyisveryneartoanend,andinthenext,andfinalchapter,we'lldiscusssometrickstoquicklyextractthesquareandcuberootsofanynumber.

XXII-Squareroot,cubicrootand...thirteenthroot!

Extraction of square root from non-perfect squares (numbers whose square root is not an integerquantity)canbeveryoftenaquitecumbersomeandfrustratingoperation.Andthat'swhy,asgoodfastcalculators,alwayslookingforthefasterandmostefficientstrategytoaccomplishourtasks,eveninthiscasewe'lldoourbesttoavoidtediousproceduresandget,rather,agoodresultinaslesstimeaspossible.Let'sstart!

Howtocheckwhetheranumberisanon-perfectsquare

First, let's give a look to a very simple "trick" we can use to immediately identify a non-perfectsquare:aperfectsquarealwaysendsin4,5,6,9,or0.

Theopposite,ofcourse,isnotalwaystrue(ifanumberendsin4,5,6,8or0wecan'tautomaticallysaywhetherit'saperfectsquareornot)butatthesametime,thankstothisrule,wheneverweseeanumberthatdoesnotendinoneoftheabove-mentioneddigits,wecanbe100%surethatitssquarerootwon'tbean integernumberandsowecanprepareourselves toworkon thecalculationof itsapproximation,aswe'llseelater.

Asamatterof fact there isanother,morecomplexrule,which letsuspositively identifywhetheranumberisaperfectsquareornot:anumberisalwaysaperfectsquareif,afterbeingdecomposedintofactors,allofitsfactorshaveanevenexponent(orarerepeatedanevennumberoftimes,whichisexactlythesamething).

IfyourememberwhatwesaidaboutfactorisationinChapterIX,you'llimmediatelyunderstandwhatI'mtalkingaboutandinwhichcasesthisrulecouldcomeinhandy.36,forexampleisaperfectsquaresinceit'sequalto3x3x2x2.Samething1764,sinceitequals7x7x3x3x2x2.Ofcourseifyou're not very confident with divisions and decomposing numbers into factors (despite thetechniquesweintroducedinthelastchapters)thiscanbeaquitecomplexoperationtoperform,somyadvicehereistouseitonlywhenstrictlynecessary.

Twowaystorapidlyapproximatetheextractionofasquareroot

Afirsttechniqueforapproximatingthecalculationofasquarerootisbasedontakingadvantageofaspecificmathematicalrelationshipinvolvinganysquareroots.Morespecifically,givenanumber"a"wewant toextract thesquare root from,andanumber"b"wealreadyknowthesquarerootof, wehavethat:

√aapproximatelyequals√b-[b-a/(2x√b)],

wheretheapproximationisasclosertotherealsquarerootas"a"and"b"areoneneartoeachother.

But since the mathematical formulas without any examples can only create confusion, let'simmediatelyshowapracticalapplicationofthetechniqueandlet'strytocalculatethesquarerootof5.Inthiscasewecanofcoursetake4asareferencenumber,sinceitsprettywidelyknownsquarerootistwo,andsowecando:

√4=2FromthepreviousresultIhavetosubtractthefraction(4-5)/(2x2)Fractiondenominator:4-5=-1Fractionnumerator:2x2=4-1 / 4 equals -0.25 (remember thata negative number divided ormultiplied by a positive onegivesanegativenumberasaresult)Now2-(-0.25)=2+0.25=2.25.Wethusobtainedthat theapproximatedsquarerootof5is2.25.Therealone isactually2.236but,asyoucansee,since4and5arequitecloseonetoeachother,theresultingapproximationisquiteacceptable.

Ifwewant a greater accuracy in our result, anyway,wemust use anothergradual strategy, whichworksasbetterasweperformmorestepsofit,andcanbeappliedthisway:

Takenthenumberwewanttoextractthesquarerootfrom,let'stakeitsnextlowerperfectsquareagain(forexample,ifwewanttoextractthesquarerootof38,let'stake36).Now let's calculate the mathematical mean between our number and the considered perfectsquare(forexample, in thiscase, (38+36) /2=37)and let'sdivide itby the"known"squareroot of the perfect square (so, in this case, we'll calculate 37 / 6 = 6.1666). This is a firstapproximationofthesquarerootof38.Butifwewanttogetanevenmoreaccurateresultwecanperformonemorestep: let's take thenumberwe'reextractingthesquarerootfromandlet'sdivideitbytheapproximationwegotinthepreviousstep.So,inthiscase,wehadthatourpreviousapproximationwas37/6=6.1666.Sonowwehavetocalculate38/6.1666=6.1621621621Dowewant themost accurate result of all? Let's add a last step and let's calculate the meanbetweenthetwopreviousresults!Here themeanbetween6.1666and6.1621621621 is6.164411,while therealsquarerootof38was6.164414.So,aswecannoticebyourselves,despitethemorestepstoperform,theresultisreallyclose to therealone.And ifwedon'tcareabout that1 /100000 imprecision level, thistechniqueisagreat,fastreplacementforthe"classic"procedureeverybodylearntatschool.

Tipsforextractingcuberootandfifthroot

Theextractionofthecuberootofanumberisdefinitelynotasimpletasktoperform,andthat'swhyhere wewill introduce just a specific trick to quickly extract the roots of perfect cubes (numberswhosecuberootisaninteger)madeupof4,5or6digits.

The first thing thatmust be done in order to apply this trick is to learn the fundamental cubes ofnumbersfrom1to10:

Cubed1=1

Cubed2=8

Cubed3=27

Cubed4=64

Cubed5=125

Cubed6=216

Cubed7=343

Cubed8=512

Cubed9=729

Cubed10=1000

So, after helping yourself in thememorization processwith somemnemonic technique rememberthat,unlikewhathappensinthesquarerootcase,thelastnumberofaperfectcubecanimmediatelyletusdeterminewhichisthelastdigitofitscuberoot.Inparticularwehavethat:

Ifthecubeendsin0,itsrootendsin0Ifthecubeendsin1,itsrootendsin1Ifthecubeendsin2,itsrootendsin8Ifthecubeendsin3,itsrootendsin7Ifthecubeendsin4,itsrootendsin4Ifthecubeendsin5,itsrootendsin5Ifthecubeendsin6,itsrootendsin6Ifthecubeendsin7,itsrootendsin3Ifthecubeendsin8,itsrootendsin2Ifthecubeendsin9,itsrootendsin9

Andthistablecanbeeasilymemorizedbyrecognisingthatthecubealwayshasthesamefinaldigitasitsroot,exceptwhenitendsin2,3,7or8.Inthesecases,infact,theroot,insteadofhavingthesamefinaldigitasitscube,hasitsdifferencefrom10.

Nowyouhaveeverythingyouneedtocalculatetherootofyourperfect4,5or6-figurecube.Infact:

Itsrootwillbeatwo-figurenumber.Theunitsdigitcanbeeasilyfoundafterlookingattheabovetable.Inordertofindthetensdigitsinstead,asafirststep,removethethreerightmostdigitsfromyourperfectcube.Nowyougotanumber"a"madeupof1,2or3digits.

Taketheperfectcubestableand,goingfromthetopdownwards,takethelast(=thelarger)cubethat'sstilllowerthanyour"a".Itscuberootwillbethetensdigityouwerelookingfor.

Butlet'shaveanexampletoclarifyhowthistechniqueworksandlet'simagineyouwanttoextractthecuberootof85184.Theunitsdigitcanbeobtainedimmediatelyandit's4.Inordertogetthetensoneinsteadwewillhavetoremovethethreerightmostdigitsofthecube,gettingan85.

So,let'stakealooktotheperfectcubestableandwe'llseethatthelastlowercubethan85is64; itscuberootis4andthenourtensdigitisnothingbut4.Wefoundthatthecuberootof85184is44!

LasttrickIwanttoteachyouaboutthecuberoot:youcanactuallyapproximateitscalculationjustbyusingany classic non-scientific calculator having only the square root button. In particular, here'swhatyouhavetodointhiscase:

Typethenumberyouwanttoextractthecuberootfrom.Pressthesquarerootbuttononce.Pressthe"multiply"button.Pressthesquarerootbuttontwice.Pressthe"multiply"button.Pressthesquarerootbuttonfourtimes.Pressthe"multiply"button.Pressthesquarerootbuttoneighttimes.Pressthe"multiply"button....

And so on, doubling the number of times you press the "square root" button until pressing the"multiplication" one does not change the result anymore.At that point, just press the square rootbuttononelasttimeandyou'llseeyourcuberoot.A curiosity: do you want to approximate the calculation of the fifth root instead of the cubeone?Repeatthesameprocedure,butbyreplacingthe"multiply"buttonwiththe"divide"one!

Thirteenthroot?

Asfarastherootdegreeincreases,therelatedoperationsstarttobemoreandmorecomplicatedandat thesame time tomove furtheraway fromthedailychores, and that'swhywe'll justconclude thisbookwithsomelittlecuriosityaboutthirteenthroot.

Theextractionofthirteenthroot,infact,isactuallyaquitepopularprocedureformentalcalculationworld records. The firstworld record ever, for example,was achieved in 1974 byHerbert B.DeGrote,whocalculatedthethirteenthrootofa100-figurenumberin23minutes.Andifthislookslikea titanic accomplishment to you, just be aware that nowadays Gert Mittring, a German mentalcalculatorwe talked about some chapters ago, is able to perform the same calculationwithin 11.8seconds.OfcourseMittring,liketheotherswhotriedtoachievethatresultbeforehim,performsthiscalculation through a specific algorithm, which is highly complex and only partially known. In

addition,itisalwaysassumedthatthenumberthepotentialrecord-breakerstrytheirchallengewithisanintegerthirteenthroot,whichslightlysimplifiestheprocedure.Infact,withoutgoingintodetail:

Theintegerthirteenthrootofanumberalwaysendsinthesamedigitasthenumberitself.Theintegerthirteenthrootofa100-figurenumberisalwaysan8-figurenumberbetween41246264and49238826.Becauseofthis,it'sgenerallyconsideredasa"quiteeasy"task,basedonthecalculationofalogarithmandsomeconsequentmultiplications.Theintegerthirteenthrootofa200-figurenumberisalwaysa16-figurenumberbetween2030917620904736and2424462017082328.This,ontheotherhand,isconsideredasa"quitecomplex"tasktoperform.Theworldrecordhereinfact,isabout5minutesforathirteenthrootcalculation,atime30timeshigherthantherecordforthe100digitsone!

Needlesstosay,somepeoplealsousetoassociatethiscalculationtomysticalandesotericmeanings,andthisseemstobeduetothestrongsymbolismbehindthenumber13,whichisassociatedastotherevolt ofLucifer against the heavens, as toHebrewword for "One", referring toGod.Again, theoccultmeanings behind numbers and calculations is a topic far beyond the actual purpose of thisbook,butIanywayhopethatthischaptermadesparkoffsomethinginyou:whoknows,maybeyou'llbethegeniusbreakingthenextspeedmath-relatedworldrecord!

"Archimedes will be remembered when Aeschylus is forgotten, because languages die andmathematicalideasdonot."Immortality"maybeasillyword,butprobablyamathematicianhasthebestchanceofwhateveritmaymean."

(G.H.Hardy)

Inconclusion...

Yourpersonaljourneyintotheworldofcreative,fastandusefulmathematicsendshere.

Ihopethatreadingthisbookhasbeenforyoucharming,funny,andmostofallthatitgaveyouthat"extraboost"thateverygoodbookshouldgivetoitsreader.

IwantalsotakethisopportunitytothankDaniloforhavinghelpedmeonceagainwiththeediting,thetranslationandthelayoutofthisedition,withgreatpatience,accuracyandprofessionalism.

But,mostofall,thankyoureader,forallthehelpandaffectionthatyouconstantlygiveto"The101bibles"publishingproject,whichisconstantlygrowingthankstothesupportofallofyou!

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Appendix:Primenumbersfrom2to5000

23571113171923

29313741434753596167

717379838997101103107109

113127131137139149151157163167

173179181191193197199211223227

229233239241251257263269271277

281283293307311313317331337347

349353359367373379383389397401

409419421431433439443449457461

463467479487491499503509521523

541547557563569571577587593599

601607613617619631641643647653

659661673677683691701709719727

733739743751757761769773787797

809811821823827829839853857859

863877881883887907911919929937

9419479539679719779839919971009

1013101910211031103310391049105110611063

1069108710911093109711031109111711231129

1151115311631171118111871193120112131217

1223122912311237124912591277127912831289

1291129713011303130713191321132713611367

1373138113991409142314271429143314391447

1451145314591471148114831487148914931499

1511152315311543154915531559156715711579

1583159716011607160916131619162116271637

1657166316671669169316971699170917211723

1733174117471753175917771783178717891801

1811182318311847186118671871187318771879

1889190119071913193119331949195119731979

1987199319971999200320112017202720292039

2053206320692081208320872089209921112113

2129213121372141214321532161217922032207

2213222122372239224322512267226922732281

2287229322972309231123332339234123472351

2357237123772381238323892393239924112417

2423243724412447245924672473247725032521

2531253925432549255125572579259125932609

2617262126332647265726592663267126772683

2687268926932699270727112713271927292731

2741274927532767277727892791279728012803

2819283328372843285128572861287928872897

2903290929172927293929532957296329692971

2999300130113019302330373041304930613067

3079308330893109311931213137316331673169

3181318731913203320932173221322932513253

3257325932713299330133073313331933233329

3331334333473359336133713373338933913407

3413343334493457346134633467346934913499

3511351735273529353335393541354735573559

3571358135833593360736133617362336313637

3643365936713673367736913697370137093719

3727373337393761376737693779379337973803

3821382338333847385138533863387738813889

3907391139173919392339293931394339473967

3989400140034007401340194021402740494051

4057407340794091409340994111412741294133

4139415341574159417742014211421742194229

4231424142434253425942614271427342834289

4297432743374339434943574363437343914397

4409442144234441444744514457446344814483

4493450745134517451945234547454945614567

4583459145974603462146374639464346494651

4657466346734679469147034721472347294733

4751475947834787478947934799480148134817

4831486148714877488949034909491949314933

4937494349514957496749694973498749934999

Bibliography

"Mentalcalculation...revealed!"-G.Golfera,2013

"Themathematicalwizard"-ASMnemonic,2013

"Howmathcansaveyourlife"-J.D.Stein,2013

http://www.vedicmaths.org

http://www.lifehacker.co.uk

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