4. Tests of Divisibility
Free study material · concepts, shortcuts & solved questions
These rules let you determine divisibility without performing the actual division — pure speed tools for MCQ exams.
| Divisor | Rule |
|---|---|
| 2 | Last digit is 0, 2, 4, 6, or 8 (even) |
| 3 | Sum of all digits is divisible by 3 |
| 4 | Number formed by the last two digits is divisible by 4 |
| 5 | Last digit is 0 or 5 |
| 6 | Number is divisible by both 2 and 3 |
| 7 | Double the last digit, subtract it from the remaining number; repeat until a small number remains; if that is divisible by 7 (or is 0), the original number is divisible by 7 |
| 8 | Number formed by the last three digits is divisible by 8 |
| 9 | Sum of all digits is divisible by 9 |
| 10 | Last digit is 0 |
| 11 | (Sum of digits at odd positions) − (Sum of digits at even positions), counted from the right, is 0 or a multiple of 11 |
| 12 | Number is divisible by both 3 and 4 |
| 13 | Multiply the last digit by 4, add it to the remaining number; repeat until a small number remains; if that is divisible by 13, so is the original |
| 25 | Number formed by the last two digits is divisible by 25 (i.e., ends in 00, 25, 50, or 75) |
Why it works (short version): Since 10 ≡ 1 (mod 3) and 10 ≡ 1 (mod 9), the value of a number mod 3 or mod 9 equals its digit-sum mod 3 or mod 9 — hence the “sum of digits” rules. Since 10² = 100 is exactly divisible by 4, 25, and (close to) by 8’s multiples, only the last few digits govern divisibility by 4, 8, 25. Since 10 ≡ −1 (mod 11), alternating digit sums determine divisibility by 11. The 7 and 13 rules come from finding a multiplier (called an “osculator”) that mimics removing a 10 without changing divisibility — for 7 it is −2 (double-and-subtract), for 13 it is +4 (quadruple-and-add).