Divisibility rules, 2 to 12
A divisibility rule is a quick test on a number's digits that tells you whether it divides exactly, with no long division needed. Learn the eleven below and you can check most numbers in your head.
Written for KS2 and KS3 (Years 5 to 8), and grades 4 to 7 in the US.
741Divisible by 3
The digit sum is a multiple of 3.
See the worked exampleHow this maps to the game. In Divisible Dash the rules unlock in tiers. Beginner covers 2, 3, 5 and 10. Intermediate is 3, 4, 6, 8, 9, 11 and 12. Advanced mixes all eleven, including 7, the trickiest of the lot. Master them here, then put them under pressure in the game.
The rules at a glance
| Number | Quick test |
|---|---|
| 2 | The last digit is even (0, 2, 4, 6 or 8). |
| 3 | The digit sum is a multiple of 3. |
| 4 | The last two digits form a multiple of 4. |
| 5 | The last digit is 0 or 5. |
| 6 | It passes both the 2 test and the 3 test. |
| 7 | Double the last digit, subtract it from the rest, look for a multiple of 7. |
| 8 | The last three digits form a multiple of 8. |
| 9 | The digit sum is a multiple of 9. |
| 10 | The last digit is 0. |
| 11 | From the last digit, add and subtract the digits in turn (+, −, +, ...). The result is 0 or a multiple of 11. |
| 12 | It passes both the 3 test and the 4 test. |
Jump to a rule:
Divisibility rule for 2
A number is divisible by 2 when its last digit is even, that is 0, 2, 4, 6 or 8. Nothing else about the number matters.
3,978 ends in 8, and 8 is even, so 3,978 is divisible by 2. (3,978 ÷ 2 = 1,989.)
Why it works: every place above the units (tens, hundreds, thousands) is already a multiple of 2, so only the last digit can leave a remainder.
Divisibility rule for 3
Add up the digits. If that digit sum is in the 3 times table, the whole number is divisible by 3. If the sum is still large, add its digits again.
741: 7 + 4 + 1 = 12. 12 is in the 3 times table (3 x 4 = 12), so 741 is divisible by 3. (741 ÷ 3 = 247.)
Why it works: 10 leaves remainder 1 when divided by 3, and so do 100, 1,000 and so on. That means each digit contributes just itself to the remainder, so only the digit sum matters.
Divisibility rule for 4
Look only at the last two digits. If the number they form is a multiple of 4, the whole number is divisible by 4.
1,916: the last two digits are 16, and 16 = 4 x 4, so 1,916 is divisible by 4. (1,916 ÷ 4 = 479.)
Common trap. It is the last two digits, not just the last one. 34 ends in an even 4 but is not a multiple of 4.
Why it works: 100 is divisible by 4, so the hundreds, thousands and everything above them are already multiples of 4. Only the last two digits can leave a remainder.
Divisibility rule for 5
A number is divisible by 5 when its last digit is 0 or 5.
2,335 ends in 5, so 2,335 is divisible by 5. (2,335 ÷ 5 = 467.)
Why it works: 10 is a multiple of 5, so every full ten is too. That leaves only the units digit to check, and just 0 and 5 keep the number in the 5 times table.
Divisibility rule for 6
6 is 2 times 3, so a number is divisible by 6 only when it passes both the 2 test and the 3 test. It must be even and have a digit sum that is a multiple of 3.
534 ends in 4, so it is even (passes the 2 test). 5 + 3 + 4 = 12, a multiple of 3 (passes the 3 test). Both pass, so 534 is divisible by 6. (534 ÷ 6 = 89.)
Common trap. One test alone is not enough. 15 has digit sum 6 (passes 3) but is odd (fails 2), so it is not divisible by 6. 14 is even (passes 2) but its digit sum 5 fails 3, so it is not either.
Divisibility rule for 7
7 is the hardest rule, and honestly it is often quicker just to divide. If you want the trick: take off the last digit, double it, and subtract it from the number that is left. If the result is a multiple of 7 (including 0), the original number is divisible by 7.
672: drop the 2, double it to get 4, and take it from the remaining 67. 67 − 4 = 63, and 63 = 7 x 9, so 672 is divisible by 7. (672 ÷ 7 = 96.)
For bigger numbers, repeat the step until you reach a number you recognise.
3,101: drop the 1, double to 2, and 310 − 2 = 308. Repeat on 308: drop the 8, double to 16, and 30 − 16 = 14. 14 = 7 x 2, so 3,101 is divisible by 7. (3,101 ÷ 7 = 443.)
Divisibility rule for 8
Look at the last three digits. If the number they form is a multiple of 8, the whole number is divisible by 8. If you prefer, halve the last three digits three times and check you land on a whole number.
5,120: the last three digits are 120, and 120 = 8 x 15, so 5,120 is divisible by 8. (5,120 ÷ 8 = 640.)
Common trap. It is the last three digits, not two. A number can pass the 4 test on its last two digits and still fail the 8 test.
Why it works: 1,000 is divisible by 8, so the thousands and everything above them are already multiples of 8. Only the last three digits can leave a remainder.
Divisibility rule for 9
Add up the digits. If that digit sum is in the 9 times table, the number is divisible by 9. As with 3, you can add the digits again if the sum is still large.
6,453: 6 + 4 + 5 + 3 = 18. 18 = 9 x 2, so 6,453 is divisible by 9. (6,453 ÷ 9 = 717.)
Common trap. Passing the 3 test does not guarantee the 9 test. 24 has digit sum 6, a multiple of 3 but not of 9, so 24 is divisible by 3 but not by 9.
Why it works: just like 3, every power of 10 leaves remainder 1 when divided by 9, so only the digit sum decides the remainder.
Divisibility rule for 10
A number is divisible by 10 when its last digit is 0.
4,290 ends in 0, so 4,290 is divisible by 10. (4,290 ÷ 10 = 429.)
Why it works: 10 divides any whole number of tens exactly, so a number sits in the 10 times table precisely when it has nothing left over in the units, that is a 0.
Divisibility rule for 11
Start with the last digit. Moving left, subtract the next digit, add the one after that, and keep taking turns. If the result is 0 or a multiple of 11, the number is divisible by 11.
2,728: 8 − 2 + 7 − 2 = 11. 11 is a multiple of 11, so 2,728 is divisible by 11. (2,728 ÷ 11 = 248.)
Common trap. Adding all the digits, as you would for 3 or 9, tells you nothing about 11. The signs must alternate. A result of 0, or a negative multiple of 11, still counts: 9,163 gives 3 − 6 + 1 − 9 = −11, so 9,163 is divisible by 11. (9,163 ÷ 11 = 833.)
Why it works: 10 is one less than 11, so the place values leave remainders that take turns between 1 and −1 (units 1, tens −1, hundreds 1, and so on). That is why the digits take turns being added and subtracted.
Divisibility rule for 12
12 is 3 times 4, so a number is divisible by 12 only when it passes both the 3 test and the 4 test. Its digit sum must be a multiple of 3 and its last two digits must form a multiple of 4.
348: 3 + 4 + 8 = 15, a multiple of 3 (passes the 3 test). The last two digits are 48, and 48 = 4 x 12 (passes the 4 test). Both pass, so 348 is divisible by 12. (348 ÷ 12 = 29.)
Common trap. 2 and 6 also multiply to 12, but they do not work as a pair. 18 passes the 2 test and the 6 test, yet 18 is not divisible by 12. The two tests must share no factor, which is why it is 3 and 4.
Put the rules to work
Reading them is one thing. Spotting them at speed is where they stick.