Number system and divisibility — Practice MCQs

1. Types of numbers

2. Divisibility rules

3. Important results

• Number of factors of N = p^a q^b → (a + 1)(b + 1).
• Unit digit cycles: 2 → 2, 4, 8, 6; 3 → 3, 9, 7, 1; 7 → 7, 9, 3, 1; 8 → 8, 4, 2, 6 (cycle of 4). 0, 1, 5, 6 repeat; 4 and 9 cycle of 2.
• Remainder of a product = product of remainders (then divide).
• Sum 1 to n = n(n + 1)/2; squares n(n + 1)(2n + 1)/6.
• Product of any n consecutive integers is divisible by n!.

4. Solved examples

Unit digit of 7⁹⁵: 95 mod 4 = 3 → 3.

Is 3,25,358 divisible by 11? (8 + 3 + 2) − (5 + 5 + 3) = 0 → Yes.

(ASO 2021) 6n² + 6n = 6n(n + 1) is always divisible by 6 and 12.

(ASO 2012) a, b odd primes → a² − b² = (a − b)(a + b) is composite.

Solved examples by question type

Each type that comes in the ASO paper, worked step by step. Try it yourself first, then read the solution.

Basic level

Example 1. Which of the following numbers is divisible by 3?
(a) 21,239   (b) 21,241   (c) 21,242   (d) 21,240
Answer: (d) 21,240
Solution: Digit sum divisible by 3. 21240 passes the test.

Example 2. Which of the following numbers is divisible by 3?
(a) 19,519   (b) 19,518   (c) 19,517   (d) 19,520
Answer: (b) 19,518
Solution: Digit sum divisible by 3. 19518 passes the test.

Example 3. The unit digit of 2^77 is:
(a) 4   (b) 2   (c) 5   (d) 8
Answer: (b) 2
Solution: Unit digits of powers of 2 repeat as 2, 4, 8, 6. 77 mod 4 → 2.

Example 4. The unit digit of 96^428 is:
(a) 6   (b) 9   (c) 1   (d) 7
Answer: (a) 6
Solution: Unit digits of 6^n repeat in a cycle of 4 (or less). 428 mod 4 = 4 → unit digit 6.

Example 5. The unit digit of 72^453 is:
(a) 7   (b) 3   (c) 2   (d) 5
Answer: (c) 2
Solution: Unit digits of 2^n repeat in a cycle of 4 (or less). 453 mod 4 = 1 → unit digit 2.

Example 6. The least number to be added to 49,245 to make it exactly divisible by 29 is:
(a) 3   (b) 28   (c) 27   (d) 26
Answer: (d) 26
Solution: 49,245 ÷ 29 leaves remainder 3. Add 29 − 3 = 26.

Example 7. The least number to be added to 78,285 to make it exactly divisible by 56 is:
(a) 53   (b) 55   (c) 4   (d) 3
Answer: (d) 3
Solution: 78,285 ÷ 56 leaves remainder 53. Add 56 − 53 = 3.

ASO exam level

Example 8. What least number must be subtracted from 2557 to get a number exactly divisible by 29?
(a) 10   (b) 24   (c) 5   (d) 6
Answer: (c) 5
Solution: 2557 ÷ 29 leaves remainder 5. Subtract 5.

Example 9. What least number must be added to 3892 to make it exactly divisible by 17?
(a) 17   (b) 1   (c) 16   (d) 2
Answer: (b) 1
Solution: 3892 ÷ 17 leaves remainder 16. Add 17 − 16 = 1.

Example 10. What least number must be added to 6128 to make it exactly divisible by 23?
(a) 14   (b) 10   (c) 13   (d) 23
Answer: (c) 13
Solution: 6128 ÷ 23 leaves remainder 10. Add 23 − 10 = 13.

Example 11. The smallest number that must be added to 2509 to make it a perfect square is:
(a) 9   (b) 92   (c) 102   (d) 94
Answer: (b) 92
Solution: √2509 lies between 50 and 51. 51² = 2601. Add 2601 − 2509 = 92.

Example 12. If the number 840xy is divisible by 90, then (x + y) = ?
(a) 6   (b) 9   (c) 3   (d) 7
Answer: (a) 6
Solution: Divisible by 90 = by 10 and by 9. By 10: y = 0. By 9: 8+4+0 + x must be a multiple of 9 → x = 6. x + y = 6.

Example 13. What is the sum of two consecutive even numbers, the difference of whose squares is 44?
(a) 24   (b) 20   (c) 12   (d) 22
Answer: (d) 22
Solution: (n+2)² − n² = 4n + 4 = 44 → n = 10. Sum = 10 + 12 = 22. (Asked in OPSC ASO 2021 and 2025.)

Example 14. What is the sum of two consecutive even numbers, the difference of whose squares is 68?
(a) 32   (b) 18   (c) 36   (d) 34
Answer: (d) 34
Solution: (n+2)² − n² = 4n + 4 = 68 → n = 16. Sum = 16 + 18 = 34. (Asked in OPSC ASO 2021 and 2025.)

Example 15. A number when divided successively by 5 and 4 leaves remainders 3 and 1 respectively, with final quotient 8. The number is:
(a) 176   (b) 168   (c) 165   (d) 173
Answer: (b) 168
Solution: Work backwards: 8×4 + 1 = 33; then ×5 + 3 = 168.

Example 16. A number when divided successively by 5 and 4 leaves remainders 1 and 2 respectively, with final quotient 9. The number is:
(a) 186   (b) 196   (c) 191   (d) 190
Answer: (c) 191
Solution: Work backwards: 9×4 + 2 = 38; then ×5 + 1 = 191.

Example 17. The remainder when 658 × 400 × 801 is divided by 8 is:
(a) 1   (b) 2   (c) 7   (d) 0
Answer: (d) 0
Solution: Take remainders: 2 × 0 × 1 = 0; divided by 8 leaves 0.

Example 18. The number of factors of 3087 is:
(a) 6   (b) 12   (c) 14   (d) 13
Answer: (b) 12
Solution: 3087 = 7^3 × 3^2. Factors = (3+1)(2+1) = 12.

Advanced level

Example 19. How many two-digit numbers are divisible by 3?
(a) 30   (b) 32   (c) 31   (d) 29
Answer: (a) 30
Solution: Multiples of 3 up to 99: 33; up to 9: 3. Count = 30.

Example 20. 1 + 2 + 3 + … + 30 = ?
(a) 900   (b) 435   (c) 495   (d) 465
Answer: (d) 465
Solution: Sum of first n natural numbers = n(n+1)/2 = 30×31/2 = 465.

Example 21. 1 + 2 + 3 + … + 20 = ?
(a) 230   (b) 400   (c) 190   (d) 210
Answer: (d) 210
Solution: Sum of first n natural numbers = n(n+1)/2 = 20×21/2 = 210.

Example 22. The greatest number that divides 363, 619 and 763 leaving the same remainder in each case is:
(a) 16   (b) 8   (c) 11   (d) 32
Answer: (a) 16
Solution: Take differences: 256, 144, 400. Their HCF = 16.

Topic-wise practice for Number system and divisibility (Quant): Divisibility, remainders, unit digit, consecutive numbers, largest and smallest number. Useful for OPSC ASO, OSSC, OSSSC and other Odisha exams.

Question types covered

• Types of numbers: natural, whole, integer, rational, irrational, prime, composite
• Divisibility rules for 2, 3, 4, 5, 6, 8, 9, 11
• Finding the missing digit for divisibility
• Remainder when a number or expression is divided
• Unit digit of a product or power
• Sum of consecutive odd, even and natural numbers
• Number of factors and sum of factors
• Largest and smallest number, ascending and descending order