101 Word Problems/Solutions

Problem #1: Antifreeze

A quart bottle contains a mixture that is 1/3 antifreeze, 2/3 water. A half gallon bottle contains a mixture that is 3/4 antifreeze, 1/4 water. The contents of the two bottles are poured into a gallon jug. What is the ratio (one integer to another, in the form a:b) of antifreeze to water in the jug?

Solution #1: Antifreeze

11:7

Problem #2: Consecutive Birthdays

A woman and her grandson have the same birthday. For six consecutive birthdays, she is an integral multiple of his age. How old is the grandmother at the sixth of these birthdays?

Solution #2: Consecutive Birthdays

66

Problem #3: Cube Numbers

Find all sets of four positive consecutive integers such that the sum of the cubes of the first three is the cube of the fourth.

Solution #3: Cube Numbers

3,4,5,6

Problem #4: Don't Spend It All At Once

If I start with $10 and spend all but $3, how much do I have left?

Solution #4: Don't Spend It All At Once

$3. If I spend all but $3, I must have $3 left.

Problem #5: Hands of a Clock

How many times do the two hands of a clock point in the same direction between 6:00 am and 6:00 pm of a single day?

Solution #5: Hands of a Clock

11. It is tempting to give the "obvious" answer 12, but the true answer is 11 as will be found by counting the individual occasions.

1. Between 6:30 am and 6:35 am. 2. Between 7:35 am and 7:40 am. 3. Between 8:40 am and 8:45 am. 4. Between 9:45 am and 9:50 am. 5. Between 10:50 am and 10:55 am. 6. At Noon. 7. Between 1:05 am and 1:10 am. 8. Between 2:10 am and 2:15 am. 9. Between 3:15 am and 3:20 am. 10. Between 4:20 am and 4:25 am. 11. Between 5:25 am and 5:30 am.

Problem #6: How Old Is My Daughter?My daughter is twice as old as my son and half as old as I am. In twenty-two years my son will be half my age. How old is my daughter?

Solution #6: How Old Is My Daughter?Let us assume my daughter is age x. We are told my daughter is twice as old as my son, so that my son must be age x/2. We are also told that I am twice as old as my daughter so my age is 2x. In 22 years time my son will be (x/2 + 22) and I will be (2x + 22). Since he will be half of my age at that time,

x/2 + 22 = 1/2 (2x+22)Multiplying both sides by 2

x + 44 = 2x + 22or

x = 22My daughter is 22 years old.

Problem #7: The Striking Clock

I woke up one night and heard my clock strike "one." I was too tired to turn on the light to see what time it was. As I lay there pondering, it occurred to me to speculate how long I would have to lie awake in order to be sure what was the exact time. My clock strikes the hours and strikes "one" each half hour. I fell asleep before I solved the problem, but can you work out what is the longest time I would have to lie awake after hearing the strike "one," to be sure of the time?

Solution #7: The Striking Clock

The strike I heard was either one o'clock or some half hour. If, after another half hour I hear two more strikes, I know what hour it is, but if I again hear one strike, I may have heard successively 12:30 am and 1 o'clock or 1 o'clock and 1:30 am. After another half hour I either hear 1 strike

(1:30 am) or two strikes (2 o'clock). I may, therefore, have to lie awake a whole hour before I can be sure of the exact time.

Problem #8: Relations

1. Is it legal for a man to marry his widow's sister? 2. Brothers and sisters I have none, but this man's father is my father's son. Who is "this man?"

Solution #8: Relations1. Only dead men have widows. So... no, a dead man cannot legally marry anyone. 2. Since I have no brothers or sisters, "my father's son" must be referring to me; I am "this man's father." So, "this man" must be my son.

Problem #9: School NotesIf 6 boys fill 6 notebooks in 6 weeks and 4 girls fill 4 notebooks in 4 weeks, how many notebooks will a class of 12 boys and 12 girls fill in 12 weeks?

Solution #9: School NotesIf 6 boys fill 6 notebooks in 6 weeks, then 12 boys fill 12 notebooks in 6 weeks, and 12 boys fill 24 notebooks in 12 weeks; If 4 girls fill 4 notebooks in 4 weeks, then 12 girls fill 12 notebooks in 4 weeks, and 12 girls fill 36 notebooks in 12 weeks; Hence 12 girls and 12 boys fill 24+36=60 notebooks in 12 weeks.

Problem #10: Simple AlgebraGiven x = 1 and y = 1, we have

x = yMultiplying each side by x

x2 = xySubtracting y2 from each side

x2 - y2 = xy - y2

Factoring each side (x + y)(x - y) = y (x - y)

Dividing out the common term ((x - y) we have x + y = y

Substituting the given values 1 + 1 = 1

Or 2 = 1

What is wrong with this proof?

Solution #10: Simple AlgebraThere was nothing wrong up through line

(x + y)(x - y) = y (x - y)If we substitute the values of x and y we have

(2)(0) = (1)(0)or

0 = 0However, when we divide both sides by (x - y), we break a fundamental rule of mathematics that we cannot divide out by a fraction which is equal to zero.

Problem #11: A Dinner PartyWe were given a formal dinner party for ten (including ourselves) which is a number I always like because the host can sit at one end of the table and the hostess at the other, and still maintain the correct alternate male and female around the table. My wife was trying to work out the seating. "Tom and Jean have not been here to dinner before so they are the guests of honor. Tom must sit on my right and Jean on your right, but I don't know how I want to seat the others.: "Well," I said, "I would like Janet on my left. I have a soft spot for her." "You can have her," replied my wife, "but I will not have her husband Jack next to me; I think he should be next to Mary Ann." Since we do not place husbands and wives next to each other, this determined the seating of everyone, including Howard's wife Lois, and Mary Ann's husband Bill. Can you work out the seating arrangement?

Solution #11: A Dinner Party

Numbering the ten places around the table 1 to 10 as shown in the figure below, if I take the head of the table (1), my wife will sit at (6), Jean will be at (10), Tom at (5), and Janet at (2).

Now my wife will not have Jack next to her so he must be at (9), because (3) would place him next to his wife. Since Mary Ann is next to Jack, she must be at (8); then Howard must be at (7), Bill at (3), and Lois at (4).

Problem #12: Dominoes on Board

A board, 9 inches by 7 inches, is marked out into 63 one-inch squares. To make the number of squares even, the middle square is blocked out by placing a chessman on it. John attempts to cover the board, other than the middle square, with dominoes which are 2 inches by 1 inch.

In the figure above he has 11 dominoes in place. Since 31 dominoes will be needed in all, more than one set will be required. Can you completely cover the board, other than the middle square, without having any domino stick out over the edge of the board, or can you prove it cannot be done?

Solution #12: Dominoes on Board

The use of a chessman to block out the middle square gives the clue to the solution of this problem. If the board is marked out as a chess board with alternate black and white squares, with the middle square black, we find that there are 32 white squares and 30 black squares (not counting the middle square). Now a domino however placed will cover one white square and one black square and hence having placed 30 dominoes we shall always be left with 2 white squares which cannot possibly be covered by a single domino. Hence, there is no possible solution to the problem.

Problem #13: An Unusual Series

Can you find the fifth term in the following series?

77, 49, 36, 18, ...

Solution #13: An Unusual Series

Each term consists of the first digit of the preceding term multiplied by the second digit of the same term. Thus,

49 = 7 x 7

36 = 4 x 9

18 = 3 x 6

Hence, the fifth term is 1 x 8 = 8.

Problem #14: Is the Rope Knotted?

A loop of rope is lying on the ground in the position show in the figure below: You are too far away to see which section of the rope is above or below at each of the crossovers at A, B and C.

If we assume that it is equally likely that either section is on top at each crossover, what is the probability that the rope is knotted?

Solution #14: Is the Rope Knotted?

We can assume without loss of completeness that at A the rope going from top left to bottom right is on top. (If it is the other way a mirror solution is produced which does not alter the probability.)

Then we have four equal possibilities. The segment BOC can be either on top or underneath at each of the points B and C.

Segment BOC

At B At C The rope is

On top On top Not knotted

On Top Under Not knotted

Under On top Knotted

Under Under Not knotted

Hence the chance that the rope is knotted is 1/4.

Problem #15: The Two Rugs

A woman has two rugs of the same material, one 10 feet by 10 feet and one a long runner 8 feet by 1 foot. How can she make a single continuous cut in the 10 foot by 10 foot rug, so as to form two pieces which, when combined with the 8 foot by 1 foot rug can be stitched into a rug 9 feet by 12 feet? The two rugs are of plain pattern and have a pile on one side only.

Solution #15: The Two Rugs

The 10 foot by 10 foot rug must be cut as shown below, each step being 2 feet long and 1 foot broad.

This gives two odd-shaped pieces together, a gap will be left in the middle which can be exactly filled by the 8 foot by 1 foot strip, which is shaded below.

Problem #16: Mexican Table Mats

My son gave me a set of table mats he bought in Mexico. They were rectangular and were made of straw circles joined together in the form shown in the figure below:

All the circles on the edges of each mat were white and the inner circles red. I noted that there were 20 white circles and 15 red circles.

I wondered if it were possible to make a mat in this form where the number of white circles equaled the number of red circles. I found there were two possible solutions. Can you find them and show that there are only two such solutions?

Solution #16: Mexican Table Mats

Let the number of circles along the top edge of the mat be x and the number of circles along the side of the mat be y. Then the total number of circles in the mat is

xy

and the number of circles around the edge is

2x + 2y - 4

Since this must be one-half the total number of circles we have

xy = 4x + 4y - 8

xy - 4x - 4y + 16 = 16 - 8

and (x - 4) (y - 4) = 8

Since x and y must be integers, so must (x - 4) and (y - 4) and these must be factors of 8. The only integer factors of 8 are (8 and 1) and (4 and 2), which give {x = 12, y = 5} and {x = 8, y = 6}.

Hence the mat must be 12 by 5 circles or 8 by 6 circles.

Problem #17: The Curious Sequence

What is the next letter in the following series?

O T T F F S S E

Solution #17: The Curious Sequence

This is on of those nasty catchy puzzles. The real solution is

O T T F F S S E N

being the initial letters of the counting numbers: One, Two, Three, Four, Five, Six, etc.

Problem #18: Colored Labels

Three intelligent women, Alice, Barb and Carol, sit down to try out a test in logical reasoning. They are so arranged that each can see the color of a label which is either red or blue, attached to the hats worn by the other two but no one of them can see the color of the label attached to her own hat. They are told that at least one of the labels is red. If any one of them can logically deduce the color of the label on her hat, she is to declare it. Carol decides to play this game with her eyes closed, knowing that the other two women have their eyes open. After a little time Carol, who has not seen the label on any of the hats declares her label is red. How can she deduce this?

Solution #18: Colored Labels

Carol argues as follows: 1. If Alice sees two blue labels she will know that her label is red and will declare this and

since a little time has passed without this happening Betty and Carol cannot both have blue labels.

2. Similarly, if Betty sees two blue labels he will know that his label is red and will declare this, and hence, Alice and Carol cannot both have blue labels.

3. Carol can now deduce that if her label is blue, both Alice and Betty must have red labels. Now Carol knows that both Alice and Betty are intelligent. If Betty and seen a red label on Alice's hat and a blue label on Carol's hat, she would know her hat did not have a blue label, because in this case Alice would have declared her label to be red.

4. Since a little time has passed and neither Alice nor Betty has declared she could deduce the color of her label, Carol knows her label cannot be blue, and therefore must be red.

Problem #19: War in the Middle AgesDuring a grim medieval battle, 85% of the warriors lost an ear, 80% lost an eye, 75% lost an arm, and 70% lost a leg. What is the smallest percentages possible of combatants who lost all of the above?

Solution #19: War in the Middle AgesSince 85% of the warriors lost an ear and 80% an eye (a total of 165%), at least 65% of them must have lost both an ear and an eye. Since at least 65% of the warriors lost both an ear an eye, and 75% lost an arm (a total of 140%), at least 40% of them must have lost an ear, an eye, and arm and a leg each. Since at least 40% of the warriors lost an ear, an eye, and an arm and 70% lost a leg (a total of 110%), at least 10% of them must have lost an ear, an eye, and arm and a leg each. (This problem was set by Lewis Carroll.)

Problem #20: How Many Are We?How many are we? You tell us the answer, knowing that the probability that at least two of us have birthdays on the same day is less than half, but that this would not be the case were we one more in number.

Solution #20: How Many Are We?Probability that two people have different birthdays: 364/365 Probability that three people have different birthdays: (364/365) x (363/365), etc. For n people P = (364/365) x (363/365) x ... x ([653-n+1]/365) It can be checked that the product of these fractions becomes less than 1/2 when n goes from 22 to 23. We are therefore 22.

Problem #21: JawsA seaside resort along the Pacific Ocean is equipped with an electronic shark detection system which sounds an alarm on the average of 1 day out of 30. There are 10 times more false alarms than there are undetected sharks. We also know that only three out of four sharks are detected. Given the above information, what is the percentage of "peaceful" resort days if that term is defined as days unmarred by alarms or sharks?

Solution #21: JawsLet x be the percentage of a shark going undetected. We know that the probability of an alarm is 1/30. Hence the probability of having neither an undetected shark nor an alarm is 1 - (1/30 + x) = 29/30 - x

We have also been told that the probability of false alarm is 10x. Hence the probability of real alarm is 1/30 = 10x. But this is equal to 3x since 3 times as many sharks are detected as go undetected 1/30 - 10x = 3x Hence x = 1/390 Thus the probability of a peaceful day is 29/30 - 1/390 = 0.964

Problem #22: MeteorologyIt rains here one day out of three. Our local meteorologists, who are pessimistic by nature, are wrong in their daily forecasts one time out of two when they should have predicted good weather but wrong only one time out of five when they should have forecasted rain. Each morning, Francine leaves home for the day. If she departs without an umbrella and it rains, she is twice as annoyed as if it had been fair and she had taken her umbrella with her. "Would I be wiser," she wonders, "to listen to the morning weather forecast and take my umbrella only if rain is predicted? Or should I consistently carry it? Or should I never take it? What would you advise Francine to do?

Solution #22: MeteorologyThere are four possible situations:

1. Fair weather predicted and occurred: probability 1/2 * 2/3 = 1/3. 2. Fair weather predicted but rain occurred: probability 1/5 * 1/3 = 1/15. 3. Rain predicted and occurred: probability 4/5 * 1/3 = 4/15. 4. Rain predicted but fair weather occurred: probability 1/2 * 2/3 = 1/3.

Let i be the measure of the inconvenience for Francine to carry her umbrella around for one whole fine day and 2i the measure of inconvenience of not having an umbrella when it rains. If she systematically carries her umbrella every day, the inconvenience will amount to i on two days out of three, hence an average inconvenience of 2i/3. If she never takes her umbrella, the inconvenience will amount to 2i one day out of three, for an average inconvenience again of 2i/3. If she decides to follow the advice of the weather forecasters and takes her umbrella only when rain is predicted why will suffer an inconvenience of i on 1 day out of 3, and ii on 1 day out of 15, which amounts to and average penalty of 7/i/15 (less than the two previous average

penalties). We recommend, therefore, that Francine follow the advice of the weather forecasters.

Problem #23: Amateur DoctorAspirin does wonders for my headache and helps my rheumatic knee but it gives me nausea and upsets my stomach. Herbal medicine cures my nausea and stomach upset but gives me hip pain. Antibiotics sooth my headache and nausea but irritate my stomach and knee and give me a stiff neck. Cortisone helps my stiff neck and rheumatic knee but aggravates my hip condition. Hot compresses work wonders for my upset stomach and stiff neck. I woke up today with a pounding headache which prevents me from figuring out how best to doctor myself. What do you think I should do?

Solution #23: Amateur Doctor

I advise you to take aspirin and antibiotics and to apply hot compresses to your neck. Aspirin relieves the headache but gives you nausea and upsets your stomach; the antibiotics relieve the nausea but add pains to your knee and give you a stiff neck. The knee is taken care of my the aspirin. As far as the upset stomach and the stiff neck are concerned, the hot compresses will take care of them.

Problem #24: Bar Flies

At the same instant that Pete was leaving Harry's Bar to go to the Shamrock Tavern, Johnny was leaving the Shamrock Tavern to go to Harry's Bar. They walked at a constant speed. When they met, Pete announced that he had covered 200 meters more than Johnny had. The latter, his mind befuddled by alcohol, took this as a personal insult and started to beat up Pete, who returned the punches. When the fighting stopped, the men embraced in tears, then each continued his original journey, but at half his original speed, as both were slightly hurt. Pete thus arrived at the Shamrock Tavern in 8 minutes while Johnny took 18 minutes to get to Harry's Bar. How far apart are the two bars.

Solution #24: Bar Flies

Let x be the distance between the two bars.Let d be the distance covered by Pete when he met Johnny.Let V be Pete's speed and v be Johnny's speed before the fight.

The sum of the distances covered at the time they met is x. d + (d-200) = x (1) Hence x = 2d - 200When they met, each had walked for the same length of time: d/V = (d - 200)v (2)After the fight, Pete walked for 8 minutes: (d - 200)(V/2) = 8

Hence V = (d - 200)/4 (3)and Johnny for 18 minutes: d(v/2) = 18

Hence v = d/9 (4)By substituting the values for V and v into equation (2), we have 4d/(d - 200) = 9(d - 200)/dHence 5d2 - 3600d + 360,000 = 0Solving this quadratic in the classic way yields d = 600 or 120 and x = 2d - 200 = 1000 or 40However, d must be smaller than x, so only one solution is possible: d = 600 and x = 1000The distance between the two bars is exactly 1000 meters.

Problem #25: Subway Escalator

I am in the habit of walking up the subway escalator while it is running. I climb 20 steps at my normal pace and it takes me 60 seconds to reach the top, whereas my wife climbs only 16 steps and it takes her 72 seconds to reach the top. If the escalator broke down tomorrow, how many steps would I have to climb?

Solution #25: Subway Escalator

Let x be the unknown number of steps. When I ride the escalator, I cover x - 20 steps in 60 seconds. When my wife uses the escalator, she covers x - 16 steps in 72 seconds. The escalator therefore moves at a rate of 4 steps every 12 seconds or 20 steps in 60 seconds. Its total height is therefore the sum of these 20 steps and of the additional 20 steps I climb: a total of 40 steps.

Problem #26: Lighthouse

How far is the horizon from the top of a 125.7-meter-high lighthouse? (The earth can be considered spherical with a circumference of 40,000 km.)

Solution #26: Lighthouse

Let S be the top of the lighthouse. Let A be a point on the horizon.

Let O be the center of the earth. Triangle OAS is a right-angle triangle since SA is tangent to the earth, which we consider to be a perfect sphere. We thus have OS2 = OA2 + AS2 but OA = 40,000,000/(2pi) meters and OS = OA + 125.7 = OA + 40pi/OS2 ~= OA*OA + 2 OA = 40pi Consequently, AS = sqrt(OS2 - OA2) ~= sqrt(2*40pi.40,000,000/2pi) = 40,000 m The horizon is therefore approximately 40 km from the top of the lighthouse.

Problem #27: The Demonstration

If we set out by ranks of 10, we will be one short. We will also be one short if we set out by ranks of 9, 8, 7, 6, 5, 4, 3, and even 2. Yet there are fewer than 5000 participants. How many are we?

Solution #27: The Demonstration

Let x be the number of demonstrators. (x + 1) must be a multiple of 2, 3, 4, ..., 9. (x + 1) is therefore a multiple of the smallest common multiple of these numbers, that is, a multiple of 2 x 2 x 3 x 3 x 5 x 7 = 2520. Let x + 1 = k(2520), k being a whole number. x = k(2520) - 1. Since there are fewer than 5000 demonstrators, k = 1. Hence x = 2519, the number of marchers.

Problem #28: Year of Birth

Take away from your year of birth the sum of the four numerals that make it up. You will end of with a number divisible by 9. Why is this?

Solution #28: Year of Birth

Let T, h, t and u be the four digits that make up your year of birth (Thousands, hundreds, tens, and units). Your hear of birth can be written: 1000T + 100h + 10t + u. If you take away T + h + t + u you are left with 999T + 99h + 9t which is divisible by 9.

Problem #29: Spanish Arithmetic

4 + 4 + 4 + 4 + 4 = 20

In Spanish, this can be written:

CUATRO+ CUATRO+ CUATRO+ CUATRO+ CUATRO-------- VEINTEHow are the 10 numerals represented by the 10 letters A, C, E, I, N, O, R, T, U and V for this sum to come out right?

Solution #29: Spanish Arithmetic 170469+ 170469+ 170469+ 170469+ 170469-------- 852345(not yet completed)

Problem #30: A Story of 1's and 2'sWrite, side by side, the numeral 1 an even number of times. Subtract from the number thus formed the number obtained by writing, side by side, a series of 2s half the length of the first number. You will always get a perfect square. For instance, 1111 - 22 = 1089 = 332 Can you say why this is?

Solution #30: A Story of 1's and 2's

11...1 - 22...2 = 11...1 11...1 - 2(11...1) ------ ------ ------ ------ ------2n times n times n times n times n times

= 11...1 00...0 - 11...1 ------ ------ ------ n times n times n times

= 11...1 x (100...0 - 1) ------ ------ n times n times

= 11...1 x 99...9 ------ ------ n times n times

= 11...1 x 9 x 11...1 ------ ------ n times n times

= 32 x 11...12

------ n times

= 33...32

------ n times

Example: 11 - 2 = 9 = 32.

Problem #31: How Many Children Were You?

How many children were you?

If you ask me such a question, I will reply that my mother dreamed of having at least 19 children but her dream did not come true, that my sisters were three times as more numerous than my first cousins, and that I had half as many brothers as I had sisters.

Solution #31: How Many Children Were You?

The number of my sisters must be a multiple of 3 and a multiple of 2. It must therefore be a multiple of 6.

Total number of children: [(multiple of 6) x (1 + 1/2)] + 1.

Since this number must be less than 19, the only possible solution for the multiple of 6 is 6 exactly. I therefore have 6 sisters and 3 brothers. We are 10 children in all.

Problem #32: Brother and Sister

"Sister, you have as many brothers as you have sisters." "Brother, you have twice as many sisters as you have brothers." Can you deduce from this conversation how many children there are in the family?

Solution #32: Brother and Sister

Let x be the number of boys and y the number of girls.

What the boy says to the girl can be written as x = y - 1.

The answer that the sister gives her brother can be written as:

y = 2(x - 1) or y = 2x - 2

Substituting x = y - 1 gives us y = 4 and hence x = 3.

There are therefore seven children in the family.

Problem #33: Ursula and the Cats

If you ask old Ursula how many cats she has at home, she answers sadly: "Four-fifths of my cats plus four-fifths of a cat." How many cats does this add up to?

Solution #33: Ursula and the Cats

Let n be the number of cats. We have n = 4/5 n + 4/5. Hence n = 4. Ursula lives with four cats.

Problem #34: New MathMy son learned how to count in a base different from 10, so that, for instance, instead of writing 136, he writes 253. In what base does he count?

Solution #34: New Math

Let b be the unknown base. When my son writes 253 in base b, this can be interpreted as 2b2 + 5b + 3 = 136 or 2b2 + 5b - 133 = 0 or (b - 7)(2b + 19) = 0 Since b is a whole, positive number, the only possible solution is 7; the unknown base is 7.

Problem #35: Chinese Numbers

Think of a number between 1 and 26. Look at the following table of six squares, one square at a time.

+----------+----------+----------+

| 1 4 7 | 2 5 8 | 3 4 5 |

| 10 13 16 | 11 14 17 | 12 13 14 |

| 19 22 25 | 20 23 26 | 21 22 23 |

+----------+----------+----------+

| 6 7 8 | 9 10 11 | 18 19 20 |

| 15 16 17 | 12 13 14 | 21 22 23 |

| 24 25 26 | 15 16 17 | 24 25 26 |

+----------+----------+----------+

Each time the number you have picked belongs to one of the squares, write down the number in the top left-hand corner. Add together all these numbers. For example, 16 is in the first, fourth, and fifth squares. If the first numbers of each of these squares are added together, we get 1 + 6 + 9 = 16, which is the original number we picked. Can you explain this?

Solution #35: Chinese NumbersAny number n equal to or smaller than 26 must certainly be smaller than 33 = 27. It can therefore be broken down (in base 3) in the following manner: n = (0 or 1 or 2) x 30 + (0 or 1 or 2) x 31 + (0 or 1 or 2) x 32 In other words n = (0 or 1 or 2) + (0 or 3 or 6) + (0 or 9 or 18) Thus all number that have a "1" in their makeup will be found in the first square, all number that have a "2" in the second square, a "3" in the third square, a "6" in the fourth square, a "9" in the fifth, and an "18" in the sixth. The number n can therefore be found by adding the numbers at the top left-hand corner of the squares where n appears.

Problem #36: The Smallest Possible

Which is the smallest number which, when divided by 2, 3, 4, 5 and 6 will give 1, 2, 3, 4 and 5 as remainders, respectively?

Solution #36: The Smallest Possible

Let n be this unknown number. Since n divided by 2 leaves a remainder of 1, n + 1 must be divisible by 2. Since n divided by 3 leaves a remainder of 2, n + 1 must be divisible by 3. Similarly, n + 1 must be divisible by 4, 5, and 6. The smallest common multiple of 1, 2, 3, 4, 5, and 6 is 60. Therefore, n + 1 = 60. Hence n = 59.

Problem #37: For Those Under 16Tell me in which columns your age appears and I will guess your age by adding the first number of the corresponding columns. Is this magic? squares, one square at a time. +----+----+----+----+| 2 | 8 | 4 | 1 || 3 | 9 | 5 | 3 || 6 | 10 | 6 | 5 || 7 | 11 | 7 | 7 || 10 | 12 | 12 | 9 || 11 | 13 | 13 | 11 || 14 | 14 | 14 | 13 || 15 | 15 | 15 | 15 |+----+----+----+----+

Solution #37: For Those Under 16In is not a matter of magic but simply the characteristics of a number written in base 2. Any whole number less than 16 can be written as x020 + x121 + x222 + x323 where each xn is either a 0 or a 1. When xn = 1, the corresponding age is in the column headed 2n. When xn = 0, the corresponding age is not in the column headed 2n. Example: If you are 13 years old, your age will appear in the last three columns because 13 = 1.20 + 1.21 + 0.22 + 1.23 = 8 + 4 + 0 + 1

Problem #38: ProductThe product of four consecutive whole numbers is 3024. What are these numbers?

Solution #38: Product3024 ends in neither a 5 nor a 0; therefore, none of the four numbers given can be divisible by 5 or by 10. If all four numbers were greater than 10, their product would exceed 10,000, which is not the case. The four numbers must either be {1,2,3,4} or {6,7,8,9}. In the first case, the product is 24. The answer can only be {6,7,8,9}, as can be easily verified.

Problem #39: Riverside DriveWould you like to have dinner with me tonight? Don't go to the wrong house. I live in one of the 11 houses along Riverside Drive. When I am at home, facing the river, and I multiply the number of houses on my left by the number of houses on my right, I get a number which is greater by 5 units than the number that my neighbor to my left would get if he did the same thing. Where along the Drive do I live?

Solution #39: Riverside DriveLet r be the number of houses to the right of my house and l the number of houses to the left of my house when I look toward the river. We have the following simultaneous equations: r + l = 10rl - (r+1)(l-1) = 5Hence, r + l = 10 and r - l = 4. thus r = 7 and l = 3. My house is therefore the fourth from the left when facing the river.

Problem #40: Summit MeetingTwo delegations are to meet at the top floor of a skyscraper whose elevators can hold up to nine people at a time. The first delegation to arrive makes up a certain number of elevator loads, filling each one except the last, which has space for five more people. The second delegation does the same, not using one-third of the last elevator load. At the start of the meeting, each member of each delegation shakes hands with each member of the other delegation, and each time, a photo is taken. Knowing that the photographer was using films with nine exposures on each, how many unexposed frames will he have left on his last film?

Solution #40: Summit MeetingLet d1 be the number of the members in the first delegation and d2 the number of the members in the second. d1 = (multiple of 9) + 4d2 = (multiple of 9) + 6Number of photographs: d1 x d2 = (multiple of 9) + (4 x 6) = (multiple of 9) + (18 + 6) = (multiple of 9) + 6The photographer therefore exposed six frames on his last film. There are three frames left.

Problem #41: A Bag Of MarblesA group of children share marbles from a bag. The first child takes one marble and a tenth of the remainder. The second child takes two marbles and a tenth of the remainder. The third child takes three marbles and a tenth of the remainder. And so on until the last child takes whatever is left. Knowing that all the children end up with the same number of marble, how many children were there and how many marbles did each one get?

Solution #41: A Bag Of MarblesLet n be the number of children, x each child's share of marbles, and N the total number of marbles. We have N = nx. First child's share: 1 + (N-1)/10 = x. Last child's share: x. Penultimate child's share: n - 1 + x/9 = (N/x) - 1 + x/9 = x. Consider the expression for the first child's share: N = 10x - 9. Substituting into the equation for the penultimate child's share gives us (10x - 9)/x - 1 + x/9 = x which can be written 8x2 - 81x + 81 = 0 or (x - 9)(8x - 9) = 0 The number of marbles taken by each child being a whole number, the only possible solution is x = 9; hence N = 81 and n = 9. There were nine children. Each got nine marbles.

Problem #42: The Five NumbersCan you find five consecutive whole number that are all positive and such that the sum of the squares of the two largest numbers is equal to the sum of the squares of the three smallest?

Solution #42: The Five NumbersLet n be then number in the middle of the series. We have the relationship (n + 1)2 + (n + 2)2 = n2 + (n - 1)2 + (n - 2)2 Expanding and simplifying gives us n(n-12) = 0 The only acceptable solution is n = 12. The five consecutive whole numbers are 10, 11, 12, 13 and 14

Problem #43: 100!How many zeros are there at the end of 100! ?

Solution #43: 100!The number of zeros at the end of a number is equal to the number of times that number is a multiple of 10. However, 5 and 2 are factors of 10. The number of zeros will therefore be equal to the smaller of the following two numbers: number of times 2 appears as a factor and number of times 5 appears as a factor in the breakdown of the original number into prime factors. Here, of course, 2 appears as a factor more often than 5. Let us calculate the number of times 5 appears as a factor: 100! has 20 numbers that are multiples of 5. Some of them (4 of them) are even multiples of 25: 25, 50, 75, and 100. In the breakdown of 100! into prime factors, one will find 5 raised to the power 24 (20 + 4 = 24). There are therefore 24 zeros at the end of 100!.

Problem #44: House of CardsDo you know how to build card houses? The first floor is easy: (2 cards) The second floor is easy, too:

(7 cards) Here's the third:

(15 cards) And so forth. How many cards does it take to build a 47-story house?

Solution #44: House of CardsNote first that in all houses of cards, every level has three cards more than the level immediately above it.

The number of cards for a 47-story house is therefore 2 + 2 + 3 + 2 + 3.2 + 2 + 3.3 + ... + 2 + 3.(n-1) + ... + 2 + 3.(47-1)----------------------47.2 + 46.47/2 x 3 = 3337

Problem #45: Lunch With FriendsTen couples meet for lunch. After a cocktail they go into the dining room in a random order, one by one. How many people must enter the dining room to assure that: 1. There is at least one married couple in the group? 2. There are at least two diners of the same sex?

Solution #45: Lunch With FriendsAs soon as 11 people enter the dining room, there will necessarily be one married couple among them. As soon as three people enter the room, there will necessarily be two diners of the same sex.

Problem #46: Thirty-two CardsAlfred, Brian, Christopher and Damon play with a deck of 32 cards. Damon deals them out unequally, then says: "If you want us to have the same number of cards, do exactly as I say. You, Alfred, divide half of your cards between Brian and Christopher. Then, Brian, you do the same with Christopher and Alfred. Finally, Christopher, you follow suit with Alfred and Brian." How did Damon distribute the cards?

Solution #46: Thirty-two CardsAt the end of the game each of the four players has 8 cards. Christopher, having just shared half his cards between Brian and Alfred, must have had 16 cards and Alfred and Brian 4 each. But Brian had just shared half his cards between Christopher and Alfred: before that share-out, Brian must have had 8, Alfred 2, and Christopher 14. But this was just after Alfred shared half his cards between Brian and Christopher. Hence at the start Alfred had 4 cards, Brian 7, Christopher 13, and Damon (who was not involved in the three various share-outs) 8.

Problem #47: Checker PatternHow many turquoise squares will be required to build the twentieth figure in this pattern?

Solution #47: Checker Pattern761. One visual representation for the pattern is n2 + (n-1)2 (outer squares + inner squares)

Problem #48: Odd or EvenDoes 124FIVE represent an odd number? How can you determine whether a number is odd by looking at its base-five representation?

Solution #48: Odd or EvenYes. The number 124FIVE is represented by the base-five pieces that follow. Each piece has an odd number of units. The case of 124FIVE has an odd number of pieces, 7, so 124FIVE is odd. In general, if the sum of the digits in base five is odd, the number is odd.

Problem #49: Five Times FiveThe digits in this addition problem have been replaced by letters. Replace each letter with a different digit to obtain a correct sum that is as large as possible. What is the value of "FIVE"? FIVE FIVE FIVE FIVE+ FIVE------ ISLE

Solution #49: Five Times Five1970.

1. F must equal 1 because anything higher would make the sum greater than four digits. 2.

(not yet completed)

Problem #50: Largest ProductUse each digit 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 exactly once to form two five-digit numbers that when multiplied produce the largest quantity.

Solution #50: Largest Product96,420 and 87,531 with a product of 8,439,739,020. A reasonable first guess is two numbers that begin 97... and 86..., however these do not actually give the largest product.

Problem #51: Multiple ChoiceGiven that one and only one answer is correct, which of the following is true?

A. All of the below B. None of the below C. One of the above D. All of the above E. None of the above F. None of the above

Solution #51: Multiple ChoiceE. Consider the choices. Contradictions between answers, such as between C and D, eliminate A as a possibility. B is false because if it were correct, then C would also be true. It follows that C is false because both A and B are false. Similarly, it follows that D is false. Aha! E is true! Of course, that result makes F false.

Problem #52: A Square NumberFor what bases does the number 121BASE represent a square number?

Solution #52: A Square NumberFor any base! In base b, 121 = 1.b2 + 2.b + 1 = (b + 1)2.

Problem #53: Reverse MultiplicationFind all five-digit numbers that are reversed when multiplied by 4.

Solution #53: Reverse Multiplication21,978 is the only such number. Suppose that ABCDE x 4 = EDCBA. Since 4A < 10 and A is even, A = 2. Therefore, E = 8. Note that E x 4 must end in 2. It follows that B <= 2 because 23 x 4 > 89. In fact, B = 1 because the two-digit number B2 must be a multiple of 4. Thus far we have found that 21CD8 x 4 = 8DC12. Observe that D >= 4 and that D8 x 4 ends in 12. Therefore, D = 7. The only possible value of C is 9.

Problem #54: Odd ProductThe product of three consecutive odd numbers is 357,627. What is the smallest of the three?

Solution #54: Odd Product69. The cube root of 357,627 is slightly less than 71. It seems that 69 x 71 x 73 is a possible product. The result can be verified. Note that 9 x 1 x 3 = 27, thus making the last digit a 7, which is a good visual check. Alternatively, solve a cubic equation by setting up a product of successive odd numbers, say, a-2, a, and a+2. This setup produces a more workable equation than does 2n-1, 2n+1, and 2n+3.

Problem #55: Carnival DiceA carnival game offers you the opportunity to bet $1 on a number from 1 through 6 on a single roll of the two dice. If your number comes up on one die, you win $2 and keep th $1 you be. If it comes on both dice, you win $5 and keep the $1 you bet. Only if the number does not appear on either die do you lose your $1 bet. Does this game favor you or the carnival?

Solution #55: Carnival DiceThe game is fair and does not favor either you or the carnival. The chart shows the eleven winning and twenty-five losing combinations for betting on the number 1. One outcome will win $5 for you, ten will win $2, and twenty-five will lose $1. Since these outcome are all equally likely, your expected winnings are: ($2 x 10) + ($5 x 1) - ($1 x 25) = 0

1 2 3 4 5 61 1,1 1,2 1,3 1,4 1,5 1,6---2 2,1 2,2 2,3 2,4 2,5 2,6 L3 3,1 3,2 3,3 3,4 3,5 3,6 O4 4,1 4,2 4,3 4,4 4,5 4,6 S5 5,1 5,2 5,3 5,4 5,5 5,6 E6 6,1 6,2 6,3 6,4 6,5 3,6 | WIN |---------LOSE------+

Problem #56: Sum Cubes23 = 23 + 23 + 13 + 13 + 13 + 13 + 13 + 13 + 13 Only one other integer requires nine cubes of positive integers to represent its value as a sum of cubes. This other integer is between 200 and 300. Can you find it?

Solution #56: Sum Cubes239. The sum can be expressed as: 23 = 43 + 43 + 33 + 33 + 33 + 33. + 13 + 13 + 13

Problem #57: Tower of SquaresFor this tower of nine squares, determine a line passing through point P that will split the area of the nine squares into two equal parts.

Solution #57: Tower of SquaresPA has the area of 5 squares below it, and PB has the area of 3 squares below it. Thus the area of /\PAB is equal to the area of 2 squares. We have to choose Q so that AQ:QB as 1:3 to get a line with the area of 4.5 squares below it.

Problem #58: Sum to 30Find different ways to express 30 as the sum of two or more consecutive integers.

Solution #58: Sum to 30The expressions 4 + 5 + 6 + 7 + 8, 6 + 7 + 8 + 9, 9 + 10 + 11, (-3) + (-2) + (-1) + ... + 6 + 7 + 8, (-5) + (-4) + (-3) + ... + 7 + 8 + 9, and (-8) + (-7) + (-6) + ... + 9 + 10 + 11 are possibilities. Three of the foregoing six sums consist of only positive integers. Any sum of consecutive positive integers (a+1) + (a+2) + ... + (a+n) is unchanged by placing the sum (-a) + (-a+1) + ... + (a-1) + a in front of it. Therefore, the three "positive only" sums ensure that three more possibilities are feasible. Consider the factors of 30: 1 x 30, 2 x 15, 3 x 10, and 5 x 6. All odd factors greater than 1 produce consecutive integral sums, as follows. Take 3 x 10, for example. Three consecutive integers with 10 in the middle will add to 30, which produces 9 + 10 + 11. Observe that 30 = 0.5 x 60. The consecutive integers 0 and 1 average 0.5. Another sum can be formed by taking thirty pairs of integers each totaling 1. The sum is given by (-29) + (-28) + (-27) + ... + 28 + 29 + 30. The result is actually the sum based on putting integers in front of 30, which comes from 30 x 1. Note that 1.5 x 40, 2.5 x 24, and 7.5 x 8 correspond to some of the other foregoing sums.

Problem #59: The King's Fortune Telling ScaleThe king has a fortune telling scale. For one penny the scale spits out a slip of paper with a fortune on it along with the weight of the king or whatever is on the scale. The king also has five bags of gold that he has given to five trusted friends for safe keeping. Rumor has it that one of his five trusted caretakers is not to be trusted and has been asking a counterfeiter to make phony gold. All the king could find out about the counterfeit gold was that one of the king's five bags is now filled with counterfeit gold and that the weight of that bag is different from the others. He does not know whether it weighs more or less than each of the others. The king calls back all of his bags of gold. Using his fortune telling scale, the king wants to find out:

which bag contains counterfeit gold, and how much it weighs, exactly.

Being a very economical king, he wants to find this out using the least number of pennies possible. His wizard says it can be done with just 3 pennies. No one else can see how. Anyone can do it with 5 pennies. Some say they can do it with 4 pennies. Can you figure out how to do it with 3 pennies? (Assume that the fortunes the scale spits out offer no clues to which bag contains the counterfeit gold.)

Solution #59: The King's Fortune Telling ScaleFive pennies: Weigh each bag once.

Four pennies:(not yet completed)Three pennies:Solution provided by Alan Fung. Let's name the five bags of gold as A, B, C, D and E. One of them is odd and we don't know if the odd bag is lighter or heavier. The first two weights are like these: A + B + C = 3X ----------------(weight #1)A + D = 2Y ----------------(weight #2)X and Y are average unit weight of the combination. That is, they are roughly the weight of each bag of gold. If X = Y, it means the weight ratio among A, B, C and D are the same. (You may check it by adjusting one of them as lighter or heavier, X will not equal to Y). Because A, B, C and D are the same, E must be odd. So the final weight will be to weigh E and that will be the answer. If X < Y, it is because "B or C is lighter" or "A or D is heavier".If X > Y, it is because "B or C is heavier" or "A or D is lighter". The final weight will be: C + D + E = 3Z ----------------(weight #3)The reasons behind picking C, D and E is because now we know E has to be normal, we need to pick E as reference and two bags from the lighter and heavier possibilities. I tried to pick A but it wouldn't work because I shouldn't pick the same thing with the same coefficient that appears in all three equations. By the same token, C and B are interchangeable. Now to make the solution easier to understand, I will start with the solution and work out the weight comparisons. If X < Y If B is lighter, X < Z and Y = ZIf C is lighter, X = Z and Y > ZIf A is heavier, X > Z and Y > ZIf D is heavier, X < Z and Y > ZIf X > Y If B is heavier, X > Z and Y = ZIf C is heavier, X = Z and Y < ZIf A is lighter, X < Z and Y < ZIf D is lighter, X > Z and Y < ZBecause all results from weights comparisons are unique, we covered all possible solutions. For the second part of the question, to know how exactly does the odd bag weigh, we can use simple substitution to work it out. For example, if we found B is odd, B = 3X - 2Y.

Problem #60: Knights of the Round TableKing Arthur liked to invite his knights over for parties around the round table. When the king had a gift that he could only give to one knight, he had them play a game that went like this: First, King Arthur numbered the chairs around the table. At the start, every chair was occupied by a knight. (King Arthur himself did not sit at the table.) Then he stood behind the knight in chair 1 and said, "You're In." Next, he moved to the knight in chair 2 and said, "You're Out," and that knight left his seat and went off to stand at the side of the room to watch the rest of the game. Next he moved to the knight in chair 3 and said, "You're In." Then he said, "You're Out" to the knight in the chair 4, and that knight left his seat and went to the side of the room.

He continued around the table in this matter. When he came back around to the knight in chair 1, he said either "You're In" or "You're Out," depending on what he had said to the last knight. (If the last knight was "In," then the first knight was now "Out, " and vice versa.) The king kept moving around and around the table, alternately saying, "You're In" or "You're Out" to the knights that remained at the table. (If a chair was now empty, he just skipped it.) He continued until only one knight was left sitting at the table. That knight was the winner. If you were a knight, which chair number would you try to sit in at King Arthur's table? The number of knights at the table was always changing, so you should come up with a general rule.

Solution #60: Knights of the Round Table(not yet completed)

Problem #61: Photographers and CannibalsThree National Geographic photographers and three cannibals are traveling together through a jungle when they come to a river. The largest boat available can carry only two people at a time. The photographers are safe only if on each side of the river there are equal numbers of photographers and cannibals or there are more photographers than cannibals; otherwise, the photographers become dinner. How can they all get across? (Tokens such as pennies and nickels may be helpful in solving this problem.)

Solution #61: Photographers and CannibalsA cannibal and a photographer cross. The photographer returns. Two cannibals cross. One cannibal returns. Two photographers cross. One cannibal and one photographer return. Two photographers cross. One cannibal returns. Two cannibals cross. One cannibal returns. Two cannibals cross.

Problem #62: HandshakesIs the number of people in the world who have shaken hands with an odd number of people odd or even? (Dead people count.)

Solution #62: HandshakesEven. If you were to ask everyone in the world how many hands he or she has shaken, the total would be an even number because each handshake would have been counted twice -- once each by the two people who shook hands. A group of numbers whose sum is even cannot contain an odd number of odd numbers.

Problem #63: BookwormA set of encyclopedias consists of volumes that have 1/8-inch covers and one inch of pages. The set is arranged in order on a shelf from left to right. If a bookworm starts at the first page of Volume I and eats its way through to the last page of Volume II, how far does it travel?

Solution #63: BookwormOnly 1/4 inch -- the thickness of the two covers. Put a paperclip on the first page of one book, and another on the last page of a second book. Put both books on a shelf together, the first on the left and the second on the right, and you will see why this surprising answer is correct.

Problem #64: Match GameUsing six identical matches, make exactly four equilateral triangles.

Solution #64: Match GameThe solution is a three-dimensional tetrahedron.

Those who have trouble with this one assume they must solve it in only two dimensions. Once you realize that the use of a third dimension is permissible, it's easy.

Problem #65: Where There's a WillA father wishes to divide a square piece of land among his five sons. One son is his favorite, and he wants to give him one quarter of the land, as shown:

How can he divide the remaining land into four plots of equal size and shape?

Solution #65: Where There's a Will

Problem #66: Rope TrickIs it possible to pick up a piece of rope, one hand holding each end, and tie a knot in the rope without letting go of either end?

Solution #66: Rope Trick not completed

Problem #67: Alphabet SoupPut the rest of the alphabet in its proper place: A EF HI KLMN B D G J C O

Solution #67: Alphabet SoupA EF HI KLMN T VWXYZ B D G J PQR U C O SThe top row contains letters that are made up of straight lines; the middle row, letters formed by combinations of straight and curved lines; the third row, letters formed only by curves.

Problem #68: All WetA recipe calls for 4 cups of water. You have only a 3-cup and a 5-cup container. How can you measure out four cups of water?

Solution #68: All WetHere are two solutions: I.

a. Fill the 3-cup container and pour it into the 5-cup container. b. Fill the 3-cup container again and from it fill the 5-cup container, leaving 1 cup of water in the 3-

cup container. c. Empty the 5-cup container and pour the remaining cup from the 3-cup container into it. d. Now fill the 3-cup container and add it to the 1 cup that is already in the 5-cup container.

The result is the desired 4 cups. II.

a. Fill the 5-cup container. b. Fill up the 3-cup container using the water in the 5-cup container, leaving 2 cups in the 5-cup

container. c. Empty the 3-cup container. d. Pour the remaining 2 cups from the 5-cup container into the 3-cup container. e. Fill the 5-cup container again. f. Pour water from the 5-cup container into the 3-cup container (that already has 2 cups of water

in it) until the 3-cup container is full. Now, 4 cups of water remain in the 5-cup container.

Problem #69: The Telltale NumberWrite a ten-digit number so that the first digit tells how many zeros there are in the number, the second how many ones, the third how many twos, and so forth.

Solution #69: The Telltale Number6,210,001,000

Problem #70: Something FishyA fish weighs ten pounds plus half of its weight. How much does it weigh?

Solution #70: Something FishyTwenty pounds. Here is an algebraic solution:

W = 10 + 1/2 WW - 1/2 W = 10 1/2 W = 10 W = 20

Problem #71: The Baffling BicycleA bicycle climbs a certain hill at 10 miles per hour and returns at 20 miles per hour. What is its average speed for the entire trip?

Solution #71: The Baffling BicycleThe apparent answer -- 15 miles per hour -- is wrong. The correct answer is 13 1/3 miles per hour because speed is determined by dividing the distance by time. Notice, incidentally, that the answer is the same no matter how long the hill is.

Problem #72: Figure EightAn eight-digit number contains two 1's, two 2's, two 3's, and two 4's. The 1's are separated by one digit, the 2's by two digits, the 3's by three digits, and the 4's by four digits. What is the number?

Solution #72: Figure Eight

41,312,432Problem #73: The Balancing BrickA brick balances evenly with three quarters of a pound and three quarters of a brick. What does the whole brick weigh?

Solution #73: The Balancing BrickThree pounds. If the brick balances with three quarters of a brick plus three quarters of a pound, then one quarter of a brick must weigh three quarters of a pound. Thus a whole brick weighs four times as much, or three pounds.

Problem #74: The Scrambled SalesmanAn egg salesman was asked how many eggs he had sold that day. He replied, "My first customer said, 'I'll buy half your eggs and half an egg more.' My second and third said the same thing. When I had filled all three orders I was sold out and I had not had to break a single egg all day."

Solution #74: The Scrambled SalesmanSeven. He sold four eggs to the first customer, two to the second, and one to the third. The problem is solved most easily if you start with the last customer and work backwards.

Problem #75: Wrong NumberThirteen per cent of the people in a certain town have unlisted phone numbers. You select three hundred names at random from the phone book. What is the expected number of people who will have unlisted numbers?

Solution #75: Wrong NumberNone. An unlisted phone number doesn't appear in the phone book.

Problem #76: Magic NumbersThis is a good way to mystify a friend. Ask someone to pick a number, any number at all, but not to tell you what it is. Have him silently multiply it by five, add five to the product, multiply the resulting number by two, add two to it, and tell you the result. Instantly you are able to tell him the number he started with. What is the secret and why does it work?

Solution #76: Magic NumbersThe secret: To find the number selected, simply delete the last digit from the number given you. Then subtract one from the remaining number. Reason: (not yet completed)

Problem #77: Case of the Counterfeit CoinsYou have ten stacks of ten silver dollars each. They are identical, except that one stack consists entirely of counterfeit dollars. You know the weight of an authentic dollar, and you also know that a counterfeit dollar weights one gram less. How many weighings are needed to reveal which stack is counterfeit?

Solution #77: Case of the Counterfeit CoinsOnly one. Weigh one coin from the first stack, two coins from the second, and so forth. The number of grams by which the total is light will correspond to the number of the counterfeit stack.

Problem #78: Strange SeriesThe following number is the only one of its kind. Can you figure out what is so special about it?

8,549,176,320

Solution #78: Strange SeriesIt is the only one that contains all the numerals in alphabetical order.

Problem #79: Boxed InThree boxes contain two coins each. One contains two nickels, one contains two dimes, and one contains a dime and a nickel. All three boxes are mislabeled. (Each contains the label of one of the other two boxes.) If you are permitted to take out only one coin at a time, how many must you take out in order to be able to label all three boxes correctly?

Solution #79: Boxed InOnly one. Take it from the box labeled "Dime and Nickel." Since you know all three boxes are mislabeled, the box contains either two nickels or two dimes. Put the correct label on that box after inspecting one of the coins from the box. Then simply switch the two remaining labels.

Problem #80: Weighty QuestionWhich weighs more, a pound of gold or a pound of lead? (You may think you have heard this one before, but you probably haven't.)

Solution #80: Weighty QuestionA pound of lead. Lead is weighed in the standard measure, in which 7,000 grains equal a pound. Gold is measured in Troy weight, in which 5,7600 grains equal a pound.

Problem #81: Early BirdA chauffeur always arrives at the train station at exactly five o'clock to pick up his boss and drive her home. One day his boss arrives an hour early, starts walking home, and is picked up by the chauffeur on the way out to the train station. They arrive at home twenty minutes earlier than usual. How long did she walk before she met her chauffeur?

Solution #81: Early BirdFor fifty minutes. She saved the chauffeur ten minutes of traveling time each way and thus was picked up at 4:50 pm rather than the usual time.

Problem #82: Take a LetterWhat three different digits are represented by X, Y, and Z in this addition problem?

XZY+XYZ---- YZX

Solution #82: Take a LetterX=4, Y=9 and Z=5. In the middle column, adding Y to Z gives a digit of Z. This means Y must be either a 0 or a 9. (It can be a 9 if the addition in the far right column causes carrying a 1 into the middle column.) Since Y is the first digit in the number on the bottom, it cannot be a zero. So, Y=9. The two X's in the left column plus a possible 1 carried from the middle column must add up to Y, which is 9. Since 9 is odd, a 1 must have been carried over from the middle column, leaving X+X=8. So, X=4. The value of Z can be determined from the right column: Since Y=9 and X=4, Z must be 5 because it is the only digit that can be added to 9 that puts a 4 in the one's place (9+5=14).

Problem #83: Water and MilkThere are two glasses on the table, one containing water and the other one milk. They both contain exactly the same amount by volume. If you take a teaspoon of water and mix it into the milk and then take a teaspoonful from the milk glass and mix it with the water, both glasses become contaminated. But which is the more contaminated? Does the water now contain more milk than the milk does water or the other way round?

Solution #83: Water and MilkThey are both equally contaminated. The water contains exactly as much milk as the milk contains water. The most elegant proof for this celebrated little puzzle is as follows: It does not matter how many transfers are made between the glasses or whether the contents are stirred. Provided that the volumes in the two glasses are equal, then any water in the water glass must be in the milk, there is nowhere else it can be. The milk that it has replaced must be in the water glass. The water glass therefore contains as much milk as the milk contains water.

Problem #84: Four SheepFarmer Giles has four sheep. One day, he notices that they are standing in such a way that they are all the same distance away from each other. That is to say, the distance between any two of the four sheep is the same. How can this be so?

Solution #84: Four SheepThe sheep are standing on the four corner points of an equal-sided pyramid (a tetrahedron). Or to put it another way, three are on the points of an equilateral triangle and the other is on a mound of earth in the center.

Problem #85: Do Helmets Increase Head Injuries?At the beginning of the first World War, the uniform of the British soldiers included a brown cloth cap. They were not provided with metal helmets. As the war went on, the army authorities and the War Office became alarmed at the high proportion of men suffering head injuries. They therefore decided to replace the cloth headgear with metal helmets. From then on, all soldiers wore the metal helmets. However, the War Office was amazed to discover that the incidence of head injuries then increased. It can be assumed that the intensity of fighting was the same before and after this change. So why should the recorded number of head injuries per battalion increase when men wore metal helmets rather than cloth caps?

Solution #85: Do Helmets Increase Head Injuries?The number of recorded head injuries increased, but the number of deaths decreased. Previously, if a soldier had been hit on the head by a piece of shrapnel, it would have pierced his cap and probably killed him. This would have been recorded as a death, not a head injury. After helmets were issued it was more likely that a fragment of shrapnel would cause an injury rather than death. Thus, the incidence of head injuries increased, while the incidence of deaths decreased.

Problem #86: SonsI. Mrs. Jones has two children. At least one is a boy. What are the chances that both are boys? II. Mrs. Brown has two children. The younger one is a boy. What are the chances that both are boys?

Solution #86: SonsFor two children, there are only four possible combinations:

Older Younger------------------A Girl GirlB Girl BoyC Boy GirlD Boy Boy

Each of these combinations is equally likely (i.e. there is a one in four chance that any two-child family will have one of the above combinations). For Mrs. Jones, the possibilities are narrowed down to A, B or C and of these, only A means that both are both boys. Therefore, the chance that both her children are boys is one in three. For Mrs. Brown, A and C are the only possibilities. There is thus a one in two chance that both her children are boys.

Problem #87: Dinner for ThreeAn ancient Arabic puzzle goes like this: A hunter met two shepherds, one of whom had three loaves of bread and the other, five loaves. All the loaves were the same size. The three men agreed to share the eight loaves equally between them. After they had eaten, the hunter gave the shepherds eight bronze coins as payment for his meal. How should the two shepherds fairly divide this money?

Solution #87: Dinner for ThreeThe shepherd who had three loaves should get one coin and the shepherd who had five loaves should get seven coins.

If there were eight loaves and three men, each man ate two and two-thirds loaves. So the first shepherd gave the hunter one third of a loaf and the second shepherd gave the hunter two and one-third loaves. The shepherd who gave one-third of a loaf should get one coin and the one who gave seven-thirds of a loaf should get seven coins.

Problem #88: One ClockIn the days before watches were invented, clocks were valuable items. There was a man who had one clock in his house. It kept good time, but one day he found that it had stopped. He had no idea what the correct time was. He walked to the next valley to visit his friend who had a clock showing the right time. He spent a little while chatting with his friend then he walked home. He did not know the exact length of the journey before he started. How did he manage to set his one clock correctly on his return?

Solution #88: One ClockHe wound his clock and set it at some particular time before he left. He noted the exact time of his arrival at and departure from his friend's house. He noted the time showing on his clock when he returned. He walked at the same pace on the two journeys. The elapsed time on his clock is the duration of the two journeys plus the length of his visit to his friend. Knowing the time he spent with his friend, he subtracts this from the elapsed time on his clock and divides the result by two in order to calculate the duration of the journey. He adds this to the exact time he left his friend's house in order to set his clock at the correct time. For example, assume he set his clock at 12, arrived at his friend's at 6:30, and left at 7:30. When he returned, his clock showed 4:00. Then his journey took one and one half hours and the correct time is 9:00.

Problem #89: The Mongolian Postal ServiceThe Mongolian Postal Service has a strict rule stating that items sent through the post must not be more than 1 meter long. Longer items must be sent by private carriers, and they are notorious for their expense, inefficiency, and high rate of loss of goods. Boris was desperate to send his valuable and ancient flute safely through the post. Unfortunately, it was 1.4 meters long and could not be disassembled as it was one long hollow piece of ebony. Eventually he hit on a way to send it through the Mongolian Postal Service. What did Boris do?

Solution #89: The Mongolian Postal ServiceBoris placed the flute diagonally in a suitcase that measured 1 meter by 1 meter. This suitcase was quite acceptable to the postal officials its sides measured 1 meter. From corner to corner, it measures 1.414 meters -- the square root of two. Incidentally, if his flute had measured 1.7 meters, he could have fitted it across the diagonal of a box whose sides were 1 meter long. The diagonal of a cube is the square root of the three times its side -- 1.73 meters. From a theoretical mathematical viewpoint, there is no reason why this process cannot be extended indefinitely. If Boris could construct a four-dimensional box with 1-meter sides, then he could get a 2-meter flute in it (square root of 4) and a 25-dimensional construction could contain a 5-meter flute while still meeting the rules of having no side longer than 1 meter.

Problem #90: The Amorous CommuterJohn Jones lives in Maidenhead. He has one girlfriend in Reading and another in Slough. He has no car and therefore takes a train whenever he goes to see them. Trains stopping in Maidenhead can go either east or west. If they are westbound, they will go to Reading. If they are eastbound, they will go to Slough. There are an equal number of trains going in each direction. John likes his two girlfriends equally. Because he finds it hard to choose between them, he decides that when he goes to the station, he will take the first arriving train, regardless of whether it is going east or west. After he has done this for a month, he finds that he has visited the girlfriend in Slough 11 times as often as he visited the girl in Reading. Assuming that he arrived an the Maidenhead station at random times, why should the poor girl in Reading have received so little attention?

Solution #90: The Amorous CommuterThe Slough trains depart from Maiden head at five past every hour. The Reading Trains depart at ten past the hour. If John arrives at any time between ten past the hour and five past the next hour, then the first train to arrive will be bound for Slough. He will catch the Reading train only if he happens to arrive between five past and ten past the hour. It is therefore 11 times more likely that he will catch the Slough train than the Reading train.

Problem #91: Short RoadsThere are four main towns in Lateralia. We will call them A, B, C and D. They lie at the corners of a ten-mile square. In order to improve communications between the towns, the Lateralian Department of Transport decided to build a new road linking all four towns together. Because they had very little money, it was decided that the new road system should be as short as possible and still allow access from any one town to any other. The engineers came up with three designs shown below.

Number one uses 40 miles of road, number two uses 30 miles of road, and number three uses 28.3 miles of road. The designers naturally recommend plan number three because it employed the smallest road area and, therefore, cost the least. However, when they submitted their plan to the Minister of Finance, he accused them of extravagance and quickly pointed out a better design that required even less total road surface. What was his superior solution?

Solution #91: Short RoadsThe shortest solution is shown. It represents some 27.3 miles of road in total and therefore saves a precious mile in road-building expense compared to the two diagonals. Someone starting from A would have a shorted journey to D but a longer one to B or C.

Problem #92: A Weighty Problem IA shopkeeper wants to be able to dispense sugar in whole pounds ranging from one pound up to 40 pounds. He has a standard, equal-arm balance weigh scale. Being of an extremely economical outlook, he wants to use the least possible number of weights to enable him to weigh any number of pounds between 1 and 40. How many weights does he need and what are they?

Solution #92: A Weighty Problem IIf the weights can be placed in either of the scale pans, then you can solve the problem with weights of 1, 3, 9, and 27 pounds only. With that combination any weight from 1 to 40 pounds can be measured. If the weights can be placed in one scale pan only, then you need the weights 1, 2, 4, 8, 16, and 32 pounds, which enables you to measure any weight up to 63 pounds. This latter solution is really an example of counting with a binary number system where any number can be expressed as the sum of 2 raised to various powers. For example, 63 expressed in binary form is 111111.

Problem #93: A Weighty Problem IIYou have an equal-arm balance scale and twelve solid balls. You are told that one of the balls has a different weight from all the others, but you do not know whether it is lighter or heavier. You can weigh the balls against each other in the scale balance. Can you find the odd ball and tell if it is lighter or heavier in only three weighings?

Solution #93: A Weighty Problem IILet us call the balls A, B, C, D, ..., L. Start by weighing four against four. If they balance, then weigh any of the remaining three against any three of the good balls. If they balance then we know the odd one is the remaining ball and we can identify whether it is heavier or lighter in the final weighing. If the three against three do not balance then we take the three containing the odd ball and weigh an one against another. If the first weighing of four against four does not produce a balance, then the second weighing involves three against three with balls switched between the two pans and a good ball introduced. So: If A + B + C + D > E + F + G + HWe try A + B + E against C + F + JIf A + B + E = C + F + J,then we know that either D is heaver or G or H is lighter, so we weight G against H. If A + B + E > C + F + J,then we know that either F is lighter or A or B is heavier, so we weight A against B.

If A + B + E < C + F + J,then we know that either E is lighter or C is heavier, so we weight either against a good ball (e.g. K against E). An alternative second weighing is A + B + E against C + D + F, which follows similar lines to the above.

Problem #94: Eight Years OldA girl was eight years old on her first birthday. How could that be?

Solution #94: Eight Years OldShe was born on February 29, 1896. The year 1900 was not a leap year (only centuries divisible by 400 are leap years), so the next February 29 fell in 1904 when she was eight. She was twelve on her second birthday.

Problem #95: Cover That HoleA manhole is a hole which allows someone to gain access to the sewers or other pipes which are below ground. Our local town council recently decided that all the town's manhole covers should be changed from square to round ones. We are used to the town council making silly decisions, but this time they were absolutely right. Why?

Solution #95: Cover That HoleA square or rectangular manhole cover can fall down the hole, while a round manhole cover cannot. The square cover will fit down the diagonal of the hole (unless the rim it sits on is very large) but no matter how you turn a circle it never measures less than its diameter. So for safety and practicality all manhole covers should be round. Alex Freuman also notes: "In addition to the advantage that a circular manhole cover cannot fall into a manhole, one can roll a circular manhole cover rather than lifting it."

Problem #96: Pond ProblemA man wishes to reach the island in the middle of an ornamental lake without getting wet. The island is 20 feet from each edge of the pond (see diagram) and he has two planks each 19 feet long. How does he get across?

Solution #96: Pond ProblemHey lays the planks as shown in this diagram:

Problem #97: River Problem IA man came to a river carrying a fox, a duck, and a bag of corn. There was a boat in which he could ferry of the three items across the river at any one time. He could not leave the fox alone with the duck, nor the duck alone with the corn, so how did he get all three across?

Solution #97: River Problem IFirst the man took the duck across, then he came back and took the fox over. He left the fox on the far side of the river and returned with the duck. He then left the duck on the near side and took the corn over. Then he returned and took the duck across. Pretty straightforward, eh?

Problem #98: River Problem IIThis time the man reached the river with a fox, a duck, and a bag of corn, but this fox ate corn as well as ducks! There was the same boat as before in which he could take only one of the three with him. He could not leave the fox with either the corn or the duck, and of course, the duck would gladly eat the corn if they were left together. How did he get all three across?

Solution #98: River Problem IIThe man tied the duck to the back of the boat wit a rope. The duck swam along behind the boat as the man ferried the fox and corn over in turn.

Problem #99: The Bicycles and the FlyTwo boys on bicycles 20 miles apart, began racing directly toward each other. The instance they started, a fly on the handle of one bicycle started flying straight toward the other cyclist. As soon as it reached the other handle bar it turned and started back. The fly went back and forth this way, from handle bar to handle bar, until the two bicycles met. If each bicycle had a constant speed of 10 miles per hour, and the fly flew at a constant speed of 15 miles per hour, how far did the fly fly?

Solution #99: The Bicycles and the FlyEach bicycle travels at 10 miles per hour, so they will meet at the center of the 20-mile distance in exactly one hour. The fly travels at 15 miles per hour, so at the end of the hour it will have gone 15 miles.

Problem #100: Where does the square go?Paste a sheet of graph paper on a piece of cardboard. Draw the square shown in figure 1, then cut along the lines to make five pieces. When you rearrange these same five pieces, in the manner shown in Figure 2, a hole will appear in the center of the square!

The square in Figure 1 is made up of 49 smaller square. The square in Figure 2 has only 48 small squares. Which small square has vanished and where did it go?

Solution #100: Where does the square go?When the two largest pieces are stitched, each small square that is cut by the diagonal line becomes a trifle higher than it is wide. This means that the large square is no longer a perfect square. In has increased in height by an area that is exactly equal to the area of the hole.

Problem #101: SiblingsTodd and Kristina (not yet completed)

Solution #101: Siblings(not yet completed)