Quantitative Aptitude

Number System: Practice Questions with Answers

21 practice questions on Number System, with answers; most also have an explanation or a worked solution. Try each one before you open its answer.

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1. Q.1

   Check whether 9,251 is divisible by 11.

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   Answer: Yes, divisible by 11

   Solution:

   1. From the right: position 1 = 1, position 2 = 5, position 3 = 2, position 4 = 9.
   2. Sum of odd positions (1st, 3rd) = 1 + 2 = 3. Sum of even positions (2nd, 4th) = 5 + 9 = 14.
   3. Difference = |3 − 14| = 11, which is divisible by 11 — so 9,251 is divisible by 11.

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2. Q.2

   Find the unit digit of 7⁹⁵.

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   Answer: 3

   Solution:

   1. Unit digits of powers of 7 cycle every 4 steps: 7¹→7, 7²→9, 7³→3, 7⁴→1, then repeat.
   2. 95 = 4 × 23 + 3, so the remainder is 3 — this points to the 3rd entry in the cycle.
   3. The 3rd entry is 3.

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3. Q.3

   A number, when divided by 15, leaves a remainder of 7. Find the remainder when the same number is divided by 5.

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   Answer: 2

   Solution:

   1. The number is N = 15Q + 7 for some quotient Q. Since 15 is a multiple of 5, the term 15Q is exactly divisible by 5.
   2. So the remainder on dividing N by 5 is just 7 mod 5 = 2. (Check with Q = 1: N = 22, and 22 ÷ 5 leaves remainder 2.)

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4. Q.4

   Find the remainder when 7¹⁰⁰ is divided by 5.

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   Answer: 1

   Solution:

   1. Find the pattern of 7ⁿ mod 5: 7¹ mod 5 = 2, 7² mod 5 = 4, 7³ mod 5 = 3, 7⁴ mod 5 = 1, and then it repeats — a cycle of length 4.
   2. 100 ÷ 4 leaves no remainder, so 7¹⁰⁰ lands on the 4th position in the cycle: remainder 1.

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5. Q.5

   Check whether 4,572 is divisible by 9.

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   Answer: Yes, divisible by 9

   Solution:

   1. Digit sum = 4 + 5 + 7 + 2 = 18.
   2. 18 is divisible by 9, so 4,572 is divisible by 9.

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6. Q.6

   Find the number of trailing zeros in 100!.

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   Answer: 24

   Solution:

   1. Count multiples of 5 up to 100: ⌊100/5⌋ = 20.
   2. Count multiples of 25 up to 100 (each contributes one extra factor of 5): ⌊100/25⌋ = 4.
   3. Multiples of 125 up to 100: 0, since 125 > 100.
   4. Total trailing zeros = 20 + 4 = 24.

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7. Q.7

   Find the number of factors of 360.

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   Answer: 24

   Solution:

   1. 360 = 2³ × 3² × 5¹.
   2. Number of factors = (3+1) × (2+1) × (1+1) = 4 × 3 × 2
   3. = 24

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8. Q.8

   Find the sum of the first 20 natural numbers, the sum of their squares, and the sum of their cubes.

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   Answer: Sum = 210, Sum of squares = 2,870, Sum of cubes = 44,100

   Solution:

   1. Sum of first n numbers = n(n+1)/2 = 20 × 21 / 2 = 210.
   2. Sum of first n squares = n(n+1)(2n+1)/6 = (20 × 21 × 41) / 6 = 17,220 / 6 = 2,870.
   3. Sum of first n cubes = [n(n+1)/2]² = 210² = 44,100.

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9. Q.9

   Check whether 61,809 is divisible by 11.

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   Answer: Yes (odd-position sum 23 minus even-position sum 1 = 22, divisible by 11)

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10. Q.10

    Find the sum of the first 15 natural numbers and the sum of their squares.

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    Answer: Sum = 120, Sum of squares = 1,240. 15×16/2 = 120; (15×16×31)/6 = 7,440/6.

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11. Q.11

    Find the number of trailing zeros in 50!.

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    Answer: 12. ⌊50/5⌋ + ⌊50/25⌋ = 10 + 2.

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12. Q.12

    Find the number of factors of 180.

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    Answer: 18. 180 = 2²×3²×5¹; (2+1)(2+1)(1+1) = 3×3×2.

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13. Q.13

    Find the unit digit of 8⁵³.

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    Answer: 8. Cycle of 8 is 8,4,2,6 (length 4); 53 = 4×13+1, so the 1st entry, which is 8.

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14. Q.14

    Check whether 3,684 is divisible by 4.

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    Answer: Yes (last two digits, 84, are divisible by 4)

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15. Q.15

    A number, when divided by 24, leaves a remainder of 11. Find the remainder when the same number is divided by 8.

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    Answer: 3

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16. Q.16

    What is the sum of the first 30 positive odd integers, 1 + 3 + 5 + ... + 59?

    1. A 870
    2. B 930
    3. C 1770
    4. D 900
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    Correct answer: (D) 900

    The sum of the first n positive odd integers is n², so the sum is 30² = 900. To check independently, pair the first and last terms: 1 + 59 = 60, 3 + 57 = 60, and so on. There are 15 such pairs, giving 15 × 60 = 900.

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17. Q.17

    The four-digit number 57d2 is divisible by 9, where d is a single digit from 0 to 9. What is d?

    1. A 3
    2. B 4
    3. C 5
    4. D 7
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    Correct answer: (B) 4

    The digit sum is 5 + 7 + d + 2 = 14 + d. It ranges from 14 to 23, so the only possible multiple of 9 is 18. Thus d = 18 − 14 = 4. Checking directly, 5742 = 9 × 638.

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18. Q.18

    What is the smallest non-negative integer that must be added to 1000 to make the result divisible by 16?

    1. A 4
    2. B 16
    3. C 8
    4. D 24
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    Correct answer: (C) 8

    Since 1000 = 16 × 62 + 8, another 16 − 8 = 8 is needed to reach the next multiple. Checking, 1000 + 8 = 1008 = 16 × 63. The consecutive multiples 992 and 1008 surround 1000, so no smaller non-negative addition works.

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19. Q.19

    A prime number is an integer greater than 1 with exactly two positive factors. Which of the following numbers is prime?

    1. A 21
    2. B 29
    3. C 39
    4. D 49
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    Correct answer: (B) 29

    The numbers 21 = 3 × 7, 39 = 3 × 13, and 49 = 7 × 7 have factors other than 1 and themselves. Any composite number has a prime factor no greater than its square root. Since √29 is between 5 and 6, only the primes 2, 3, and 5 need testing; none divides 29, so 29 is prime.

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20. Q.20

    The units digits of successive positive powers of 7 repeat in the cycle 7, 9, 3, 1, starting with 7¹. What is the units digit of 7^43?

    1. A 3
    2. B 7
    3. C 9
    4. D 1
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    Correct answer: (A) 3

    The cycle has length 4, and 43 = 4 × 10 + 3, so the required digit is the third entry, 3. Checking from the cycle boundary, 7^40 ends in 1, and multiplying by 7³ = 343 gives a units digit of 3.

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21. Q.21

    What is the largest four-digit positive integer divisible by 36?

    1. A 9972
    2. B 9936
    3. C 9996
    4. D 10008
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    Correct answer: (A) 9972

    The largest four-digit integer is 9999. Since 9999 = 36 × 277 + 27, subtracting 27 gives 9972 = 36 × 277. Checking the next multiple, 9972 + 36 = 10008, which has five digits, so 9972 is the largest four-digit multiple.

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