You can tell whether a number divides evenly by 2 up to 12 just by looking at its digits. Learn every rule, why it works, and how to use it on real problems.
Is 4,536 divisible by 12? You can answer in about five seconds without a calculator or any long division, by looking only at its digits. Divisibility rules are short checks that tell you whether a number divides exactly, and they save time in fractions, factorisation, HCF and LCM, and everyday sharing problems. This guide gives you every rule from 2 to 12, shows why each rule works, and finishes with problems from real life.
A number is divisible by another number if dividing them leaves no remainder. 12 is divisible by 3, because 12 ÷ 3 = 4 exactly. 13 is not divisible by 3, because 13 ÷ 3 leaves a remainder of 1.
The same fact can be said three ways: 3 is a factor of 12, 12 is a multiple of 3, and 12 is divisible by 3. (See Factors & Multiples.) Two small facts are worth knowing: every number is divisible by 1, and 0 is divisible by every number (0 ÷ 5 = 0 with nothing left over).
Is 84 divisible by 6? 84 ÷ 6 = 14 exactly, so yes. And 85 ÷ 6 = 14 remainder 1, so no. A divisibility rule gets you the same yes/no answer without doing the division.
These three rules only need the last digit, because 10 = 2 × 5, so every multiple of 10 is already divisible by both 2 and 5.
246 ends in 6, which is even, so it is divisible by 2 (246 ÷ 2 = 123). 3,125 ends in 5, so it is divisible by 5 (3,125 ÷ 5 = 625). 8,710 ends in 0, so it is divisible by 10 (8,710 ÷ 10 = 871). Notice that 8,710 is also divisible by 2 and by 5.
For 4 and 8, the number of digits you check depends on a simple fact: 100 = 4 × 25, so every hundred (and every thousand) is already divisible by 4. Only the last two digits can change the answer. In the same way 1,000 = 8 × 125, so for 8 only the last three digits matter.
(a) 1,728: the last two digits are 28, and 28 ÷ 4 = 7, so 1,728 is divisible by 4 (1,728 ÷ 4 = 432).
(b) 1,728 again: the last three digits are 728, and 728 ÷ 8 = 91, so 1,728 is divisible by 8 (1,728 ÷ 8 = 216).
(c) 7,380: the last two digits are 80, and 80 ÷ 4 = 20, so 7,380 is divisible by 4 (7,380 ÷ 4 = 1,845).
More rules in the same family:
For 3 and 9 you add up all the digits. If the digit sum is divisible by 3, so is the number. If the digit sum is divisible by 9, so is the number.
4,536: 4 + 5 + 3 + 6 = 18. Since 18 is divisible by both 3 and 9, 4,536 is divisible by 3 (4,536 ÷ 3 = 1,512) and by 9 (4,536 ÷ 9 = 504).
Why it works. Each place value is one more than a multiple of 9: 10 = 9 + 1, 100 = 99 + 1, 1,000 = 999 + 1, and so on. So 4,536 = 4 × 1,000 + 5 × 100 + 3 × 10 + 6 = 4 × 999 + 5 × 99 + 3 × 9 + (4 + 5 + 3 + 6). The first three parts are multiples of 9, so the whole number is a multiple of 9 exactly when the digit sum is. Since 3 divides 9, the same trick works for 3.
Divisibility by 3 does not guarantee divisibility by 9. Nine is a stricter check.
The fix: 96,432 has a digit sum of 9 + 6 + 4 + 3 + 2 = 24. Since 24 is divisible by 3, the number passes the rule for 3. But 24 ÷ 9 = 2 remainder 6, so it fails the rule for 9. Always check the rule for the number you actually need.
The number 5,7■2 is divisible by 9. Find the missing digit. The known digits add to 5 + 7 + 2 = 14. The digit sum must be a multiple of 9, and the next multiple of 9 after 14 is 18. So the missing digit is 18 − 14 = 4. Check: 5,742 ÷ 9 = 638. ✓
A number is divisible by a bigger number if it is divisible by two smaller numbers that share no common factor and multiply to give it. That lets you build new rules from old ones.
| Divisible by | Passes both | Because |
|---|---|---|
| 6 | 2 and 3 | 6 = 2 × 3 |
| 12 | 3 and 4 | 12 = 3 × 4 |
| 15 | 3 and 5 | 15 = 3 × 5 |
| 18 | 2 and 9 | 18 = 2 × 9 |
| 72 | 8 and 9 | 72 = 8 × 9 |
(a) Is 4,536 divisible by 6? It is even (passes 2) and its digit sum 18 is divisible by 3 (passes 3), so yes: 4,536 ÷ 6 = 756.
(b) Is 4,536 divisible by 12? Digit sum 18 passes 3, and the last two digits 36 pass 4 (36 ÷ 4 = 9). So yes: 4,536 ÷ 12 = 378.
(c) Is 36,288 divisible by 72? Last three digits 288 ÷ 8 = 36, so it passes 8. Digit sum 3 + 6 + 2 + 8 + 8 = 27, so it passes 9. Since 8 and 9 share no factor, 36,288 is divisible by 72: 36,288 ÷ 72 = 504.
The two numbers must share no common factor. Both 4 and 6 are even, so they overlap.
The fix: 4 and 6 share the factor 2, so passing both rules does not prove divisibility by 24. Test it: 36 is divisible by 4 and by 6, but 36 ÷ 24 = 1.5, which is not a whole number. The same happens with 12. For 24, use 3 and 8 instead, since 3 and 8 share no factor and 3 × 8 = 24.
Seven has no single-digit shortcut, but there is a reliable multi-step rule: double the last digit, subtract it from the rest of the number, and repeat until you have a small number you can judge. If the result is 0 or a multiple of 7, the original number is divisible by 7.
(a) 812: double the last digit, 2 × 2 = 4. Subtract from the rest: 81 − 4 = 77. Since 77 = 7 × 11, 812 is divisible by 7 (812 ÷ 7 = 116).
(b) 2,401: double the last digit, 1 × 2 = 2. Subtract: 240 − 2 = 238. Repeat: 8 × 2 = 16, and 23 − 16 = 7. Since 7 is divisible by 7, so is 2,401 (it is 7 × 7 × 7 × 7).
Why it works. Write the number as 10a + b, where b is the last digit and a is everything before it. Then 10a + b = 10 × (a − 2b) + 21b. The 21b part is always a multiple of 7, and 10 shares no factor with 7, so the whole number is divisible by 7 exactly when (a − 2b) is.
A neat shortcut for 7, 11 and 13 together. Since 7 × 11 × 13 = 1,001, you can split a number into groups of three digits from the right and take an alternating sum of the groups. For 7,007 the groups are 7 and 007, and 7 − 7 = 0. So 7,007 is divisible by 7, 11 and 13 at once (7,007 = 7 × 7 × 11 × 13).
Starting from the right, alternately add and subtract the digits. If the answer is 0 or a multiple of 11, the number is divisible by 11.
(a) 8,712: 2 − 1 + 7 − 8 = 0, so it is divisible by 11 (8,712 ÷ 11 = 792).
(b) 143: 3 − 4 + 1 = 0, so 143 is divisible by 11 (143 = 11 × 13). It is not prime.
(c) 3,157: 7 − 5 + 1 − 3 = 0, so it is divisible by 11 (3,157 ÷ 11 = 287).
(d) 2,013: 3 − 1 + 0 − 2 = 0, so it is divisible by 11 (2,013 ÷ 11 = 183).
Why it works. Because 10 is one less than 11, 10 leaves remainder −1 when divided by 11, 100 leaves +1, 1,000 leaves −1, and so on. So each digit is counted with alternating signs.
| Divisible By | Rule | Quick Example |
|---|---|---|
| 2 | Last digit is even (0, 2, 4, 6, 8) | 246 |
| 3 | Digit sum divisible by 3 | 4,536 (18) |
| 4 | Last two digits divisible by 4 | 1,728 (28) |
| 5 | Last digit is 0 or 5 | 3,125 |
| 6 | Passes both 2 and 3 | 4,536 |
| 7 | Double the last digit, subtract from the rest, repeat | 812 → 77 |
| 8 | Last three digits divisible by 8 | 1,728 (728) |
| 9 | Digit sum divisible by 9 | 4,536 (18) |
| 10 | Last digit is 0 | 8,710 |
| 11 | Alternating digit sum (from the right) is 0 or a multiple of 11 | 8,712 (0) |
| 12 | Passes both 3 and 4 | 4,536 |
Tap any number below to instantly see every number from 2 to 12 it divides evenly by.
Splitting a restaurant bill of Rs 840 evenly among a group: knowing that 84 is divisible by 2, 3, 4, 6, 7 and 12 (but not by 5, 8 or 9) tells you straight away which group sizes divide 84 evenly, and 840 behaves in the same way, with 10 added as a factor. A group of 7 pays Rs 120 each, a group of 12 pays Rs 70 each, but a group of 8 cannot split it into whole rupees.
A farm has 360 eggs. Which tray sizes (2 to 12) hold them with none left over? 360 ends in 0 (so 2, 5 and 10 work). Its digit sum is 9, so 3 and 9 work, and together with 2 that makes 6 work too. The last two digits 60 are divisible by 4, and 3 and 4 together make 12 work. The last three digits 360 are divisible by 8 (360 ÷ 8 = 45). So every size from 2 to 12 works except 7 and 11. Check: 360 ÷ 7 ≈ 51.4 and 360 ÷ 11 ≈ 32.7. ✓
Simplify 378/504. Both are even, so divide by 2: 189/252. Both digit sums are divisible by 9 (1 + 8 + 9 = 18 and 2 + 5 + 2 = 9), so divide by 9: 21/28. Both are divisible by 7: 3/4. So 378/504 = 3/4. (The HCF used was 2 × 9 × 7 = 126.) See HCF & LCM and Fraction Fix-It.
Is 143 prime? It is odd and does not end in 5. Its digit sum is 8, so it is not divisible by 3. Its alternating sum 3 − 4 + 1 = 0 means it is divisible by 11. So 143 = 11 × 13, and it is not prime. Divisibility rules are the quickest way to rule out small factors when testing for primes.
You can use the digit sum to check a multiplication. To check 27 × 13 = 351: the digit sums are 27 → 9, 13 → 4, and 351 → 9. The digit sum of the product of 9 and 4 is 36 → 9. Both sides give 9, so the answer passes the check. If the digit sums differ, the answer is definitely wrong. If they match, the answer is probably right but not guaranteed (for example, swapped digits are not caught).
The old 10-digit ISBN numbers on books use a check digit based on divisibility by 11. The digits are multiplied by weights and added, and the total must be a multiple of 11. If someone mistypes a digit, the total fails the test, and the error is caught.
Tap the correct answer for each question. Your score is tracked as you go, and your best score is saved on this device.
1. 7,380: the last two digits are 80, and 80 ÷ 4 = 20, so yes. 5,193: the digit sum is 5 + 1 + 9 + 3 = 18, so yes (5,193 ÷ 9 = 577).
2. 3 − 1 + 0 − 2 = 0, so yes (2,013 ÷ 11 = 183).
3. 1,344 is even (2), digit sum 12 (3 but not 9), last two digits 44 (4), last three digits 344 = 8 × 43 (8). So 2, 3, 4, 6, 8 and 12 work. 7 also works: 134 − 8 = 126, and 126 = 7 × 18. It fails 5, 9, 10 and 11. So the answer is 2, 3, 4, 6, 7, 8 and 12.
1. 2,346 ends in 6, so it is even and divisible by 2. Its digit sum is 2 + 3 + 4 + 6 = 15, which is divisible by 3. A number divisible by both 2 and 3 is divisible by 6. (2,346 ÷ 6 = 391.)
2. 5 + 7 + 2 = 14, and the digit sum must be 18, so the missing digit is 4 (5,742 ÷ 9 = 638).
3. From the right: 7 − 5 + 1 − 3 = 0, which is a multiple of 11. So 3,157 is divisible by 11 (3,157 ÷ 11 = 287).
4. 1,000 ÷ 12 = 83 remainder 4, so 1,000 is 4 more than a multiple of 12. Add 12 − 4 = 8 to get 1,008. Check: digit sum 9 (divisible by 3) and last two digits 08 (divisible by 4), so 1,008 is divisible by 12.
5. The last three digits 288 ÷ 8 = 36, so it is divisible by 8. The digit sum is 3 + 6 + 2 + 8 + 8 = 27, so it is divisible by 9. Since 8 and 9 share no factor, it is divisible by 8 × 9 = 72. (36,288 ÷ 72 = 504.)
Do not teach these as eleven separate tricks. Group them by what they check: last digits (2, 5, 10, 4, 8), the digit sum (3, 9), the alternating sum (11), and combinations (6, 12 and so on). Ask learners to explain why 100 being divisible by 4 means only the last two digits matter. That one question connects the rules to place value. Let them use their own checking habit: apply a rule, then divide on a calculator to confirm. Learners who test their own guesses on real numbers (phone numbers, house numbers, prices) remember the rules much better.
Divisibility rules are the foundation of number theory. They speed up simplifying fractions, finding factors, testing for primes, and finding the HCF and LCM. They also make everyday sharing and grouping problems fast to solve in your head, and they give you a free way to check your own arithmetic.
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