Divisibility Rules from 2 to 12

Updated 27 October 2026

A divisibility rule tells you whether one number divides into another exactly, without doing the division. Each one below comes with the reason it works, because a rule you can see the reason for is a rule you can rebuild when you forget it.

What are the divisibility rules?

A divisibility rule tells you whether a number divides exactly, without doing the division. A number divides by 2 when its last digit is even, by 5 when it ends in 0 or 5, and by 10 when it ends in 0. It divides by 4 when its last two digits do, and by 8 when its last three do. It divides by 3 or by 9 when its digits add to a multiple of 3 or of 9, and by 11 when the digits, tagged + and − alternately from the right, add to 0 or a multiple of 11. Take 5,832: it ends in 2, so it divides by 2, and its digits add to 5 + 8 + 3 + 2 = 18, so it divides by 3 and by 9. The sections below explain why each rule works. For 6, 12 and 15, combine two rules whose numbers share no factor. Dividing confirms a result: 5,832 ÷ 9 = 648.

The rest of this page is the reasons. They fall into three families: the last digits, the digit sum, and the alternating sum.

Why do the last digits decide 2, 5, 10, 4 and 8?

The last digits decide because every other part of the number is already a multiple of the divisor. Ten is a multiple of 2, 5 and 10, so all the tens in a number divide by those three, and only the ones digit is left to check. A hundred is a multiple of 4, since 100 = 4 × 25, so every hundred divides by 4 and only the last two digits matter. Take 9,136. It is 91 hundreds and 36, the hundreds divide by 4 whatever 91 is, and 36 = 4 × 9, so 9,136 divides by 4. A thousand is a multiple of 8, since 1,000 = 8 × 125, so the test for 8 looks at the last three digits. Two digits are not enough for 8, because a hundred is not a multiple of 8. For example, 116 ends in 16, which divides by 8, but 116 does not. A division confirms the test for 4: 9,136 ÷ 4 = 2,284.

Write 3,746 as 374 tens and 6. Every one of those tens is 10, and 10 divides by 2, so the whole block of tens divides by 2 whatever 374 happens to be. Only the 6 can break it, and 6 is even, so 3,746 divides by 2.

So 4,312 is 4 thousands and 312, the thousands take care of themselves, and 312 = 8 × 39 settles it.

Why do the digit sums work for 3 and 9?

The digit sum works because every place value is one more than a multiple of 9: 10 is 9 + 1, 100 is 99 + 1, and 1,000 is 999 + 1. Take 513. Its 5 hundreds are 5 × 99 and 5 more, its 1 ten is 1 × 9 and 1 more, and then there are 3 ones. The parts built from 99 and 9 are multiples of 9, so they cannot affect the remainder, and what is left is 5 + 1 + 3 = 9, the digit sum. So 513, like 9, divides by 9: 513 = 9 × 57. The rule for 3 comes free, because every multiple of 9 is also a multiple of 3. When the digit sum is itself large, add its digits again: 8,976 has digit sum 30, and 3 + 0 = 3, so it divides by 3 but not by 9. Division checks this: 8,976 ÷ 3 = 2,992.

Take 471. It is 4 hundreds, 7 tens and 1. Write each hundred as 99 + 1 and each ten as 9 + 1:

1. 471 = 4 × 100 + 7 × 10 + 1
2. = 4 × (99 + 1) + 7 × (9 + 1) + 1
3. = (4 × 99 + 7 × 9) + (4 + 7 + 1)

The first bracket is built entirely from 99 and 9, so it is a multiple of 9 and contributes nothing to the remainder. What is left is 4 + 7 + 1 = 12 — the digit sum. So 471 and 12 leave the same remainder against 9, and 471 divides by 3 exactly when 12 does.

For 8,946 the digits add to 27, and 27 adds to 9, so 8,946 divides by 9.

How do you test for 11?

To test for 11, tag the digits + and − alternately, starting with + on the ones digit, and add them. If the total is 0 or a multiple of 11, the number divides by 11. For 3,729, the total is 9 − 2 + 7 − 3 = 11, so 3,729 divides by 11, and 3,729 = 11 × 339. The test works because 10 is one less than 11, so the places take turns counting backward and forward. A total of 0 passes too: 5,654 gives 4 − 5 + 6 − 5 = 0, and 5,654 = 11 × 514. A negative multiple of 11 passes as well: for 8,294, the total is 4 − 9 + 2 − 8 = −11, so 8,294 divides by 11. Any other total, such as 3 or 14, means the number does not divide. A short division confirms it: 8,294 ÷ 11 = 754.

Ten leaves −1 against 11, a hundred leaves +1, a thousand leaves −1 again, and so on down the places.

How do you test for 6, 12 and 15?

To test for 6, 12 or 15, split the number into two factors that share no common factor, and check that both rules pass. Six is 2 × 3, twelve is 3 × 4, and fifteen is 3 × 5. Take 2,316. Its digits add to 12, so it divides by 3, and its last two digits make 16, so it divides by 4. It passes both tests, so it divides by 12, and 2,316 = 12 × 193. For 15, the number 4,035 ends in 5 and has digits adding to 12, so it divides by 15. This works because a number that divides by 3 and by 4, which share no factor, must contain 3 × 4 as a factor. Twelve is also 2 × 6, but that pair fails: 18 divides by 2 and by 6 and not by 12, because 2 and 6 share the factor 2. A division check settles any doubt: 4,035 ÷ 15 = 269.

The condition that the two factors share no common factor is the whole of it, and it is where this goes wrong. Twelve is also 2 × 6, but 2 and 6 are both even, so passing both proves less than it looks:

Is there a rule for 7?

There is a rule for 7: remove the last digit, double it, and subtract it from the number that remains. If the result is a multiple of 7, including 0 or a negative multiple, so is the original number. Repeat the step until the number is small enough to recognize. Take 1,358. Removing the 8 leaves 135, and 135 − 16 = 119. Removing the 9 leaves 11, and 11 − 18 = −7, which is a multiple of 7, so 1,358 divides by 7. The step never changes whether a number divides by 7, because it only takes away a multiple of 7 and divides by 10, and 10 has no factor of 7. When the result is not a multiple of 7, the original is not one either: 123 gives 12 − 6 = 6, so 123 does not divide by 7. Dividing confirms the first example: 1,358 ÷ 7 = 194.

It works because multiplying by 10 cannot create or destroy a factor of 7. If the shortened number a − 2b divides by 7 then so does ten times it, and ten times it differs from the original 10a + b by exactly 21b — three sevens for every b.

Worth saying plainly: this is the one rule here that often costs more than the division it saves. For 861 it is quick. For a six-digit number, dividing by 7 is usually faster.

Where this fits in the ladder

Divisibility is the tool you reach for when simplifying a fraction, finding a common denominator, or hunting for prime factors — every one of those starts with "what divides this?".

Related guides: Factors, Multiples and Number Theory, Division and Fractions.