Divisibility Rules for Aptitude Tests: 2026 Guide

Divisibility rules let you test whether one number divides another evenly without running the full division, and that speed advantage is what placement aptitude rounds are designed to reward.

Why Placement Aptitude Tests Reward Divisibility Shortcuts

Quantitative Ability sections in campus placement drives follow a consistent pattern. Questions ask you to find the largest N-digit number divisible by X, identify which of five given numbers is divisible by Y, or find a missing digit that makes a number divisible by Z. These questions are not conceptually hard. They are time-hard. Under a 60-second-per-question pace, the student who checks divisibility by 7 in four steps beats one who runs long division every time.

Placement rounds at mass recruiters (AMCAT, TCS NQT, and similar campus screening tests) include at least one Quantitative Ability section where number-theory questions appear alongside ratios and averages. The rules below cover every divisor from 2 to 16 that placement papers use. Each rule is stated once, with a single verified example. For a related shortcut used by the same question designers, see the article on multiplying by 111.

Divisibility Rules for 2, 3, 4, 5, and 6

These five rules each reduce to a single check or a pair of checks.

Divisor	Rule	Verified Example
2	Last digit is 0, 2, 4, 6, or 8	4568 ends in 8: divisible
3	Sum of all digits is divisible by 3	65178: 6+5+1+7+8 = 27, 27 / 3 = 9: divisible
4	Last two digits form a number divisible by 4	83764: last two digits = 64, 64 / 4 = 16: divisible
5	Last digit is 0 or 5	8475 ends in 5: divisible
6	Divisible by both 2 AND 3	18: even, digit sum = 9, 9 / 3 = 3: divisible by 6

The rule for 4 works because 100 is already divisible by 4, so any contribution from the hundreds place and above cancels. Only the last two digits carry the divisibility information for 4, and for the same reason only the last three digits carry it for 8.

Divisibility Rules for 7, 8, 9, and 10

Divisibility by 7

The standard method: remove the last digit, subtract twice that digit from the remaining number, and check whether the result is divisible by 7. Repeat as needed.

• Example: 3563119.
• Step 1: Last digit = 9. Remaining = 356311. Compute: 356311 - (2 x 9) = 356293.
• Step 2: Last digit = 3. Remaining = 35629. Compute: 35629 - (2 x 3) = 35623.
• Step 3: Last digit = 3. Remaining = 3562. Compute: 3562 - (2 x 3) = 3556.
• Step 4: Last digit = 6. Remaining = 355. Compute: 355 - (2 x 6) = 343.
• Step 5: 343 / 7 = 49. Divisible.

For numbers with more than six digits, the block method is faster. Group the digits in threes from the right, then alternately subtract and add those blocks (starting with + for the rightmost). If the result is divisible by 7, so is the original number.

• Example: 6,517,739,025. Blocks from right: 025, 739, 517, 6.
• Alternating sum: 025 - 739 + 517 - 6 = 25 - 739 + 517 - 6 = -203.
• 203 / 7 = 29. Divisible.

Divisibility by 8

A number is divisible by 8 if its last three digits form a number divisible by 8. Since 1000 is divisible by 8, everything above the hundreds place cancels.

• Example: 83864. Last three digits: 864. 864 / 8 = 108. Divisible.

Divisibility by 9

Digit-sum rule, same logic as 3 but stricter: the sum of all digits must be divisible by 9.

• Example: 65178. Digit sum: 6 + 5 + 1 + 7 + 8 = 27. 27 / 9 = 3. Divisible.

Divisibility by 10

Last digit must be 0.

Divisibility Rules for 11, 12, 13, and Their Composites

Divisibility by 11

Assign alternating signs to each digit, left to right: add odd-position digits (1st, 3rd, 5th…) and subtract even-position digits (2nd, 4th, 6th…). If the result is 0 or divisible by 11, the number is divisible by 11.

• Example A: 83721. Digits: 8, 3, 7, 2, 1.

• Alternating sum: 8 - 3 + 7 - 2 + 1 = 11. Divisible.

• Verification: 83721 / 11 = 7611. Confirmed.

• Example B: 506. Digits: 5, 0, 6.

• Alternating sum: 5 - 0 + 6 = 11. Divisible.

• Verification: 506 / 11 = 46. Confirmed.

• Example C: 1234. Digits: 1, 2, 3, 4.

• Alternating sum: 1 - 2 + 3 - 4 = -2. Not divisible by 11.

• Verification: 1234 / 11 = 112.18… Confirmed not divisible.

The algebraic proof that this works appears in Wolfram MathWorld’s coverage of divisibility tests, which also lists the formal proofs for the block methods below.

Divisibility by 12

A number is divisible by 12 if it is divisible by both 3 and 4. Run both checks: digit sum divisible by 3, and last two digits divisible by 4.

Divisibility by 13

The block method applies here with the same three-digit grouping used for 7. If the alternating sum of the three-digit blocks is 0 or divisible by 13, the original number is divisible by 13.

• Example: 53872. Blocks from right: 872, 53.
• Alternating sum: 872 - 53 = 819. 819 / 13 = 63. Divisible.
• Verification: 53872 / 13 = 4144. Confirmed.

This shared structure across 7, 11, and 13 is not coincidence. As Khan Academy’s divisibility review explains, all three work because 1001 = 7 x 11 x 13. Since 1001 is congruent to 0 modulo each of these three divisors, alternating blocks of three digits give the correct remainder test for all three.

Composite Rules for 14, 15, 16, and 18

Rules for composite divisors combine checks for their prime factors.

Divisor	Combination Rule
14	Divisible by 2 AND divisible by 7
15	Divisible by 3 AND divisible by 5
16	Last four digits form a number divisible by 16
18	Divisible by 2 AND divisible by 9

The rule for 16 follows the same positional logic as 4 and 8: 10000 is divisible by 16, so only the last four digits carry the remainder.

Solved Placement-Style Examples

Largest three-digit number divisible by 7

• Step 1: Start with 999, the largest three-digit number.
• Step 2: 999 / 7 = 142 remainder 5.
• Step 3: 999 - 5 = 994.
• Answer: 994. Verify: 994 / 7 = 142. Confirmed.

Largest six-digit number divisible by 12

• Step 1: Start with 999999.
• Step 2: 999999 / 12 = 83333 remainder 3.
• Step 3: 999999 - 3 = 999996.
• Answer: 999996. Verify: digit sum = 9+9+9+9+9+6 = 51, divisible by 3; last two digits 96 / 4 = 24. Confirmed.

Check 833 for divisibility by 2, 3, 7, and 9

• Divisible by 2: 833 ends in 3 (odd). No.
• Divisible by 3: digit sum = 8 + 3 + 3 = 14. Not divisible by 3. No.
• Divisible by 7: remove last digit 3, remaining = 83; compute 83 - (2 x 3) = 83 - 6 = 77; 77 / 7 = 11. Yes.
• Divisible by 9: digit sum = 14. Not divisible by 9. No.

These question types also appear in Calendar and Clock aptitude sections, which use the same modular arithmetic logic. The articles on calendar problems in aptitude tests and clock problems for competitive exams cover the overlapping methods.

The block method for 7, 11, and 13 all run on the single fact that 1001 = 7 x 11 x 13. One algebraic identity replacing three separate memorised procedures is the kind of structural shortcut that placement aptitude tests are designed to reward, and it is also the kind of pattern-matching instinct that shows up when working with token sequences in language models. TinkerLLM runs real LLM experiments at a ₹299 entry point, and the divisibility-rule mindset transfers more directly than you might expect.