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481. The difference of a number consisting of two digits and the number formed by interchanging the digits is always divisible by
- A. 5
- B. 7
- C. 9
- D. 11
Answer: 9
Explanation: Let the ten's digit be x and the unit's digit be y. Then, (10x + y) – (10y + x) = 9x – 9y = 9(x–y), which is divisible by 9.
482. The number 6n2 + 6n for natural number n is always divisible by
- A. 6 only
- B. 6 and 12
- C. 12 only
- D. 18 only
Answer: 6 and 12
Explanation: 6n2 + 6n = 6n (n + 1) (n = 1 ⇒ 6n2 + 6n = 12), (n = 2 ⇒ 6n2 + 6n = 36), (n = 3 ⇒ 6n2 + 6n = 72), each one of which is divisible by 6 and 12 both.
483. The sum of two numbers is 7 and their product is 12. What is the sum of their reciprocals?
- A. 712
- B. 815
- C. 713
- D. 127
Answer: 712
Explanation: Let the two number be a and b According to question, a + b = 7 a×b = 12 Sum of reciprocals 1 1a b= + a b 7ab 12+= =
484. If a and b are two odd positive integers, by which of the following integers is (a4 – b4) always divisible?
- A. 3
- B. 6
- C. 8
- D. 12
Answer: 8
Explanation: Let a = 2k + 1 & b = 2m + 1. Then, a4 – b4 = (a2 – b2) (a2 + b2) = (a + b) (a – b) (a2 + b2) = (2k + 2m + 2) (2k – 2m) (4k2 + 4m2 + 2 + 4k + 4m) = 8(k + m + 1) (k – m) (2k2 + 2m2 + 1 + 2k + 2m), which is divisible by 8.
485. If n be any natural number then by which largest number (n3 – n) is always divisible?
- A. 3
- B. 6
- C. 12
- D. 18
Answer: 6
Explanation: (13 – 1) = 0, (23 – 2) = 6, (33 – 3) = 24, (43 – 4) = 60 and so on, each one of which is divisible by 6.
486. If x and y are positive integers such that (3x + 7y) is a multiple of 11, then which of the following is also a multiple of 11?
- A. 5x – 3y
- B. 9x + 4y
- C. 4x + 6y
- D. x + y + 6
Answer: 5x – 3y
Explanation: Let 3x + 7y = 11k. Then, y = (11 – 3 ) 7 k x . Then,5x – 3y = 5x – 3 (11 – 3 ) 35 – 33 9 7 7 kx x kx + = = = 44 – 33 11 (4 – 3 ), which is divisible by 11. 77 x k xk
487. How many of the following numbers are divisible by 132? 264, 396, 462, 792, 968, 2178, 5184, 6336
- A. 4
- B. 5
- C. 6
- D. 7
Answer: 4
Explanation: 132 = 11 × 3 × 4. So, the required number must be divisible by 3, 4 and 11. 264 → divisible by 3, 4 and 11. 396 → divisible by 3, 4 and 11. 462 → not divisible by 4. 792 → divisible by 3, 4 and 11. 968 → not divisible by 3 2178 → not divisible by 4 5184 → not divisible by 11. 6336 → divisible by 3, 4 and 11. Hence, out of the given numbers 4 are divisible by 3, 4 and 11.
488. If a number 774958A96B is divisible by 8 and 9, the respective values of A and B will be
- A. 5 and 8
- B. 7 and 8
- C. 8 and 0
- D. None of these
Answer: 8 and 0
Explanation: Since 774958A96B is divisible by 8, so the number 96B must be divisible by 8. So, B = 0, as 960 is divisible by 8. Now, 774958A960 is divisible by 9. So, (55 + A) must be divisible by 9. This happens when A = 8. Hence, A = 8 and B = 0.
489. If the seven-digit number 876p37q is divisible by 225, then the values of p and q respectively are
- A. 0 and 0
- B. 9 and 0
- C. 0 and 5
- D. 9 and 5
Answer: 0 and 5
Explanation: 225 = 9 × 25, where 9 and 25 are co-primes. So, a number is divisible by 225 if it is divisible by both 9 and 25. Given number is divisible by 25, only if 7q is divisible by 25, i.e. if q = 5. Sum of digits of given number = (8 + 7+ 6 + p + 3 + 7 + 5) = 36 + p, which must be divisible by 9. This is possible if p = 0. Hence, p = 0, q = 5.
490. If 37X3 is a four-digit natural number divisible by 7, then the place marked as X must have the value
- A. 0
- B. 3
- C. 5
- D. 9
Answer: 0
Explanation: 37 × 3 is divisible by 7 ⇒ (7 × 3 – 3) is either 0 or divisible by 7) ⇒ 7 × 0 is divisible by 7. ⇒ X = 0 or X = 7.
491. The digits indicated by * in 3422213** so that this number is divisible by 99 are
- A. 1, 9
- B. 3, 7
- C. 4, 6
- D. 5, 5
Answer: 1, 9
Explanation: Let the unit's digit be x and ten's digit be y. Then, the number is 3422213yx. Also, 99 = (11 × 9), where 11 and 9 are co-primes. Since the given number is divisible by 9, it follows that (3 + 4 + 2 + 2 + 2 + 1 + 3 + y + x) = (17 + y + x) must be divisible by 9. So, y + x = 1 or y + x = 10. Again, the given number is divisible by 11. So, (x + 3 + 2 + 2 + 3) – (y + 1 + 2 + 4) = x – y + 3 is either 0 or 11. ∴ (x – y + 3 = 0 or x – y + 3 = 11) ⇒ (y – x = 3 or x – y = 8) Now, (y + x = 1 and y – x = 3) ⇒ y = 2 and x = – 1. (y + x = 1 and x – y = 8) ⇒ x = 9 2 . (y + x = 10 and y – x = 3) ⇒ y = 13 2 . (y + x = 10 and x – y = 8) ⇒ x = 9 and y = 1. Thus, x = 9, y = 1. So, the required number is 342221319.
492. Which of the following numbers is divisible by 99?
- A. 114345
- B. 913464
- C. 135792
- D. 3572404
Answer: 114345
Explanation: We have, 99 = (11 × 9), where 11 and 9 are co-primes. Consider the number 114345. Clearly, (5 + 3 + 1) – (4 + 4 + 1) = 0. So, 114345 is divisible by 11. Also, sum of its digits = (1 + 1 + 4 + 3 + 4 + 5) = 18, which is divisible by 9. ∴ 114345 is divisible by 9. Hence, it is divisible by (11 × 9), i.e. 99.
493. The six-digit number 5ABB7A is a multiple of 33 for non-zero digits A and B. Which of the following could be possible value of A + B?
- A. 8
- B. 9
- C. 10
- D. 14
Answer: 9
Explanation: We know that 33 = 11 × 3, where 11 and 3 are co-primes. So, the given number must be divisible by both 11 and 3. Since 5ABB7A is divisible by 11, we have (A + B + A) – (7 + B + 5) = (2A – 12) is either 0 or 11. ⇒ 2A – 12 = 0 or 2A – 12 = 11 ⇒ A = 6 23 2 A [] ≠ ∵ So, the number becomes 56BB76, which is divisible by 3. ∴ (5 + 6 + B + B + 7 + 6) = (24 + 2B) must be divisible by 3. ∴ 2B = 6 ⇒ B = 3 3 0 and 2 BB [] ≠≠ ∵ Hence, (A + B) = (6 + 3) = 9.
494. If x and y are two digits of the number 653xy such that the number is divisible by 80, then x + y is equal to
- A. 3
- B. 4
- C. 5
- D. 6
Answer: 6
Explanation: Since 653xy is divisible by 80, we must have y = 0. Now, 653x0 must be divisible by both 5 and 16. Clearly, it is divisible by 5 for all values of x. Now, the number 53x0 must be divisible by 16. The least value of x is clearly 6. So, x + y = 6 + 0 = 6.
495. Which of the following numbers are completely divisible by 7? I. 195195 II. 181181 III. 120120 IV. 891891
- A. Only I and II
- B. Only II and III
- C. Only I and IV
- D. All are divisible
Answer: All are divisible
Explanation: I. We have(195 – 195) 0 ∴ 195195 is divisible by 7. II. We have (181 – 181) = 0 ∴ 181181 is divisible by 7. III. We have (120 – 120) = 0 ∴ 120120 is divisible by 7. IV. We have (891 – 891) = 0 ∴ 891891 is divisible by 7. Hence, all are divisible by 7.
496. If a number is divisible by both 11 and 13, then it must be necessarily
- A. 429
- B. divisible by (11 × 13)
- C. divisible by (11 + 13)
- D. divisible by (13 – 11)
Answer: divisible by (11 × 13)
Explanation: We know that 11 and 13 are co-prime. So, a number divisible by both 11 and 13 will be divisible by (11 × 13).
497. If the seven-figure number 30X0103 is a multiple of 13, then X is
- A. 1
- B. 6
- C. 7
- D. 8
Answer: 8
Explanation: We first divide the number into groups of 3 digits from the right → 3 0X0 103 Difference of sum of numbers at odd and even places = (103 + 3) – 0X0 = 106 – 0X0, which must be divisible by 13. 106 – 0X0 is divisible by 13 only for X = 8.
498. 325325 is a six-digit number. It is divisible by
- A. 7 only
- B. 11 only
- C. 13 only
- D. all 7, 11 and 13
Answer: all 7, 11 and 13
Explanation: (325 – 325) = 0, which is divisible by 7. So, the given number is divisible by 7. (5 + 3 + 2) – (2 + 5 + 3) = 0. So, the given number is divisible by 11. And, 325325 25025. 13 = So, 325325 is divisible by 13. Hence, it is divisible by all 7, 11 and 13.
499. The number 311311311311311311311 is
- A. neither divisible by 3 nor by 11
- B. divisible by 11 but not by 3
- C. divisible by 3 but not by 11
- D. divisible by both 3 and 11
Answer: neither divisible by 3 nor by 11
Explanation: Sum of the digits of the given number = (7 × 3) + (14 × 1) = (21 + 14) = 35, which is not divisible by 3. So the given number is not divisible by 3. Also, (1 + 3 + 1 + 1 + 3 + 1 + 1 + 3 + 1 + 1 + 3) – (1 + 1 + 3 + 1 + 1 + 3 + 1 + 1 + 3 + 1) = (19 – 16) = 3, which is neither 0 nor divisible by 11. So, the given number is not divisible by 11 also. Hence, it is divisible by neither 3 nor 11.
500. Which of the following numbers is exactly divisible by 24?
- A. 35718
- B. 63810
- C. 537804
- D. 3125736
Answer: 3125736
Explanation: We have 24 = 3 × 8, where 3 and 8 are co-primes. Clearly, 718 is not divisible by 8. So, 35718 is not divisible by 8. 810 is not divisible by 8. So, 63810 is not divisible by 8. 804 is not divisible by 8. So, 537804 is not divisible by 8. 736 is divisible by 8. So, 3125736 is divisible by 8. Also, sum of its digits = (3 + 1 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 3. So, 3125736 is divisible by 3 also. Hence, it is divisible by 24.
501. 6897 is divisible by
- A. 11 only
- B. 19 only
- C. both 11 and 19
- D. neither 11 nor 19
Answer: both 11 and 19
Explanation: Clearly, (7 + 8) – (9 + 6) = 0. So, 6897 is divisible by 11. Also, 6897 363 19 = . So, 6897 is divisible by 19. Hence, 6897 is divisible by both 11 and 19.
502. If the number 357*25* is divisible by both 3 and 5, then the missing digits in the unit's place and the thousandth's place respectively are
- A. 0, 6
- B. 5, 1
- C. 5, 4
- D. None of these
Answer: None of these
Explanation: Let the unit's place be x and the thousand's place be y. Then, 357y25x is divisible by 5 only when x = 0 or x = 5. Also, this number is divisible by 3 only when sum of its digits is divisible by 3. So, (22 + x + y) must be divisible by 3. ∴ x + y = 2 Taking x = 0, we get y = 2. So, the unit place = 0 and thousand's place = 2.
503. How many of the following numbers are divisible by 3 but not by 9? 2133, 2343, 3474, 4131, 5286, 5340, 6336, 7347, 8115, 9276
- A. 5
- B. 6
- C. 7
- D. None of these
Answer: 6
Explanation: Sum of the digits in respective numbers is: 9, 12, 18, 9, 21, 12, 18, 21, 15 and 24. Out of these 12, 21, 12, 21, 15, 24 are divisible by 3 but not by 9. So, the number of required numbers is 6.
504. A number 476**0 is divisible by both 3 and 11. The non-zero digits in the hundred's and ten's place respectively are
- A. 7, 4
- B. 5, 3
- C. 5, 2
- D. None of these
Answer: 5, 2
505. Which of the following numbers is divisible by 3, 7, 9 and 11?
- A. 639
- B. 2079
- C. 3791
- D. 37911
Answer: 2079
Explanation: Clearly, 639 is not divisible by 7. Consider 2079. Sum of its digits = (2 + 0 + 7 + 9) = 18. So, it is divisible by both 3 and 9. Also, (79 – 2) = 77, which is divisible by 7. So, 2079 is divisible by 7. Also, (9 + 0) – (7 + 2) = 0. So, 2079 is divisible by 11. Hence, 2079 is divisible by each one of 3, 7, 9 and 11.
506. An integer is divisible by 16 if and only if its last X digits are divisible by 16. The value of X would be
- A. 3
- B. 4
- C. 5
- D. 6
Answer: 4
Explanation: An integer is divisible by 16, if the number formed by last 4 digits is divisible by 16.
507. If m and n are integers divisible by 5, which of the following is not necessarily true?
- A. m + n is divisible by 10
- B. m – n is divisible by 5
- C. m2 – n2 is divisible by 25
- D. None of these
Answer: m + n is divisible by 10
Explanation: Take m = 15 and n = 20. Then, each one of m and n is divisible by 5. But, (m + n) is not divisible by 10. Hence, (m + n) is divisible by 10 is not true.
508. If 78*3945 is divisible by 11, where * is a digit, then * is equal to
- A. 0
- B. 1
- C. 3
- D. 5
Answer: 5
Explanation: Here (5 + 9 + x + 7) – (4 + 3 + 8) = 6 + x. So, we must have x = 5.
509. Which of the following numbers is a multiple of 8?
- A. 923872
- B. 923972
- C. 923862
- D. 923962
Answer: 923872
Explanation: We know that 872 is divisible by 8. Hence 923872 is divisible by 8.
510. Consider the following statements: The numbers 24984, 26784 and 28584 are (1) divisible by 3 (2) divisible by 4 (3) divisible by 9 Which of these are correct?
- A. 1 and 2
- B. 2 and 3
- C. 1 and 3
- D. 1, 2 and 3
Answer: 1, 2 and 3
Explanation: (2 + 4 + 9 + 8 + 4) = 27, (2 + 6 + 7 + 8 + 4) = 27 and (2 + 8 + 5 + 8 + 4) = 27. So, each one is divisible by 3 and 9 both. Also, 84 is divisible by 4. So, each one is divisible by 4. Hence, each given number is divisible by 3, 9 and 4. So, the statements 1, 2 and 3 are all correct.
511. 7386038 is divisible by
- A. 3
- B. 4
- C. 9
- D. 11
Answer: 11
Explanation: Given number is 7386038. Sum of its digits = 35, which is not divisible by any of 3 and 9. So, the given number is not divisible by any of 3 and 9. Also, 38 is not divisible by 4. So, the given number is not divisible by 4. Also, (8 + 0 + 8 + 7) – (3 + 6 + 3) = (23 – 12) = 11. So, the given number is divisible by 11.
512. Which one of the following numbers is divisible by 15?
- A. 17325
- B. 23755
- C. 29515
- D. 30560
Answer: 17325
Explanation: Consider the number 17325. Its unit digit is 5. So, it is divisible by 5. Sum of its digits = (5 + 2 + 3 + 7 + 1) = 18, which is divisible by 3. So, the given number is divisible by 3. And, since 5 and 3 are co-primes, so the given number is divisible by (5 × 3), i.e. 15.
513. Which one of the following numbers is divisible by 11?
- A. 4823718
- B. 4832718
- C. 8423718
- D. 8432718
Answer: 4832718
Explanation: For 4823718 we have (8 + 7 + 2 + 4) – (1 + 3 + 8) = (21 – 12) = 9, which is not a multiple of 11. ∴ 4823718 is not divisible by 11. Consider the number 4832718. We have (8 + 7 + 3 + 4) – (1 + 2 + 8) = (22 – 11) = 11, which is a multiple of 11. Hence, 4832718 is divisible by 11.
514. A number is divisible by 11 if the difference between the sums of the digits in odd and even places respectively is
- A. a multiple of 3
- B. a multiple of 5
- C. zero or a multiple of 7
- D. zero or a multiple of 11
Answer: zero or a multiple of 11
Explanation: Clearly, (d) is true.
515. Which one of the following numbers is divisible by 3?
- A. 4006020
- B. 2345678
- C. 2876423
- D. 9566003
Answer: 4006020
Explanation: Sum of the digits in 4006020 is 12, which is divisible by 3. Hence, 4006020 is divisible by 3.
516. The number 89715938* is divisible by 4. The unknown non-zero digit marked as * will be
- A. 2
- B. 3
- C. 4
- D. 6
Answer: 4
Explanation: Let the missing digit be x. Then, (80 + x) must be divisible by 4. Hence, x = 4.
517. Which of the following numbers is not divisible by 18?
- A. 34056
- B. 50436
- C. 54036
- D. 65043
Answer: 65043
Explanation: We have 18 = 2 × 9, where 2 and 9 are co-primes. But, 65043 is not divisible by 2. So, it is not divisible by 18.
518. 111,111,111,111 is divisible by
- A. 3 and 37 only
- B. 3, 11 and 37 only
- C. 3, 11, 37 and 111 only
- D. 3, 11, 37, 111 and 1001
Answer: 3, 11, 37, 111 and 1001
Explanation: Sum of all digits = 12, which is divisible by 3. So, the given number is divisible by 3. (Sum of digits at odd places) – (Sum of digits at even places) = 6 – 6 = 0. So, the given number is divisible by 11. The given number when divided by 37 gives 3003003003. So, the given number is divisible by 37. The given number when divided by 111 gives 1001001001. Clearly, it is divisible by 111 as well as by 1001. Hence, the given number is divisible by each one of 3, 11, 37, 111 and 1001.
519. Find the multiple of 11 in the following numbers.
- A. 112144
- B. 447355
- C. 869756
- D. 978626
Answer: 978626
Explanation: In 978626, we have (6 + 6 + 7) – (2 + 8 + 9) = 0. Hence, 978626 is completely divisible by 11.
520. What least value must be given to n so that the number 6135n2 becomes divisible by 9?
- A. 1
- B. 2
- C. 3
- D. 4
Answer: 1
Explanation: 6135n2 is divisible by 9 if (6 + 1 + 3 + 5 + n + 2) = (17 + n) is divisible by 9. This happens when the least value of n is 1.
521. Let a number of three digits have for its middle digit the sum of the other two digits. Then it is a multiple of
- A. 10
- B. 11
- C. 18
- D. 50
Answer: 11
Explanation: We know that a number having x,y,z as its digits, is a multiple of 11, if z + x – y = 0 Hence, y = z + x.
522. Let x be the product of two numbers 3,659,893,456,789,325,678 and 342,973,489,379,256. The number of digits in x is
- A. 32
- B. 34
- C. 35
- D. 36
Answer: 34
Explanation: Sum of digits in the two numbers = 19 + 15 = 34. So, the product will have 33 or 34 digits. Since 36 × 34 = 1224 (i.e. product has 2 + 2 = 4 digits), so the number of digits in x is 34.
523. The number formed from the last two digits (ones and tens) of the expression 212n – 64n, where n is any positive integer is
- A. 10
- B. 00
- C. 30
- D. 02
Answer: 00
Explanation: We have: (212n – 64n) = (212n – 24n × 34n) = 24n (28n – 34n). Putting n = 1, we get the number 24 (28 – 34) = 16 (256 – 81) = (16 × 175) = 2800 Hence, the number formed by last two digits is 00.
524. If m and n are positive integers, then the digit in the unit's place of 5n + 6m is always
- A. 1
- B. 5
- C. 6
- D. n + m
Answer: 1
Explanation: In 5n we have 5 as unit digit and in 6m we have 6 as unit digit. ∴ Unit digit in (5n + 6m) = Unit digit in (5 + 6) = Unit digit in 11 = 1.
525. If x is an even number, then x4n, where n is a positive integer, will always have
- A. zero in the unit's place
- B. 6 in the unit's place
- C. either 0 or 6 in the unit's place
- D. None of these
Answer: either 0 or 6 in the unit's place
Explanation: Let x = 2y. Then, x4n = (2y)4n = {(2y)4}n = (16 y4)n. y = 1, 2, 3, 4, 5, 6, 7, 8, 9 gives unit digit as 6 in (16y4) n. But, y = 5, gives unit digit 0 in (16 y4) n. Hence, the unit digit is 0 or 6.
526. The digit in the unit's place of the product (2464)1793 × (615)317 × (131)491 is
- A. 0
- B. 2
- C. 3
- D. 5
Answer: 0
Explanation: Unit digit in the given product = Unit digit in (41793 × 5317 × 1491) Unit digit in 42 is 6 and so the unit digit in (42)896 is 6. ∴ Unit digit of 41793 = Unit digit in (6 × 4) = Unit digit in 24, which is 4. Unit digit in 5317 is 5 and the unit digit in 1491 is 1. ∴ Unit digit in the given product = Unit digit in (4 × 5 × 1), which is 0.
527. The digit in the unit's place of [(251)98 + (21)29 – (106)100 + (705)35 – 164 + 259] is
- A. 1
- B. 4
- C. 5
- D. 6
Answer: 4
Explanation: Unit digit in [(251)98 + (21)29 – (106)100 + (705)35 – (16)4 + 259] = Unit digit in (1 + 1 – 6 + 5 – 6 + 9) = 4.
528. Unit's digit in (784)126 + (784)127 is
- A. 0
- B. 4
- C. 6
- D. 8
Answer: 0
Explanation: Unit digit of 42 is 6. So, unit digit in (42)63 is 6. ∴ Unit digit of (784)126 = Unit digit in 4126, which is 6. Unit digit in 4127 = Unit digit in (4126 × 4) = Unit digit in (6 × 4), which is 4. ∴ Unit digit in (784)127 is 4. Hence, unit digit of {(784)126 + (784)127} = Unit digit of (6 + 4) = Unit digit of 10, which is 0.
529. The digit in the unit place of the number represented by (795 – 358) is
- A. 0
- B. 4
- C. 6
- D. 7
Answer: 4
Explanation: Unit digit of 74 is 1. So, the unit digit of (74)23 is 1. ∴ Unit digit of 795 = Unit digit of (792 × 73) = (1 × 3) = 3. Unit digit of 34 is 1. So, the unit digit of (34) 14 is 1. ∴ Unit digit of 358 = Unit digit of (356 × 32) = (1 × 9) = 9. Hence, the unit digit of (795 – 358) = (13 – 9) = 4.
530. In the product 459 × 46 × 28* × 484, the digit in the unit place is 8. The digit to come in place of * is
- A. 3
- B. 5
- C. 7
- D. None of these
Answer: 3
Explanation: Unit digit in the given product = 8. Unit digit of (9 × 6 × x × 4) is 8. So, x = 3.
531. The unit's digit in the product 274 × 318 × 577 × 313 is
- A. 2
- B. 3
- C. 4
- D. 5
Answer: 2
Explanation: Unit digit in the given product = Unit digit of (4 × 8 × 7 × 3), which is 2.
532. The digit in the unit's place of the number (67)25 – 1 must be
- A. 0
- B. 6
- C. 8
- D. None of these
Answer: 6
Explanation: Unit digit of (67)25 = Unit digit of 725. Unit digit of 74 is 1 and so the unit digit of (74)6 is 1. ∴ Unit digit of 725 = (1 × 7) = 7. Hence, the unit digit of (725 – 1) is (7 – 1) = 6.
533. Match List I with List II and select the correct answer: List I List II (Product) (Digit in the unit's place) A. (1827)16 (1) 1 B. (2153)19 (2) 3 C. (5129)21 (3) 5 (4) 7 (5) 9 A B C A B C
- A. 1 4 3
- B. 4 2 3 A B C A B C
- C. 1 4 5
- D. 4 2 5
Answer: 1 4 5
Explanation: (A) 74 gives unit digit 1. So, 716 = (74) 4 gives unit digit 1. ∴ (1827)16 gives unit digit 1. So, A → (1). (B) 34 gives unit digit 1. So, (34) 4 gives unit digit 1. ∴ 319 = 316 × 33 gives unit digit = (1 × 7) = 7. ∴ (2153)19 gives unit digit 7. So, B → (4). (C) 92 gives unit digit 1. So, (92)10 gives unit digit 1. ∴ 921 = (920 × 9) gives unit digit = (1 × 9) = 9. ∴ (5129)21 gives unit digit = 9. So, C → (5). Hence, 14 5 ABC is the correct result.
534. The digit in the unit's place of the number 12399 is
- A. 1
- B. 4
- C. 7
- D. 8
Answer: 7
Explanation: 34 gives unit digit 1. So, (34)24 gives unit digit 1. And, 33 gives unit digit 7. ∴ 399 = (34) 24 × 33 gives unit digit (1 × 7) i.e. 7.
535. The unit's digit of 132003 is
- A. 1
- B. 3
- C. 7
- D. 9
Answer: 7
Explanation: 34 gives unit digit 1. So, (34)500 gives unit digit 1. And, 33 gives unit digit 7. ∴ (13)2003 gives unit digit = (1 × 7) = 7.
536. Find the number of zeros at the end of the result 3 × 6 × 9 × 12 × 15 × ......... × 99 × 102.
- A. 4
- B. 6
- C. 7
- D. 10
Answer: 7
537. Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all the elements of S. With how many consecutive zeros will the product end?
- A. 1
- B. 4
- C. 5
- D. 10
Answer: 1
538. The numbers 2, 4, 6, 8, ............, 98, 100 are multiplied together. The number of zeros at the end of the product must be
- A. 10
- B. 11
- C. 12
- D. 13
Answer: 12
539. The numbers 1, 3, 5, 7, ........., 99 and 128 are multiplied together. The number of zeros at the end of the product must be
- A. Nil
- B. 7
- C. 19
- D. 22
Answer: 7
Explanation: Let N = (1 × 3 × 5 × 7 × .... × 99) × 128. Clearly, N contains 10 multiples of 5 (5, 15, 25, 35, ....., 95) and only one multiple of 2 i.e. 128 or 27. Clearly, highest power of 5 in N is greater than that of 2. ∴ Number of zeros in N = Highest power of 2 in N = 7.
540. The number of zeros at the end of 60! is
- A. 12
- B. 14
- C. 16
- D. 18
Answer: 14
541. The number of zeros at the end of the product 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 is
- A. 5
- B. 7
- C. 8
- D. 10
Answer: 8
542. First 100 multiples of 10 i.e. 10, 20, 30, ............, 1000 are multiplied together. The number of zeros at the end of the product will be
- A. 100
- B. 111
- C. 124
- D. 125
Answer: 124
543. The numbers 1, 2, 3, 4, .........., 1000 are multiplied together. The number of zeros at the end (on the right) of the product must be
- A. 30
- B. 200
- C. 211
- D. 249
Answer: 249
544. The numbers 1, 3, 5, ........., 25 are multiplied together. The number of zeros at the right end of the product is
- A. 0
- B. 1
- C. 2
- D. 3
Answer: 0
Explanation: Clearly, multiples of 2 and 5 together yield 0. Since the product of odd numbers contains no power of 2, so the given product does not give 0 at the unit place.
545. A boy multiplies 987 by a certain number and obtains 559981 as his answer. If in the answer both 9's are wrong but the other digits are correct, then the correct answer will be
- A. 553681
- B. 555181
- C. 555681
- D. 556581
Answer: 555681
Explanation: Clearly, 7261 must be replaced by 2961, which is possibvle if 6648 is replaced by 6218, which in turn is possible if 5599 is replaced by 5556. Thus, the correct number is 555681.
546. The number of digits in the smallest number, which when multiplied by 7 yields all nines, is
- A. 3
- B. 4
- C. 5
- D. 6
Answer: 6
547. On multiplying a number by 7, all the digits in the product appear as 3's. The smallest such number is
- A. 47619
- B. 46719
- C. 48619
- D. 47649
Answer: 47619
Explanation: We keep on dividing 33333... by 7 till we get 0 as remainder. ∴ Required number = 47619
548. If m and n are natural numbers such that 2m – 2n = 960, what is the value of m?
- A. 10
- B. 12
- C. 15
- D. Cannot be determined
Answer: 10
Explanation: 2m > 960 when the least value of m is 10. Then, 210 = 1024 and 1024 – 960 = 64 = 26. ∴ m = 10 and n = 6.
549. The value of 52 + 62 + .... + 102 + 202 is
- A. 755
- B. 760
- C. 765
- D. 770
Answer: 755
550. Given n = 1 + x and x is the product of four consecutive integers. Then which of the following is true? I. n is an odd integer. II. n is prime. III. n is a perfect square.
- A. Only I is correct
- B. Only III is correct
- C. Both I and II are correct
- D. Both I and III are correct
Answer: Both I and III are correct
Explanation: Out of four consecutive integers two are even and therefore, their product is even and on adding 1 to it, we get an odd integer. So, n is odd. Some possible values of n are as under: n = 1 + (1 × 2 × 3 × 4) = (1 + 24) = 25 = 52, n = 1 + (2 × 3 × 4 × 5) = (1 + 120) = 121 = (11)2, n = 1 + (3 × 4 × 5 × 6) = (1 + 360) = 361 = (19)2, n = 1 + (4 × 5 × 6 × 7) = 841 = (29)2 and so on. Hence, n is odd and a perfect square.
551. A young girl counted in the following way on the fingers of her left hand. She started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8, thumb 9 and then back to the index figure for 10, middle finger for 11, and so on. She counted upto 1994. She ended on her
- A. thumb
- B. index finger
- C. middle finger
- D. ring finger
Answer: index finger
Explanation: Number on thumbs = 1, 9, 17, 25, ............. This is an AP in which a = 1 and d = (9 – 1) = 8. ∴ Tn = a + (n – 1)d = 1 + 8(n – 1) = (8n – 7). 8n – 7 = 1994 ⇒ 8n = 2001 ⇒ n = 250. T250 = 1 + (250 – 1) × 8 = 1 + 249 × 8 = 1993. So, 1993 lies on thumb and 1994 on index finger.
552. The sum of four consecutive even numbers A, B, C and D is 180. What is the sum of the set of next four consecutive even numbers?
- A. 196
- B. 204
- C. 212
- D. 214
Answer: 212
Explanation: Let the four consecutive even numbers be a, a + 2, a + 4 and a + 6. Then, a + a + 2 + a + 4 + a + 6 = 180 ⇒ 4a = 168 ⇒ a = 42. So, these numbers are 42, 44, 46 and 48. Sum of next four consecutive even numbers = (50 + 52 + 54 + 56) = 212.
553. Symbiosis runs a Corporate Training Programme. At the end of running the first programme, its total takings were ₹ 38950. There were more than 45 but less than 100 participants. What was the participant fee for the programme?
- A. ₹ 410
- B. ₹ 450
- C. ₹ 500
- D. ₹ 510
Answer: ₹ 410
Explanation: 38950 is completely divisible by 410 only. Hence ₹ 410 is the correct answer.
554. 7 is added to a certain number ; the sum is multiplied by 5 ; the product is divided by 9 and 3 is subtracted from the quotient. Thus, if the remainder left is 12, what was the original number?
- A. 20
- B. 30
- C. 40
- D. 60
Answer: 20
555. A positive number, which when added to 1000, gives a sum which is greater than when it is multiplied by 1000. This positive integer is
- A. 1
- B. 3
- C. 5
- D. 7
Answer: 1
Explanation: (1000 + x) > (1000 × x). Clearly, x = 1.
556. What is the number of prime factors contained in the product 307 × 225 × 3411?
- A. 49
- B. 51
- C. 52
- D. 53
Answer: 53
557. The number of prime factors in the expression 610 · 717 · 1127 is equal to
- A. 54
- B. 64
- C. 71
- D. 81
Answer: 64
558. For an integer n, n! = n(n – 1) (n – 2)............ 3.2.1. (P.C.S., 2008) Then, 1! + 2! + 3! + ......... + 100! when divided by 5 leaves remainder
- A. 0
- B. 1
- C. 2
- D. 3
Answer: 3
559. The highest power of 9 dividing 99! completely is
- A. 11
- B. 20
- C. 22
- D. 24
Answer: 24
560. * * * × * 8 * * 1 In the above multiplication problem, * is equal to
- A. 1
- B. 3
- C. 7
- D. 9
Answer: 9
Explanation: Clearly, 111 × 1 = 111 ≠ 8111. But, 999 × 9 = 8991. Hence, we have 9 in place of *.
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