Number System MCQ – Part 4 - Topic Guide and MCQs

1. Which expression is always odd for every integer n?

   1. n² + n
   2. 2n²
   3. n² − n
   4. n² + n + 1
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   Answer: (D) n² + n + 1

   Explanation: n(n + 1) is even because one of two consecutive integers is even. Thus n² + n + 1 is always odd.

2. How many positive integers at most 600 are divisible by 12 or 18, but not by both?

   1. 67
   2. 51
   3. 83
   4. 17
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   Answer: (B) 51

   Explanation: There are 50 multiples of 12, 33 of 18 and 16 of their LCM 36. Excluding the common multiples from both counts gives 50 + 33 − 2 × 16 = 51.

3. What is the sum of the place values of the two occurrences of 8 in 8,03,280?

   1. 800080
   2. 800008
   3. 800800
   4. 880000
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   Answer: (A) 800080

   Explanation: The two place values are 800000 and 80. Their sum is 800000 + 80 = 800080.

4. Find the greatest perfect cube that divides 2^8 × 3^5 × 5^4 exactly.

   1. 108000
   2. 36000
   3. 216000
   4. 432000
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   Answer: (C) 216000

   Explanation: For a cube divisor, take the greatest multiple of 3 not exceeding each exponent. Thus the divisor is 2^6 × 3³ × 5³ = (4 × 3 × 5)³ = 216000.

5. What is the unit digit of 7^24 − 3^17?

   1. 2
   2. 4
   3. 6
   4. 8
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   Answer: (D) 8

   Explanation: 7^24 ends in 1 and 3^17 ends in 3. The difference has unit digit 1 − 3 + 10 = 8.

6. Find the least positive integer k such that 90k is a perfect square and 60k is a perfect cube.

   1. 450
   2. 56250
   3. 11250
   4. 33750
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   Answer: (B) 56250

   Explanation: For 90k to be a square and 60k a cube, the least exponents of 2, 3 and 5 in k are 1, 2 and 5 respectively. Thus k = 2 × 3² × 5^5 = 56250; indeed, 90k = 2250² and 60k = 150³.

7. Find the least positive integer n such that n! is divisible by 10^29.

   1. 125
   2. 120
   3. 124
   4. 130
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   Answer: (A) 125

   Explanation: 124! has 24 + 4 = 28 trailing zeros, while 125! has 25 + 5 + 1 = 31. The count jumps past 29 at 125, so the least n is 125.

8. Find the remainder when 16^51 is divided by 17.

   1. 0
   2. 1
   3. 16
   4. 8
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   Answer: (C) 16

   Explanation: 16 is equivalent to −1 modulo 17. An odd power therefore leaves remainder −1 + 17 = 16.

9. The sum of 8 consecutive even positive integers is 344. Find the smallest integer.

   1. 34
   2. 38
   3. 43
   4. 36
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   Answer: (D) 36

   Explanation: Their average is 344 ÷ 8 = 43. The smallest is 7 below the average, so it is 43 − 7 = 36.

10. A positive integer is divided by 15 and leaves remainder 11. What remainder does its square leave when divided by 5?

    1. 1
    2. 0
    3. 2
    4. 4
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    Answer: (A) 1

    Explanation: The integer leaves remainder 1 on division by 5 because 11 leaves remainder 1. Its square therefore leaves remainder 1² = 1.

11. How many two-digit numbers have digit product 18?

    1. 2
    2. 4
    3. 3
    4. 6
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    Answer: (B) 4

    Explanation: The possible ordered digit pairs are (2, 9), (3, 6), (6, 3) and (9, 2). Hence there are 4 such numbers.

12. How many positive factors of 1,080 are divisible by 6 but not by 12?

    1. 12
    2. 8
    3. 6
    4. 18
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    Answer: (C) 6

    Explanation: 1080 = 2³ × 3³ × 5. Such a factor must have exponent exactly 1 for 2, exponent 1, 2 or 3 for 3, and exponent 0 or 1 for 5, giving 1 × 3 × 2 = 6.

13. What is the least non-negative integer that must be subtracted from 7,000 to make the result divisible by both 24 and 35?

    1. 8
    2. 560
    3. 40
    4. 280
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    Answer: (D) 280

    Explanation: LCM of 24 and 35 is 840. Since 7000 = 840 × 8 + 280, subtract the remainder 280.

14. What is the value of √3249?

    1. 57
    2. 53
    3. 67
    4. 59
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    Answer: (A) 57

    Explanation: 57² = (50 + 7)² = 2500 + 700 + 49 = 3249. Hence √3249 = 57.

15. Find the remainder when 4^31 + 4^32 + 4^33 is divided by 7.

    1. 1
    2. 0
    3. 3
    4. 6
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    Answer: (B) 0

    Explanation: Factor out 4^31: the expression becomes 4^31 × (1 + 4 + 16) = 4^31 × 21. It is divisible by 7, so the remainder is 0.

16. A sum of ₹1,080 is shared equally among n students. If 3 more students join, each share decreases by ₹30. Find n.

    1. 6
    2. 12
    3. 9
    4. 7
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    Answer: (C) 9

    Explanation: The decrease gives 1080 × 3 ÷ [n(n + 3)] = 30, so n(n + 3) = 108 = 9 × 12. Hence n = 9.

17. How many trailing zeros are there in 75! ÷ 25!?

    1. 18
    2. 6
    3. 15
    4. 12
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    Answer: (D) 12

    Explanation: 75! contains 15 + 3 = 18 factors of 5, while 25! contains 5 + 1 = 6. The quotient has 18 − 6 = 12 factors of 5 and enough factors of 2, so it has 12 trailing zeros.

18. Find the last two digits of 99^51 + 1.

    1. 00
    2. 01
    3. 98
    4. 99
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    Answer: (A) 00

    Explanation: For the last two digits, 99 acts as −1. Thus 99^51 + 1 has remainder (−1)^51 + 1 = 0 on division by 100, so it ends in 00.

19. Find the sum of all three-digit numbers formed using each of the digits 3, 5 and 8 exactly once.

    1. 1776
    2. 3552
    3. 5328
    4. 3330
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    Answer: (B) 3552

    Explanation: Each digit appears twice in each place. The sum is 2 × (3 + 5 + 8) × (100 + 10 + 1) = 3552.

20. Find the sum of all positive factors of 98 that are perfect squares.

    1. 49
    2. 57
    3. 50
    4. 98
    Show answer

    Answer: (C) 50

    Explanation: 98 = 2 × 7². Its square factors are 1 and 49, whose sum is 1 + 49 = 50.

21. The sum of the first 15 terms of an arithmetic progression is 600. Find its 8th term.

    1. 35
    2. 45
    3. 80
    4. 40
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    Answer: (D) 40

    Explanation: For an odd number of AP terms, the average is the middle term. Thus the 8th term is 600 ÷ 15 = 40.

22. If a + b = 13 and ab = 36, find a² + b².

    1. 97
    2. 169
    3. 133
    4. 241
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    Answer: (A) 97

    Explanation: a² + b² = (a + b)² − 2ab = 13² − 2 × 36 = 97.

23. What is the greatest four-digit number that is a multiple of 45 and has digit sum 18?

    1. 9990
    2. 9900
    3. 9855
    4. 9810
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    Answer: (B) 9900

    Explanation: A digit sum of 18 ensures divisibility by 9; the number must also end in 0 or 5. Taking the first two digits as 99 leaves digit sum 0 for the last two, giving the greatest number 9900.

24. What is the unit digit of the product 1³ × 2³ × 3³ × 4³ × 5³ × 6³?

    1. 2
    2. 5
    3. 6
    4. 0
    Show answer

    Answer: (D) 0

    Explanation: The product is (1 × 2 × 3 × 4 × 5 × 6)³ = 720³. Since the base ends in 0, its cube also has unit digit 0.

25. Find the remainder when (23² − 1) ÷ 8 is divided by 5.

    1. 0
    2. 2
    3. 1
    4. 4
    Show answer

    Answer: (C) 1

    Explanation: Use the difference of squares: (23² − 1) ÷ 8 = 22 × 24 ÷ 8 = 66. Dividing 66 by 5 leaves remainder 1.

Number System MCQ – Part 4 is factual rather than analytical: you either recall it in seconds or you lose the mark to somebody who did. That makes it one of the cheapest sections to convert - a handful of short sittings usually settles the whole cluster.

Expect one-line recall questions - definitions, stand-for expansions, identifiers and the ordering of things. Practise the wording as well as the concept: these papers test terminology, and the set above uses the same four-option shape they do.

Past papers reuse clusters like this almost verbatim across years and posts. Run the same set until every distractor is familiar - previous-year style questions stop being recall and start being free marks.