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401. (325 + 326 + 327 + 328) is divisible by
- A. 11
- B. 16
- C. 25
- D. 30
Answer: 30
Explanation: (325 + 326 + 327 + 328) = 325 (1 + 3 + 32 + 33) = 325 (1 + 3 + 9 + 27) = 325 × 40 = (3 × 10) × (324 × 4) = 30 × (324 × 4), which is divisible by 30.
402. If (1012 + 25)2 – (1012 – 25)2 = 10n, then the value of n is
- A. 5
- B. 10
- C. 14
- D. 20
Answer: 14
Explanation: (1012 + 25)2 – (1012 – 25)2 = 4 × 1012 × 25 [∵ (a + b)2 – (a – b)2 = 4ab] = 1012 × 100 = 1012 × 102 = 1014. Hence, n = 14.
403. (719 + 2) is divided by 6. The remainder is
- A. 1
- B. 2
- C. 3
- D. 5
Answer: 3
Explanation: (xn – an) is divisible by (x – a) for all values of n. ∴ (719 – 119) is divisible by (7 – 1) ⇒ (719 – 1) is divisible by 6 ⇒ On dividing (719 + 2) by 6, remainder obtained = 3.
404. Consider the following statements: (1) an + bn is divisible by a + b if n = 2k + 1, where k is a positive integer. (2) an – bn is divisible by a – b if n = 2k, where k is a positive integer. Which of the statements given above is/are correct?
- A. 1 only
- B. 2 only
- C. Both 1 and 2
- D. Neither 1 nor 2
Answer: Both 1 and 2
Explanation: Clearly, (1) when n is odd, (an + bn) is divisible by (a + b). So, (1) is true. (2) (an – bn) is divisible by (a – b) for all values of n. So, (2) is true.
405. What is the remainder when 231 is divided by 5?
- A. 1
- B. 2
- C. 3
- D. 4
Answer: 3
Explanation: 231 = 2 × 230 = 2 × (22) 15 = 2 × 415. When n is odd, (xn + an) is divisible by (x + a). ∴ (415 + 115) is divisible by (4 + 1) ⇒ (415 + 1) is divisible by 5 ⇒ (230 + 1) is divisible by 5 ⇒ On dividing 230 by 5, we get (5 – 1) i.e. 4 as remainder. ∴ Remainder obtained on dividing 231 by 5 = Remainder obtained on dividing (2 × 4) i.e. 8 by 5 = 3.
406. If 17200 is divided by 18, the remainder is
- A. 1
- B. 2
- C. 16
- D. 17
Answer: 1
Explanation: When n is even, (xn – an) is divisible by (x + a). ∴ (17200 – 1200) is divisible by (17 + 1) ⇒ (17200 – 1) is divisible by 18 ⇒ On dividing 17200 by 18, we get 1 as remainder.
407. The numbers from 1 to 29 are written side by side as follows: 1234567891011121314...........2829 If this number is divided by 9, then what is the remainder?
- A. 0
- B. 1
- C. 3
- D. None of these
Answer: 3
Explanation: Sum of digits of numbers from 1 to 10 = 46. Sum of digits of numbers from 11 to 20 = 56. Sum of digits of numbers from 21 to 29 = 63. Sum of digits of the given number = 46 + 56 + 63 = 165. So, the required remainder is the remainder obtained on dividing 165 by 9, which is 3.
408. When the square of any odd number, greater than 1, is divided by 8, it always leaves remainder
- A. 1
- B. 6
- C. 8
- D. Cannot be determined
Answer: 1
Explanation: Let the number be N = 2x + 1. N2 = (2x + 1)2 = 4x2 + 1 + 4x = 4x (x + 1) + 1. Clearly, 4x (x + 1) is always divisible by 8 since one of x and (x + 1) is even which when multiplied by 4 is always divisible by 8. Hence, required remainder = 1.
409. A number when divided by 3 leaves a remainder 1. When the quotient is divided by 2, it leaves a remainder 1. What will be the remainder when the number is divided by 6?
- A. 2
- B. 3
- C. 4
- D. 5
Answer: 4
410. A number is successively divided by 8, 7 and 3 giving residues 3, 4 and 2 respectively and quotient 31. The number is
- A. 3555
- B. 5355
- C. 5535
- D. 5553
Answer: 5355
411. After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number?
- A. 41
- B. 53
- C. 75
- D. 80
Answer: 53
412. A number when divided by three consecutive numbers 9, 11, 13 leaves the remainders 8, 9 and 8 respectively. If the order of divisors is reversed, the remainders will be
- A. 10, 8, 9
- B. 10, 1, 6
- C. 8, 9, 8
- D. 9, 8, 8
Answer: 10, 1, 6
Explanation: 9 x z = 13 × 1 + 8 = 21. 11 y – 8 y = 11 × z + 9 = 11 × 21 + 9 = 240. 13 z – 9 x = 9 × y + 8 = 9 × 240 + 8 = 2168. 1 – 8 Now, when order of divisors is reversed, we have: 13 2168 11 166 – 10 9 15 – 1 1 – 6 ∴ Respective remainders are 10, 1 and 6.
413. The numbers 2272 and 875 are divided by a three-digit number N, giving the same remainder. The sum of the digits of N is
- A. 10
- B. 11
- C. 12
- D. 13
Answer: 10
Explanation: Clearly, (2272 – 875) = 1397, is exactly divisible by N. Now, 1397 = 11 × 127. ∴ The required 3-digit number is 127, the sum of whose digits is 10.
414. A number divided by 13 leaves a remainder 1 and if the quotient, thus obtained, is divided by 5, we get a remainder of 3. What will be the remainder if the number is divided by 65?
- A. 16
- B. 18
- C. 28
- D. 40
Answer: 40
Explanation: The number is of the form (13k + 1), where k is of the form (5m + 3). ∴ Number = 13k + 1 = 13 (5m + 3) + 1 = 65m + 40. Clearly when the number is divided by 65, we get 40 as remainder.
415. Two numbers when divided by a certain divisor leave the remainders 4375 and 2986 respectively but when the sum of two numbers is divided by the same divisor, the remainder is 2361. The divisor in question is
- A. 4675
- B. 4900
- C. 5000
- D. None of these
Answer: 5000
Explanation: As proved in the above question, divisor = 4375 + 2986 – 2361 = 5000.
416. When a certain positive integer P is divided by another positive integer, the remainder is r1. When a second positive integer Q is divided by the same integer, the remainder is r2 and when (P + Q) is divided by the same divisor, the remainder is r3. Then the divisor may be
- A. r1 r2 r3
- B. r1 + r2 + r3
- C. r1 – r2 + r3
- D. r1 + r2 – r3
Answer: r1 + r2 – r3
Explanation: Let P = x + r1 and Q = y + r2, where each of x and y are divisible by the common divisor. Then, P + Q = (x + r1) + (y + r2) = (x + y) + (r1 + r2). (P + Q) leaves remainder r3 when divided by the common divisor. ⇒ [(x + y) + (r1 + r2) – r3] is divisible by the common divisor. Since (x + y) is divisible by the common divisor, so divisor = r1 + r2 – r3.
417. In a division sum, the remainder was 71. With the same divisor but twice the dividend, the remainder is 43. Which one of the following is the divisor?
- A. 86
- B. 93
- C. 99
- D. 104
Answer: 99
Explanation: Let the dividend be (x + 71) and the divisor be y. Then, [2(x + 71) – 43] is divisible by y ⇒ (2x + 142 – 43) is divisible by y ⇒ (2x + 99) is divisible by y. ∴ Divisor = 99 Shortcut Method: Divisor = (2 × 71 – 43) = (142 – 43) = 99.
418. When a number is divided by 13, the remainder is 11. When the same number is divided by 17, the remainder is 9. What is the number?
- A. 339
- B. 349
- C. 369
- D. Data inadequate
Answer: 349
419. When n is divided by 4, the remainder is 3. What is the remainder when 2n is divided by 4?
- A. 1
- B. 2
- C. 3
- D. 6
Answer: 2
Explanation: Let n = 4k + 3. Then, 2n = 2(4k + 3) = 8k + 6 = 4 × 2k + 4 × 1 + 2 = 4 (2k + 1) + 2. Thus, on dividing 2n by 4, we get 2 as remainder.
420. The students of a class donated a sum of ₹2,809 to the Fund. Each student donated as many rupees as the number of students in the class. The number of students in the class is
- A. 51
- B. 47
- C. 53
- D. 49
Answer: 53
421. A number when divided by 5 leaves the remainder 3. What is the remainder when the square of the same number is divided by 5?
- A. 0
- B. 3
- C. 4
- D. 9
Answer: 4
Explanation: Let the number be x and on dividing by 5, we get q as quotient and 3 as remainder. Then, x = 5q + 3 ⇒ x2 = (5q + 3)2 = (25q2 + 30q + 9) = 5(5q2 + 6q + 1) + 4. Thus, on dividing x2 by 5, we get 4 as remainder.
422. A certain number when divided by 899 gives a remainder 63. What is the remainder when the same number is divided by 29?
- A. 5
- B. 25
- C. 27
- D. None of these
Answer: 5
Explanation: Let the number be x and the quotient be q. Then, x = 899 q + 63 = (29 × 31q) + (29 × 2) + 5 = 29 (31q + 2) + 5. So, the given number when divided by 29 gives 5 as remainder.
423. A number when divided by 195 leaves a remainder 47. If the same number is divided by 15, the remainder will be
- A. 1
- B. 2
- C. 3
- D. 4
Answer: 2
Explanation: Let the number be x and the quotient be q. Then, x = 195 q + 47 = (15 × 13 q) + (15 × 3) + 2 = 15(13q + 3) + 2. So, the given number when divided by 15 gives 2 as remainder.
424. A number when divided by 136 leaves remainder 36. If the same number is divided by 17, the remainder will be
- A. 2
- B. 3
- C. 7
- D. 9
Answer: 2
Explanation: Let the number be x. Let x when divided by 136 give q as quotient and 36 as remainder. Then, x = 136 q + 36 = (17 × 8q) + (17 × 2) + 2 = 17 × (8q + 2) + 2. So, the given number when divided by 17 gives 2 as remainder.
425. On dividing a number by 19, the difference between quotient and remainder is 9. The number is
- A. 352
- B. 361
- C. 370
- D. 371
Answer: 371
Explanation: Since the required number is a 3-digit number, so on dividing by 19, it would yield a two-digit quotient which means that the quotient is greater than the remainder. Let the remainder be x. Then, quotient = x + 9. So, number N = 19 (x + 9) + x = 20x + 171. ∴ (N – 171) must be divisible by 20. Clearly, (371 – 171) = 200, which is divisible by 20. Hence, the required number = 371.
426. In doing a question of division with zero remainder, a candidate took 12 as divisor instead of 21. The quotient obtained by him was 35. The correct quotient is
- A. 0
- B. 12
- C. 13
- D. 20
Answer: 20
427. A number when divided by the sum of 555 and 445 gives two times their difference as quotient and 30 as the remainder. The number is
- A. 1220
- B. 1250
- C. 22030
- D. 220030
Answer: 220030
428. The divisor is 25 times the quotient and 5 times the remainder. If the quotient is 16, then the dividend is
- A. 400
- B. 480
- C. 6400
- D. 6480
Answer: 6480
429. In a division sum, the divisor is 12 times the quotient and 5 times the remainder. If the remainder is 48, then the dividend is
- A. 2404
- B. 3648
- C. 4808
- D. 4848
Answer: 4848
430. In a division sum, the quotient, dividend and remainder are 15, 940 and 25 respectively. The divisor is
- A. 31
- B. 50
- C. 60
- D. 61
Answer: 61
431. The number 534677 is divided by 777. The difference of divisor and remainder is
- A. 577
- B. 676
- C. 687
- D. 789
Answer: 676
432. Match List I with List II and select the correct answer: List I List II (a, b as given in (Values of q and r) Euclidean algorithm a = bq + r) A. a = – 112, b = – 7 1. q = – 13, r = 1 B. a = 118, b = – 9 2. q = 14, r = 3 C. a = – 109, b = 6 3. q = – 19, r = 5 D. a = 115, b = 8 4. q = 16, r = 0 A B C D A B C D
- A. 3 1 4 2
- B. 3 2 4 1
- C. 4 1 3 2
- D. 4 2 3 1
Answer: 4 1 3 2
433. Which of the following prime numbers while dividing 2176 leaves 9 as remainder?
- A. 17
- B. 29
- C. 167
- D. 197
Answer: 197
Explanation: (2176 – 9) = 2167 = (11 × 197). So, the required number is 197.
434. A number divided by 68 gives the quotient 260 and remainder zero. If the same number is divided by 65, the remainder is
- A. 0
- B. 1
- C. 2
- D. 3
Answer: 0
435. If 11,109,999 is divided by 1111, then what is the remainder?
- A. 1098
- B. 1010
- C. 1110
- D. 1188
Answer: 1110
436. A 9-digit number in which zero does not appear and no digits are repeated has the following properties: The number comprising the left most two digits is divisible by 2, that comprising the left most three digits is divisible by 3, and soon. The number is
- A. 183654729
- B. 381654729
- C. 983654721
- D. 981654723
Answer: 381654729
Explanation: (a) In 183654729, 1836547 is not divisible by 7. (b) In 381654729, 38 is divisible by 2, 381 is divisible by 3, 3816 is divisible by 4, 38165 is divisible by 5, 381654 is divisible by 6, 3816547 is divisible by 7, 38165472 is divisible by 8 and 381654729 is divisible by 9. (c) In 983654721, 983 is not divisible by 3. (d) In 981654723, 9816547 is not divisible by 7.
437. While writing all the numbers from 700 to 1000, how many numbers occur in which the first digit is greater than the second digit, and the second digit is greater than the third digit?
- A. 61
- B. 64
- C. 78
- D. 85
Answer: 85
Explanation: When the second digit is 1, third digit can be 0, i.e. there is one such number. When the second digit is 2, third digit can be 0 or 1, i.e. there are 2 such numbers. When the second digit is 3, third digit can be 0, 1 or 2 i.e. there are 3 such numbers, and so on. When the first digit is 7, second digit can be 1, 2, 3, 4, 5 or 6. So, there are (1 + 2 + 3 + 4 + 5 + 6) = 21 such numbers between 700 and 799. When the first digit is 8, second digit can be 1, 2, 3, 4, 5, 6 or 7. So, there are (1 + 2 + 3 + 4 + 5 + 6 + 7) = 28 such numbers between 800 and 899. When the first digit is 9, second digit can be 1, 2, 3, 4, 5, 6, 7 or 8. So, there are (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36 such numbers between 900 and 999. Hence, the required number = (21 + 28 + 36) = 85.
438. The total number of integers between 200 and 400, each of which either begins with 3 or ends with 3 or both is
- A. 10
- B. 100
- C. 110
- D. 120
Answer: 110
Explanation: Such numbers are 203, 213, 223, 233, 243, 253, 263, 273, 283, 293 and all numbers from 300 to 399. Clearly, number of such numbers = 10 + 100 = 110.
439. How many 3-digit numbers are there in between 100 and 300, having first and the last digit as 2?
- A. 9
- B. 10
- C. 11
- D. 12
Answer: 10
Explanation: Such numbers are 202, 212, 222, 232, 242, 252, 262, 272, 282, 292. There are 10 such numbers.
440. If all the numbers from 501 to 700 are written, what is the total number of times the digit 6 appears?
- A. 138
- B. 139
- C. 140
- D. 141
Answer: 140
Explanation: Numbers from 501 to 599 which have 6 as digit are 506, 516, 526, 536, 546, 556, 560, 561, 562, 563, 564, 565, 566, 567, 568, 569, 576, 586 and 596, i.e. 6 occurs 20 times. Number of times 6 occurs from 600 to 699 = 100 + 20 = 120. ∴ Total number of times 6 occurs = 20 + 120 = 140.
441. How many integers between 100 and 150, both inclusive, can be evenly divided by neither 3 nor 5?
- A. 26
- B. 27
- C. 28
- D. 33
Answer: 27
Explanation: Number of integers between 100 and 150 (including both) = 51. Numbers divisible by 3 are 102, 105, 108, ........, 150. Let Tm = 150. Then, a + (m – 1) d = Tm. ∴ 102 + (m – 1) × 3 = 150 ⇒ (m – 1) × 3 = 48 ⇒ m – 1 = 16 ⇒ m = 17. Numbers divisible by 5 are 100, 105, 110, 115, .........., 150. Let Tn = 150. Then, a + (n – 1) d = Tn. ∴ 100 + (n – 1) × 5 = 150 ⇒ (n – 1) × 5 = 50 ⇒ (n – 1) = 10 ⇒ n = 11. Numbers divisible by both 3 and 5 are 105, 120, 135, 150. They are four in number. Number of numbers divisible by 3 or 5 = (17 + 11 – 4) = 24. Number of numbers neither divisible by 3 nor by 5 are (51 – 24) = 27.
442. How many numbers less than 1000 are multiples of both 10 and 13?
- A. 6
- B. 7
- C. 8
- D. 9
Answer: 7
Explanation: Required numbers are multiples of (10 × 13), i.e. 130. These numbers are 130, 260, 390, 520, 650, 780 and 910. They are 7 in number.
443. Out of the numbers divisible by 3 between 14 and 95 if the numbers with 3 at unit's place are removed, then how many numbers will remain?
- A. 22
- B. 23
- C. 24
- D. 25
Answer: 24
Explanation: Numbers between 14 and 95 and divisible by 3 are 15, 18, 21 24, .........., 93. Tm = 93 ⇒ 15 + (n – 1) × 3 = 93 ⇒ (n – 1) × 3 = 78 ⇒ (n – 1) = 26 ⇒ n = 27. Numbers to be deleted are 33, 63, 93. They are 3 in number. Required number of numbers = (27 – 3) = 24.
444. The number of terms between 11 and 200 which are divisible by 7 but not by 3 are
- A. 18
- B. 19
- C. 27
- D. 28
Answer: 18
Explanation: Multiples of 7 between 11 and 200 are 14, 21, 28, 35, 42, ........., 189, 196. Tm = 196 ⇒ 14 + (m – 1) × 7 = 196 ⇒ (m – 1) × 7 = 182 ⇒ (m – 1) = 26 ⇒ m = 27. Multiples of 7 and 3 both, i.e. that of 21 are 21, 42, 63, ............, 189 Tn = 189 ⇒ 21 + (n – 1) × 21 = 189 ⇒ (n – 1) × 21 = 168 ⇒ (n – 1) = 8 ⇒ n = 9. ∴ Required number of terms = (27 – 9) = 18.
445. How many 3-digit numbers are completely divisible by 6?
- A. 149
- B. 150
- C. 151
- D. 166
Answer: 150
Explanation: 3-digit numbers divisible by 6 are 102, 108, 114, ......., 996. This is an A.P. in which a = 102, d = 6 and Tn = 996. ∴ Tn = a + (n – 1) d ⇒ 102 + (n – 1) × 6 = 996 ⇒ (n – 1) × 6 = 894 ⇒ (n – 1) = 149 ⇒ n = 150. Hence, there are 150 such numbers.
446. The sum of all 2-digit numbers divisible by 5 is
- A. 945
- B. 1035
- C. 1230
- D. 1245
Answer: 945
447. The smallest 6-digit number exactly divisible by 111 is
- A. 111111
- B. 110011
- C. 100011
- D. 110101
Answer: 100011
448. The smallest number by which 66 must be multiplied to make the result divisible by 18 is
- A. 3
- B. 6
- C. 9
- D. 18
Answer: 3
Explanation: 66 = 11 × 6 In order to get a number divisible by 18, the above product must be multiplied by 3. Hence 66 must be multiplied by 3.
449. The number of times 99 is subtracted from 1111 so that the remainder is less than 99 is
- A. 10
- B. 11
- C. 12
- D. 13
Answer: 11
450. The smallest number that must be added to 803642 in order to obtain a multiple of 11 is
- A. 1
- B. 4
- C. 7
- D. 9
Answer: 7
Explanation: On dividing 803642 by 11, we get 4 as remainder. Required number to be added = (11 – 4) = 7.
451. The smallest number which must be subtracted from 8112 to make it exactly divisible by 99 is
- A. 91
- B. 92
- C. 93
- D. 95
Answer: 93
452. The nearest integer to 58701 which is exactly divisible by 567 is
- A. 55968
- B. 58068
- C. 58968
- D. None of these
Answer: 58968
453. The least number more than 5000 which is divisible by 73 is
- A. 5009
- B. 5037
- C. 5073
- D. 5099
Answer: 5037
454. The least number, which must be added to the greatest 6-digit number so that the sum may be exactly divisible by 327 is
- A. 194
- B. 264
- C. 292
- D. 294
Answer: 294
455. Which is the greatest 5-digit number exactly divisible by 279?
- A. 99603
- B. 99550
- C. 99882
- D. None of these
Answer: 99882
456. Find the least 6-digit number which is exactly divisible by 349.
- A. 100163
- B. 101063
- C. 160063
- D. None of these
Answer: 100163
457. Which of the following numbers should be added to 8567 to make it exactly divisible by 4?
- A. 3
- B. 4
- C. 5
- D. None of these
Answer: None of these
Explanation: On dividing 8567 by 4, the remainder is 3. To make it divisible by 4, we must add 1 to it.
458. What least number must be added to 1056 to get a number exactly divisible by 23?
- A. 2
- B. 3
- C. 21
- D. 25
Answer: 2
459. The smallest 4-digit number exactly divisible by 7 is
- A. 1001
- B. 1007
- C. 1101
- D. 1108
Answer: 1001
Explanation: The smallest 4-digit number is 1000. This when divided by 7 leaves 6 as remainder. ∴ 1001 is the smallest 4-digit number exactly divisible by 7.
460. The students of a class donated a sum of ₹2,209. If each student donated a many rupees as the number of students in the class, then the number of students in the class is
- A. 47
- B. 49
- C. 53
- D. 51
Answer: 47
Explanation: Let the number of students in the class = x Then the amount donated by each student = ₹x According to the question, x × x = 2209 x2 = 2209 x = 2209 x = 47 Hence, the number of student in the class = 47
461. The difference of the squares of two consecutive even integers is divisible by
- A. 3
- B. 4
- C. 6
- D. 7
Answer: 4
Explanation: Two consecutive even integers are 2n and (2n + 2). ∴ (2n + 2)2 – (2n) 2 = (4n2 + 4 + 8n – 4n2) = 4 + 8n = 4(1 + 2n), which is always divisible by4.
462. The largest number that exactly divides each number of the sequence 15 – 1, 25 – 2, 35 – 3, ..., n5 – n,.... is
- A. 1
- B. 15
- C. 30
- D. 120
Answer: 30
Explanation: Required number = (25 – 2) = (32 – 2) = 30.
463. If the square of an odd natural number is divided by 8, then the remainder will be
- A. 1
- B. 2
- C. 3
- D. 4
Answer: 1
Explanation: Let the odd natural number be (2n + 1). n = 1 gives (2n + 1)2 = (2 + 1)2 = 9. This when divided by 8 gives 1 as remainder. n = 2 gives (2n + 1)2 = 25. This when divided by 8 gives 1 as remainder and so on.
464. If n is any positive integer, 34n – 43n is always divisible by
- A. 7
- B. 12
- C. 17
- D. 145
Answer: 17
Explanation: Putting n = 1, we get (34n – 43n) = (34 – 43) = (81 – 64) = 17, which is divisible by 17.
465. Consider the following statements: For any positive integer n, the number 10n – 1 is divisible by (1) 9 for n = odd only (2) 9 for n = even only (3) 11 for n = odd only (4) 11 for n = even only Which of the above statements are correct?
- A. 1 and 3
- B. 2 and 3
- C. 1 and 4
- D. 2 and 4
Answer: 2 and 4
Explanation: We know that (xn – an) is divisible by (x – a), when n is even and (xn – an) is divisible by (x + a), when n is even. ∴ (10n – 1) is divisible by (10 – 1) = 9, when n is even. And, (10n – 1) is divisible by (10 + 1) = 11, when n is even. ∴ Statements 2 and 4 are correct.
466. Which one of the following is the number by which the product of 8 consecutive integers is divisible?
- A. 4 ! / 4!
- B. 6 ! / 6!
- C. 7 ! / 7!
- D. All of these
Answer: All of these
Explanation: The product of 8 consecutive numbers is divisible by each one of 8!, 7!, 6!, 5!, 4!, 3! and 2!
467. (xn – an) is divisible by (x – a)
- A. for all values of n
- B. only for even values of n
- C. only for odd values of n
- D. only for prime values of n
Answer: for all values of n
Explanation: We know that (xn – an) is always divisible by (x – a) for all values of n.
468. The sum of the digits of a natural number (10n – 1) is 4707, where n is a natural number. The value of n is
- A. 477
- B. 523
- C. 532
- D. 704
Answer: 523
Explanation: 10n has (n + 1) digits. Then, 9 will appear n times in (10n – 1). So, sum of digits in (10n – 1) = 9n. ∴ 9n = 4707 ⇒ n = 4707 523. 9 =
469. A 6-digit number is formed by repeating a 3-digit number; for example, 256256 or 678678 etc. Any number of this form is always exactly divisible by
- A. 7 only
- B. 11 only
- C. 13 only
- D. 1001
Answer: 1001
Explanation: Numbers like 2525, 3636 etc. are divisible by 101. Numbers like 256256, 678678 etc. are divisible by 1001. Numbers like 32163216, 43754375 etc. are divisible by 10001 and so on.
470. The difference between the squares of two consecutive odd integers is always divisible by
- A. 3
- B. 6
- C. 7
- D. 8
Answer: 8
Explanation: Let the two consecutive odd integers be (2m + 1) and (2m + 3). Then, (2m + 3)2 – (2m + 1)2 = [(2m + 3) + (2m + 1)] [(2m + 3) – (2m + 1)] = (4m + 4) × 2 = 8m + 8 = 8(m + 1), which is always divisible by 8.
471. If p is a prime number greater than 3, then (p2 – 1) is always divisible by
- A. 6 but not 12
- B. 12 but not 24
- C. 24
- D. None of these
Answer: 24
Explanation: p = 5 ⇒ (p2 – 1) = (25 – 1) = 24, which is divisible by 24. p = 7 ⇒ (p2 – 1) = (49 – 1) = 48, which is divisible by 24. p = 11 ⇒ (p2 – 1) = (121 – 1) = 120, which is divisible by 24. Hence, (p2 – 1) is always divisible by 24.
472. The greatest number by which the product of three consecutive multiples of 3 is always divisible is
- A. 54
- B. 81
- C. 162
- D. 243
Answer: 162
Explanation: Three consecutive multiples of 3 are 3m, 3(m + 1) and 3(m + 2). Their product = 3m × 3(m + 1) × 3(m + 2) = 27 × m × (m + 1) × (m + 2). Putting m = 1, this product is (27 × 1 × 2 × 3) = 162. So, this product is always divisible by 162.
473. The sum of three consecutive odd numbers is always divisible by I. 2 II. 3 III. 5 IV. 6
- A. Only I
- B. Only II
- C. Only I and II
- D. Only I and III
Answer: Only II
Explanation: Let the required odd numbers be n, (n + 2) and (n + 4). Then n + (n + 2) + (n + 4) = 3n + 6 = 3(n + 2), which is always divisible by 3. But n = 1 ⇒ 3(n + 2) = 3 × 3 = 9 which is not divisible by any of 2, 5 and 6. Hence only II is true.
474. The product of any three consecutive natural numbers is always divisible by
- A. 3
- B. 6
- C. 9
- D. 15
Answer: 6
475. A number is multiplied by 11 and 11 is added to the product. If the resulting number is divisible by 13, the smallest original number is
- A. 12
- B. 22
- C. 26
- D. 53
Answer: 12
Explanation: Let the required number be x. Then + = 11 11 a 13x whole number. So, (11x + 11) must be divisible by 13. By hit and trial, we get x = 12. Hence, the smallest original number is 12.
476. The sum of the digits of a 3-digit number is subtracted from the number. The resulting number is always
- A. not divisible by 9
- B. divisible by 9
- C. not divisible by 6
- D. divisible by 6
Answer: divisible by 9
Explanation: Let the hundreds, tens and unit digits be x, y and z respectively. Then, (100x + 10y + z) – (x + y + z) = 99x + 9y = 9 (11x + y). So, the resulting number is divisible by 9.
477. If n is any odd number greater than 1, then n (n2 – 1) is
- A. divisible by 24 always
- B. divisible by 48 always
- C. divisible by 96 always
- D. None of these
Answer: divisible by 24 always
478. If n is a whole number greater than 1, then n2 (n2 – 1) is always divisible by
- A. 8
- B. 10
- C. 12
- D. 16
Answer: 12
479. The largest natural number, which exactly divides the product of any four consecutive natural numbers, is
- A. 6
- B. 12
- C. 24
- D. 120
Answer: 24
480. The sum of a number consisting of two digits and the number formed by interchanging the digits is always divisible by
- A. 7
- B. 9
- C. 10
- D. 11
Answer: 11
Explanation: Let the ten's digit be x and the unit's digit be y. Then, (10x + y) + (10y + x) = 11(x + y), which is divisible by 11.
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