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BaseA two – digit number when read from left to right is 4.5 times less than the same number read from right to left.What is the first digit of the number?a) 1 b) 2 c) 3 d) 4Let the number be 10x + y10y + x = 4.5(10x + y)10y + x = 45x + 4.5y5.5y = 44x

COUNTINGIf all the numbers between 7 and 100 are written on a paper, how many times will the number “4” be used?a) 19 b) 18 c) 9 d) 1214,24,34,….94 = 940,41,42,43,44, … 49 = 10Total = 19

CountingA boy counted in the following way on the fingers in his hand.He started numbering the thumb 1, index finger 2, middle finger 3, ring finger 4, little finger 5.Then he reversed the direction calling ring finger 6 middle finger 7, index finger 8, thumb 9 and then back to index finger 10, middle finger 11 and so on.If he count upto 1985, on which finger did he end with?a) Little finger b) Middle fingerc) Thumb d) None of theseAfter 9, it is a multiple of 8 to end on thumb.Hence 17,25,… 8n+1 will end on thumb.1985 = 8n + 1.

CountingHow many times does the digit 6 appear from 11 to 100?a) 16 b) 17 c) 18 d) 1916,26,36,…96 = 960,61,62,…69 = 10Total = 19

CountingHow many numbers are there between 7000 and 8000 containing at least one 9?a) 300 b) 295 c) 288 d) 2717000 to 7009 contains one 9 = 17000 to 7099 contains 20 9’s = 207000 to 7899 contains 9 * 20 = = 1807900 to 7999 contains 20 + 100 = 120 = 120

total = 300

CountingHow many odd numbers are there between 7000 and 10000 containing atleast one 8?a) 910 b) 813 c) 690 d) None of theseFrom 8001 to 8999 there are 500 ODD numbers containing atleast one 8.Now from 7001 to 7999:When 8 is in unit digit, the number is even and hence count = 0When 8 is in ten’s digit, count = 10*1*5 = 50 (unit digit contain only 1,3,5,7,9)When 8 is in hundred’s digit count = 1*10*5 = 50Here 7881,7883,7885,7887,7889 will be repeated twice in the counting.Hence total count between 7001 and 7999 = 95.Count of odd number between 9001 and 1000 is also 95.Hence total numbers = 500 + 95 + 95 = 690

COUNTINGIf you wrote all of the numbers from 300 to 400 on a piece of paper, how many times would you have written the number 3?a) 118 b) 119 c) 120 d) 121The hundredth place between 300 to 399 = 100The ten’s place between 330 to 339 = 10The unit’s place 303,313,323,… 393 = 10Total = 120

COUNTA four digit number is a multiple of 9 and can have a digit repeated exactly three times consecutively.How many such numbers are possible?a) 8 b) 12 c) 16 d) 20As all four digits cannot be same, the maximum sum of digits should be either 9 or 18 or 27.Let the number be in the format mmmk, kmmmCase 3m + k = 9 Possible cases are 3330,2223,3222, 1116,6111, 9000Case 3m + k = 18 Possible cases are 6660, 5553, 3555, 4446, 6444, 9333, 3999Case 3m + k = 27 Possible cases are 9990, 8883, 3888, 7776, 6777, 6669, 9666Total possible cases = 6 + 7 + 7 = 20.

DivisibilityThe sum of four consecutive odd numbers is always divisible bya) 3 b) 4 c) 7 d) 2Let the four consecutive odd numbers be 2n – 3 , 2n – 1, 2n + 1, 2n + 3.Sum of the numbers = 8nIt is always divisible by 8 and hence by 4 and 2.

Divisibility164 + 220 is divisible by a) 7 b) 11 c) 13 d) 17164 + 220 = 216 + 220 = 216(1 + 24) = 216(1 + 16)= 17k.

Divisibility• If the sum of the digits of an even number is divisible by 9, then that

number is always divisible by a) 12 b) 27 c) 24 d) 18An even number is always divisible by 2.Hence the number which is both divisible by 9 and 2 is divisible by 18.

DivisibilityIf p is a prime number greater than 3, then which of the following numbers will always divide p2 – 1?a) 20 b) 22 c) 24 d) 26p2 – 1 = (p – 1)(p +1)When p is greater than 3 then both p – 1 and p + 1 are consecutive even numbers. (2n) and 2(n + 1)One of the number (p – 1) or (p + 1) is divisible by 2 and another by 4.Since one of consecutive three numbers (p – 1), p, (p + 1) should be divisible by 3.Since p is prime number, either p – 1 or p + 1 should be divisible by 3.p2 – 1 is always divisible by 2,3,4 and hence by 24 (when p is greater than 3)

EquationInstead of multiplying a number by 53, Suresh multiplied it by 35and got the answer which was 1206 less than the expected answer.What is the number?a) 51 b) 48 c) 61 d) 6753x = 35x + 120618x = 1206x = 67

EQUATIONThe product of a two digit number by a number consisting of same digits written in reverse order is 1300. Find the smaller number.a) 45 b) 65 c) 25 d) 35(10x + y) * (10y + x) = 1300

= 13 * 100 = 13 * 2 * 50 = 13 * 2 * 5 * 10 = 13 * 4 * 25 = 52 * 25The smaller number is 25.

Even & OddIf m is an odd integer and n an even integer, which of the following is even?A) (2m + n) ( m - n) B) (m + n2)(m – n2)C) m2 + mn + n2 D) m + n2m is even and n is also even. Addition of two even is also even. Hence 2m + n is even.Product of even with odd is also even.Hence A is even.

Even or OddIf x is an integer, which of the following CANNOT be an even integer?a) 2x + 2 b) x – 5 c) 2x + 3 d) 5x + 2Whether x is even or odd, 2x is always even. 2x + 3 is CANNOT be an even integer.

ExpressionWhat is the value of the below expression?

a) b) c) d)

= = = =

FactorialWhat highest power of 8 will divide 26! exactly?a) 3 b) 5 c) 7 d) None of these = 3 = 0Answer is 3.Verification: the numbers 8,16,24 are the only three numbers less than 26! divisible by 8.

FactorsThere are two numbers, one of which is twice the other.Both had the same number of prime factorsWhile the larger number had four factors more than the smaller one.Choose the pair containing those two numbers.a) 40,80 b) 20,40 c) 30,60 d) 50,100d(n) denote the number of factors.d(40) = 4* 2 = 8 and d(80) = 5 * 2 = 10d(20) = 3 * 2 = 6 and d(40) = 4 * 2 = 8d(30) = 2 * 2 *2 = 8 and D(60) = 3 * 2 * 2 = 12d(50) = 2 * 3 = 6 and d(100) = 3 * 3 = 9Hence answer is 30 and 60.

FractionsOne-fourth of one-third of two-fifth of a number is 25.What will be 60% of the number?a) 450 b) 400 c) 900 d) 72060% of the number is three-fifth of the number.

= 450

FRACTIONWhich number when added to gives the same result as when it is multiplied by ?a) 9 b) 7 c) 5 d) 3x + = = x = X = 5.

HCFFind the HCF of 2014 – 1 and 2012 – 1.a) 543 b) 481 c) 427 d) 399HCF of 14 and 12 is 2Both (202)7 -1 and (202)6 – 1 are divisible by 202 – 1 = 400 – 1 = 399.

Highest Power in FactorialFind the highest power of 10 in 120!a) 22 b) 24 c) 27 d) 28Half of the numbers in 120! is even. To find the highest power of 5 in 120!

= 4Hence the highest power is 28.

Number SystemIn a number system with base b, 12 * 25 = 333, the value of b isa) 9 b) 8 c) 7 d) 6(b + 2)(2b +5) = 3b2 + 3b + 32b2 + 9b + 10 = 3b2 + 3b + 3b2 – 6b – 7 = 0b = = 7, -1The base is 7.

PartitionIf X + Y = 6, then XY = ? a) 2 b) 4 c) 6 d) 9The possible values of X and Y (assuming X and Y are positive integers) are(1,5),(2,4),(3,3). Only possible choice is 9.

Percentage in NumbersIf 50% of the number is added to 50, the result is number itself.The number is a) 50 b) 200 c) 150 d) 10050% of the number should be equal to 50.Hence the number is 100.In other words,(0.5)x + 50 = x 0.5x = 50 x = 100.

Powers & IndicesWhat is the ten’s digit in 29999?a) 2 b) 4 c) 6 d) 8Ten’s digit is the remainder when divided by 100.29999 mod 100210(999)* 29 mod 100210 raised to even power has last two digits 76 and when raised to odd power has last two digits 2424 * 29 mod 100288 mod 100= 88.Hence ten’s digit is 8

Powers & IndicesWhat will be the last digit of the multiplication 2222 * 3333 * 4444 ?The unit digit of 2222 = 2220 * 222 = 6 * 4 = 24 4The unit digit of 3333 = 3332 * 331 = 1 * 3 = 3The unit digit of 4444 = 6The unit digit in the multiplication= 4 * 3 * 6 = 72 2

Power and IndicesDetermine the digit in the unit position of 1121 * 1717 * 2121 .Unit digit of 1121 and 2121 = 1Unit digit of 1717 is 1716 * 171 = 1 * 7 = 7The unit digit in the product = 7.

Powers and indicesWhat are the last two digits of 72008?a) 49 b) 43 c) 01 d) 0772008 = 74(502) = (2401)502.The last two digits of (01)n is always 01.

Powers and indicesWhat is the unit digit in the product 365 x 659 x 771?a) 1 b) 2 c) 4 d) 6Unit digit in 365 = 364 * 3 = 1 x 3 = 3Unit digit in 659 = 6Unit digit in 771 = 768 * 73 = 1 * 3 = 3Unit digit in the product = 4

Powers and IndicesFind the last two digits in 259166 .a) 11 b) 41 c) 71 d) None of theseThe last digit in 259166 is 1. (since all number ending in 1 raised to any power is 1)To find the second last digit.The ten’s place in the base is 9The unit place in the power is 6Product = 54.Hence the last two digits is 41.

Powers and IndicesWhat is the value of x if 9x = 9 ÷ 3x .a) 1/3 b) 2 c) 2/3 d) 1/2 32x = 32 ÷ 3x.2x = 2 – x 3x = 2x = 2/3

Powers and IndicesFind the last two digits in 476125.a) 21 b) 41 c) 81 d) None of these.The last digit is 1.To find the second last digit,The ten’s digit in the base is 6The unit digit in the power is 5Product is 30.Hence the last two digits is 01.

ProgressionABCD is a popular software company and hence for the hiring process 2557 applicants were standing in the queue.Between every two females there were five males in the queue.The maximum number of females could bea) 427 b) 426 c) 408 d) 407It is in A.P with a = 1 , d = 6 and last term is 2557.N = = 426 + 1 = 427.

Prime NumbersThe highest prime number that can be stored in a 8-bit microprocessor isA) 247 B) 253 C) 317 D) 25128 = 256.Numbers less than 256 are 247 and 251247 is not prime. 247 divisible by 13 and 19251 is the highest prime less than 256

RemainderWhat is the remainder when 2256 is divided by 17?a) 1 b) 2 c) 5 d) None of these2256 = 24*64 = 1664 = (17 – 1)64.The remainder is (-1). That is 17 – 1 = 16.

RemainderOn dividing a number by 209, we get 50 as remainder.What will be the remainder when dividing the same number by 19?a) 15 b) 13 c) 17 d) 12Let the number be NN = 209q + 50

= 19*11q + 19*2 + 12Dividing by 19, the remainder is 12.

RemainderA number is divided by 5,2 and3 successively to get remainder of 0,1 and 2 respectively.What will be the remainder if the same number is divided by 2,3,5 successively.a) 0,2,4b) 1,0,4c) 1,1,2 d) 0,1,3Let the final quotient be q when divided by 3Divided by 5 remainder is 0 = 5xDivided by 2 remainder is 1 = x = 2k + 1 = 10k + 5Note the last digit should be 5.Divided by 3 remainder is 2 = 10k + 3 = 3n 5x = 10k + 5 5x - 2 it should be a multiple of 3. Hence when divided by 3 the remainder should be 0.

RemainderWhat is the remainder when 50! is divided by 168?a) 1 b) 3 c) 5 d) None of these168 = 232.Let us find the highest power of 2 in 50![50/2] = 25.[50/4] = 12[50/8] = 6[50/16] = 3[50/32] = 1Highest power of 2 which divides 50! is 47.232 divides 241 without remainder. Hence the remainder is zero.

RemainderWhat is the remainder when -4x3 + 8x2 + 12x + 16 is divided by x + 2?a) 8 b) 24 c) 32 d) 56P(-2) = -4(-8) + 8(4) + 12(-2) + 16

= 32 + 32 – 24 + 16= 56

RemainderWhat is the remainder when 5163 – 7593 + 11593 – 1163 is divided by 4?a)3 b) 2 c) 1 d) 05163 – 1163 is divisible by 411593 – 7593 = 400k is also divisible by 4 ( a3 – b3) = (a – b)kThe remainder is zero.

RemainderWhat is the remainder when 5555 + 55 is divided by 56?a) 0 b) 1 c) 54 d) 555555 = (56 – 1)55 When divided by 56, except the last term which is -1 because the power is odd, all other terms in the binomial expansion is divisible by 56, the remainder is -1.5555 + 55 when divided by 56, the remainder is -1 + 55 = 54.

RemainderWhich of the following divides the difference between cubes of two consecutive positive even integers without leaving a remainder? a) 16 b) 8 c) 3 d) None of these.(2n + 2)3 – (2n)3 = 23(n + 1)3 – 23(n)3 = 23[(n+1)3 – n3]It is always divisible by 8

RemainderWhat is the remainder when 9113 * 7110 is divided by 31?a) 1 b) 5 c) 12 d) 169113 * 7110 = (23 + 1)113 * (23 – 1)110 = [(23 + 1)(23 – 1)]110 * 93 =( 26 – 1)110 * 93

= (32 + 31)110 * 93.The expression (32 + 31)110 divided by 31 leaves 1 as remainder.93 = 729 divided by 31 leaves 16 as remainder.Hence the final remainder is 1 * 16 = 16.

RemainderThe numbers from 1 to 29 are written continuously like 1234567891011…272829 and if the big number formed thus is divided by 9, what is the remainder?a) 1 b) 3 c) 5 d) 71 to 9 appear three times and their sum is divisible by 9.Ten one’s and ten two’s 10 1’s and 10 2’s = 9 1’s and 9 2’s plus one 1 and one 2When divided by 9 leaves the remainder 1 + 2 = 3.

RemainderA number when divided by 765 leaves a remainder 42.What will be the remainder if the number is divided by 17?a) 8 b) 12 c) 13 d) 9Let the number be NN = 765k + 42N = (17)(45)k + (17)(2) + 8The remainder is 8

RemainderWhat is the remainder when 3126 is divided by 8?a) 1 b) 3 c) 5 d) 73126 = 32(61)

961 = (8 + 1)61 = 1(mod 8)Hence the remainder is 1.

RemainderWhich of the following option does not divide 56 – 1 completely? (i.e., rem = 0)a) 18 b) 24 c) 27 d) 3156 – 1 = (53)2 – 1 = (53 – 1)(53 + 1)= 124 * 126= 2*2*31 * 2 * 3 * 3 * 7It is not completely divisible by 27.

RemainderWhat is the remainder when 482 is divided by 6?a) 1 b) 2 c) 3 d) 4482 = (6 – 2)82. Except the last term all the terms are divisible by 6.Hence the remainder is 2 ( power is even. Hence positive)

Square• Which of the following cannot be the square of a natural number?a) 32761 b) 81225 c) 42437d) 20164Clearly no perfect square will end in 7.Hence 42437 is not a perfect square.

Squares & CubesHow many positive numbers less than 50000 exist which are both perfect squares and perfect cubes?a) 12 b) 10 c) 8 d) 6Numbers which are both perfect square and perfect cubes should be of the form x6.106 = 100000066 = 26 * 36 = 64 * 729 ≈ 43000. Hence Answer is 6.86 = 218 = 210 * 28 = 1024 * 256

Squares and CubesHow many positive numbers less than 10000 exist so that they are perfect squares but not perfect cubes?a)108 b) 104 c) 99 d) None of these Total numbers less than 10000 and perfect squares = 99 (1002 = 10000)Numbers which are perfect squares and perfect cubes = (a2)3 or (a3)2 = a6 Such numbers are 16, 26, 36, 46. ( Since 56 = 15625 > 10000)Hence total numbers which are perfect squares but not cubes = 99 – 4 = 95

Sum of the seriesFind the sum upto 20 terms in the series1 + (1 + 3) + ( 1 + 3 + 5) + (1 + 3 + 5 + 7) +…a) 2870 b) 3021c) 2920 d) 3186Since sum of n odd numbers is n2, each term given in the series is the square of nth term.Summation of squares upto 20 terms =

= 70 * 41= 2870

Summation• The sum of the odd numbers between 1 and n is 11025, where n is an

even number.• What is the value of n?• a) 210 b) 202 c) 204 d) 208• Sum of odd numbers from 1 to n ( n is even) = (n/2)2 = 11025• n/2 = 105• n = 210

Test of divisibilityWhich one of the following is divisible by 99?a)913464 b) 345694 c) 342342 d) 123654a is divisible by 9. but not divisible by 11.b is not divisible by 9c is divisible by both 9 and 11.Shortcut : a number of the form abcabc is always divisible by 7,11 and 13.342 is also divisible by 9.

Test of DivisibilityThe difference between the squares of two consecutive odd integers is always divisible by a) 3 b) 6 c) 7 d) 8Let the numbers be 2k – 1 and 2k + 1.(2k + 1)2 – (2k -1)2 = 4*2k = 8k.It is always divisible by 8

Test of divisibilityIf a number 774958A96B is to be divisible by 8 and 9, the values of A and B respectively will bea) 8,0 b) 2,8 c) 6,8 d) None of theseSum of the digits = 1 + A + B = 9k96B should be 8nB should be either 0 or 8. If B is 0 then A is 8If B is 8 then A is 0.Only correct choice is 8,0

Test of DivisibilityIf the number 109236345978x is divisible by 13, what is the value of x?a)1 b) 2 c) 5 d) 6Divide the number into groups of 3 digits from right and find the difference between sum of alternate digits.001, 092, 363, 459, 78x001 + 363 + 78x = 092 + 459364 + 78x = 55178x + 364 - 551 = 13k78x – 187 = 13k78x – 182 – 5 = 13kHence x should be equal to 5

Total number of digitsFind the total number of digits in the product 41111 * 52222.a)2520 b) 1600 c) 1642 d) 222322222 *52222 = 102222 . Hence it has 2223 digits.Alternative: the usual procedure to find the number of digits.Taking log,(1111)log 4 + (2222)log5= (1111 * 0.602) + (2222 * 0.699)= 2222Hence 2222 + 1 = 2223 digits.