Numbers standard level
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The document presents a series of mathematical problems and their solutions, ranging from basic arithmetic to more complex number theory concepts. It includes questions about digit occurrences in specific ranges, properties of odd and even numbers, divisibility rules, and factorial calculations. The document also provides various methods for solving these problems, illustrating a diverse range of mathematical principles.
![Remainder
What is the remainder when 50! is divided by 168?
a) 1 b) 3 c) 5 d) None of these
168 = 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] = 1
Highest power of 2 which divides 50! is 47.
232 divides 241 without remainder. Hence the remainder is zero.](https://image.slidesharecdn.com/numbersstandardlevel-170127012030/85/Numbers-standard-level-41-320.jpg)
![Remainder
Which 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](https://image.slidesharecdn.com/numbersstandardlevel-170127012030/85/Numbers-standard-level-45-320.jpg)
![Remainder
What is the remainder when 9113 * 7110 is divided by 31?
a) 1 b) 5 c) 12 d) 16
9113 * 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.](https://image.slidesharecdn.com/numbersstandardlevel-170127012030/85/Numbers-standard-level-46-320.jpg)
Numbers standard level
- 2. Base A 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) 4 Let the number be 10x + y 10y + x = 4.5(10x + y) 10y + x = 45x + 4.5y 5.5y = 44x 5.5 44 = 𝑥 𝑦 55 440 = 1 8 = 𝑥 𝑦
- 3. COUNTING If 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) 12 14,24,34,….94 = 9 40,41,42,43,44, … 49 = 10 Total = 19
- 4. Counting A 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 finger c) Thumb d) None of these After 9, it is a multiple of 8 to end on thumb. Hence 17,25,… 8n+1 will end on thumb. 1985 = 8n + 1.
- 5. Counting How many times does the digit 6 appear from 11 to 100? a) 16 b) 17 c) 18 d) 19 16,26,36,…96 = 9 60,61,62,…69 = 10 Total = 19
- 6. Counting How many numbers are there between 7000 and 8000 containing at least one 9? a) 300 b) 295 c) 288 d) 271 7000 to 7009 contains one 9 = 1 7000 to 7099 contains 20 9’s = 20 7000 to 7899 contains 9 * 20 = = 180 7900 to 7999 contains 20 + 100 = 120 = 120 total = 300
- 7. Counting How many odd numbers are there between 7000 and 10000 containing atleast one 8? a) 910 b) 813 c) 690 d) None of these From 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 = 0 When 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 = 50 Here 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
- 8. COUNTING If 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) 121 The hundredth place between 300 to 399 = 100 The ten’s place between 330 to 339 = 10 The unit’s place 303,313,323,… 393 = 10 Total = 120
- 9. COUNT A 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) 20 As 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, kmmm Case 3m + k = 9 Possible cases are 3330,2223,3222, 1116,6111, 9000 Case 3m + k = 18 Possible cases are 6660, 5553, 3555, 4446, 6444, 9333, 3999 Case 3m + k = 27 Possible cases are 9990, 8883, 3888, 7776, 6777, 6669, 9666 Total possible cases = 6 + 7 + 7 = 20.
- 10. Divisibility The sum of four consecutive odd numbers is always divisible by a) 3 b) 4 c) 7 d) 2 Let the four consecutive odd numbers be 2n – 3 , 2n – 1, 2n + 1, 2n + 3. Sum of the numbers = 8n It is always divisible by 8 and hence by 4 and 2.
- 11. Divisibility 164 + 220 is divisible by a) 7 b) 11 c) 13 d) 17 164 + 220 = 216 + 220 = 216(1 + 24) = 216(1 + 16) = 17k.
- 12. 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) 18 An even number is always divisible by 2. Hence the number which is both divisible by 9 and 2 is divisible by 18.
- 13. Divisibility If 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) 26 p2 – 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)
- 14. Equation Instead of multiplying a number by 53, Suresh multiplied it by 35 and got the answer which was 1206 less than the expected answer. What is the number? a) 51 b) 48 c) 61 d) 67 53x = 35x + 1206 18x = 1206 x = 67
- 15. EQUATION The 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 * 25 The smaller number is 25.
- 16. Even & Odd If 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 + n 2m 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.
- 17. Even or Odd If x is an integer, which of the following CANNOT be an even integer? a) 2x + 2 b) x – 5 c) 2x + 3 d) 5x + 2 Whether x is even or odd, 2x is always even. 2x + 3 is CANNOT be an even integer.
- 18. Expression What is the value of the below expression? 552∗552+160000 −220800 552∗552 ∗552+64000000 a) 1 552 b) 552 1600 c) 1 952 d) 1 1000 552∗552+400∗400 −552 ∗ 400 552∗552 ∗552+64000000 = a2+b2−ab a3+b3 = 1 a+b = 1 552+400 = 1 952
- 19. Factorial What highest power of 8 will divide 26! exactly? a) 3 b) 5 c) 7 d) None of these 26 8 = 3 26 64 = 0 Answer is 3. Verification: the numbers 8,16,24 are the only three numbers less than 26! divisible by 8.
- 20. Factors There are two numbers, one of which is twice the other. Both had the same number of prime factors While 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) 50,100 d(n) denote the number of factors. d(40) = 4* 2 = 8 and d(80) = 5 * 2 = 10 d(20) = 3 * 2 = 6 and d(40) = 4 * 2 = 8 d(30) = 2 * 2 *2 = 8 and D(60) = 3 * 2 * 2 = 12 d(50) = 2 * 3 = 6 and d(100) = 3 * 3 = 9 Hence answer is 30 and 60.
- 21. Fractions One-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) 720 60% of the number is three-fifth of the number. 1 4 ∗ 1 3 ∗ 2 5 ∗ 𝑥 = 25. 3 5 ∗ 𝑥 = 3 5 ∗ 25 ∗ 4 1 ∗ 3 1 ∗ 5 2 = 450
- 22. FRACTION Which number when added to 5 4 gives the same result as when it is multiplied by 5 4 ? a) 9 b) 7 c) 5 d) 3 x + 5 4 = 5x 4 5 4 = 5x 4 − x 5 4 = x 4 X = 5.
- 23. HCF Find the HCF of 2014 – 1 and 2012 – 1. a) 543 b) 481 c) 427 d) 399 HCF of 14 and 12 is 2 Both (202)7 -1 and (202)6 – 1 are divisible by 202 – 1 = 400 – 1 = 399.
- 24. Highest Power in Factorial Find the highest power of 10 in 120! a) 22 b) 24 c) 27 d) 28 Half of the numbers in 120! is even. To find the highest power of 5 in 120! 120 5 = 24 120 25 = 4 Hence the highest power is 28.
- 25. Number System In a number system with base b, 12 * 25 = 333, the value of b is a) 9 b) 8 c) 7 d) 6 (b + 2)(2b +5) = 3b2 + 3b + 3 2b2 + 9b + 10 = 3b2 + 3b + 3 b2 – 6b – 7 = 0 b = 6± 36+28 2 = 7, -1 The base is 7.
- 26. Partition If X + Y = 6, then XY = ? a) 2 b) 4 c) 6 d) 9 The 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.
- 27. Percentage in Numbers If 50% of the number is added to 50, the result is number itself. The number is a) 50 b) 200 c) 150 d) 100 50% 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.
- 28. Powers & Indices What is the ten’s digit in 29999? a) 2 b) 4 c) 6 d) 8 Ten’s digit is the remainder when divided by 100. 29999 mod 100 210(999)* 29 mod 100 210 raised to even power has last two digits 76 and when raised to odd power has last two digits 24 24 * 29 mod 100 288 mod 100 = 88. Hence ten’s digit is 8
- 29. Powers & Indices What will be the last digit of the multiplication 2222 * 3333 * 4444 ? The unit digit of 2222 = 2220 * 222 = 6 * 4 = 24 4 The unit digit of 3333 = 3332 * 331 = 1 * 3 = 3 The unit digit of 4444 = 6 The unit digit in the multiplication= 4 * 3 * 6 = 72 2
- 30. Power and Indices Determine the digit in the unit position of 1121 * 1717 * 2121 . Unit digit of 1121 and 2121 = 1 Unit digit of 1717 is 1716 * 171 = 1 * 7 = 7 The unit digit in the product = 7.
- 31. Powers and indices What are the last two digits of 72008? a) 49 b) 43 c) 01 d) 07 72008 = 74(502) = (2401)502. The last two digits of (01)n is always 01.
- 32. Powers and indices What is the unit digit in the product 365 x 659 x 771? a) 1 b) 2 c) 4 d) 6 Unit digit in 365 = 364 * 3 = 1 x 3 = 3 Unit digit in 659 = 6 Unit digit in 771 = 768 * 73 = 1 * 3 = 3 Unit digit in the product = 4
- 33. Powers and Indices Find the last two digits in 259166 . a) 11 b) 41 c) 71 d) None of these The 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 9 The unit place in the power is 6 Product = 54. Hence the last two digits is 41.
- 34. Powers and Indices What 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 = 2 x = 2/3
- 35. Powers and Indices Find 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 6 The unit digit in the power is 5 Product is 30. Hence the last two digits is 01.
- 36. Progression ABCD 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 be a) 427 b) 426 c) 408 d) 407 It is in A.P with a = 1 , d = 6 and last term is 2557. N = 2557 −1 6 + 1 = 426 + 1 = 427.
- 37. Prime Numbers The highest prime number that can be stored in a 8-bit microprocessor is A) 247 B) 253 C) 317 D) 251 28 = 256. Numbers less than 256 are 247 and 251 247 is not prime. 247 divisible by 13 and 19 251 is the highest prime less than 256
- 38. Remainder What is the remainder when 2256 is divided by 17? a) 1 b) 2 c) 5 d) None of these 2256 = 24*64 = 1664 = (17 – 1)64. The remainder is (-1). That is 17 – 1 = 16.
- 39. Remainder On 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) 12 Let the number be N N = 209q + 50 = 19*11q + 19*2 + 12 Dividing by 19, the remainder is 12.
- 40. Remainder A 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) 0,2,4 b) 1,0,4 c) 1,1,2 d) 0,1,3 Let the final quotient be q when divided by 3 Divided by 5 remainder is 0 = 5x Divided by 2 remainder is 1 = x = 2k + 1 = 10k + 5 Note 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.
- 41. Remainder What is the remainder when 50! is divided by 168? a) 1 b) 3 c) 5 d) None of these 168 = 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] = 1 Highest power of 2 which divides 50! is 47. 232 divides 241 without remainder. Hence the remainder is zero.
- 42. Remainder What is the remainder when -4x3 + 8x2 + 12x + 16 is divided by x + 2? a) 8 b) 24 c) 32 d) 56 P(-2) = -4(-8) + 8(4) + 12(-2) + 16 = 32 + 32 – 24 + 16 = 56
- 43. Remainder What is the remainder when 5163 – 7593 + 11593 – 1163 is divided by 4? a)3 b) 2 c) 1 d) 0 5163 – 1163 is divisible by 4 11593 – 7593 = 400k is also divisible by 4 ( a3 – b3) = (a – b)k The remainder is zero.
- 44. Remainder What is the remainder when 5555 + 55 is divided by 56? a) 0 b) 1 c) 54 d) 55 5555 = (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.
- 45. Remainder Which 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
- 46. Remainder What is the remainder when 9113 * 7110 is divided by 31? a) 1 b) 5 c) 12 d) 16 9113 * 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.
- 47. Remainder The 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) 7 1 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 2 When divided by 9 leaves the remainder 1 + 2 = 3.
- 48. Remainder A 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) 9 Let the number be N N = 765k + 42 N = (17)(45)k + (17)(2) + 8 The remainder is 8
- 49. Remainder What is the remainder when 3126 is divided by 8? a) 1 b) 3 c) 5 d) 7 3126 = 32(61) 961 = (8 + 1)61 = 1(mod 8) Hence the remainder is 1.
- 50. Remainder Which of the following option does not divide 56 – 1 completely? (i.e., rem = 0) a) 18 b) 24 c) 27 d) 31 56 – 1 = (53)2 – 1 = (53 – 1)(53 + 1) = 124 * 126 = 2*2*31 * 2 * 3 * 3 * 7 It is not completely divisible by 27.
- 51. Remainder What is the remainder when 482 is divided by 6? a) 1 b) 2 c) 3 d) 4 482 = (6 – 2)82. Except the last term all the terms are divisible by 6. Hence the remainder is 2 ( power is even. Hence positive)
- 52. Square • Which of the following cannot be the square of a natural number? a) 32761 b) 81225 c) 42437 d) 20164 Clearly no perfect square will end in 7. Hence 42437 is not a perfect square.
- 53. Squares & Cubes How many positive numbers less than 50000 exist which are both perfect squares and perfect cubes? a) 12 b) 10 c) 8 d) 6 Numbers which are both perfect square and perfect cubes should be of the form x6. 106 = 1000000 66 = 26 * 36 = 64 * 729 ≈ 43000. Hence Answer is 6. 86 = 218 = 210 * 28 = 1024 * 256
- 54. Squares and Cubes How 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
- 55. Sum of the series Find the sum upto 20 terms in the series 1 + (1 + 3) + ( 1 + 3 + 5) + (1 + 3 + 5 + 7) +… a) 2870 b) 3021 c) 2920 d) 3186 Since 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 = n(n+1)(2n+1) 6 = 20 ∗21 ∗41 6 = 70 * 41 = 2870
- 56. 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
- 57. Test of divisibility Which one of the following is divisible by 99? a)913464 b) 345694 c) 342342 d) 123654 a is divisible by 9. but not divisible by 11. b is not divisible by 9 c 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.
- 58. Test of Divisibility The difference between the squares of two consecutive odd integers is always divisible by a) 3 b) 6 c) 7 d) 8 Let the numbers be 2k – 1 and 2k + 1. (2k + 1)2 – (2k -1)2 = 4*2k = 8k. It is always divisible by 8
- 59. Test of divisibility If a number 774958A96B is to be divisible by 8 and 9, the values of A and B respectively will be a) 8,0 b) 2,8 c) 6,8 d) None of these Sum of the digits = 1 + A + B = 9k 96B should be 8n B should be either 0 or 8. If B is 0 then A is 8 If B is 8 then A is 0. Only correct choice is 8,0
- 60. Test of Divisibility If the number 109236345978x is divisible by 13, what is the value of x? a)1 b) 2 c) 5 d) 6 Divide the number into groups of 3 digits from right and find the difference between sum of alternate digits. 001, 092, 363, 459, 78x 001 + 363 + 78x = 092 + 459 364 + 78x = 551 78x + 364 - 551 = 13k 78x – 187 = 13k 78x – 182 – 5 = 13k Hence x should be equal to 5
- 61. Total number of digits Find the total number of digits in the product 41111 * 52222. a)2520 b) 1600 c) 1642 d) 2223 22222 *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) = 2222 Hence 2222 + 1 = 2223 digits.