Divisibility Rules from 2 to 12

Updated 16 September 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 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.

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?

Because ten is already a multiple of 2, 5 and 10, so every ten in a number divides by all three and only the units digit is left to check. A hundred is a multiple of 4 and a thousand is a multiple of 8, so for those two the check moves along to the last two and the last three digits.

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.

Eight needs one more place, for the same reason: 1000 = 8 × 125, and 100 is not a multiple of 8. 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?

Because every power of ten is one more than a multiple of 9: 10 is 9 + 1, 100 is 99 + 1, and 1000 is 999 + 1. Unbundling a number that way leaves a pile of 9s, 99s and 999s — all multiples of 9 — plus the digits themselves, so the digits carry the whole remainder.

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.

Keep folding if the sum is still large. 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?

Tag the digits + and − alternately from the right and add them. If the total is 0 or a multiple of 11, so is the number. For 2,728 that is 8 − 2 + 7 − 2 = 11, so 2,728 divides by 11, and indeed 2,728 = 11 × 248.

The signs alternate because 10 is one less than 11, not one more. Ten leaves −1 against 11, a hundred leaves +1, a thousand leaves −1 again, and so on down the places. Each digit therefore counts either forward or backward, and the pattern flips at every step to the left.

How do you test for 6, 12 and 15?

Split the number you are testing for into two factors that share nothing, and apply both rules. Six is 2 × 3, so test for 2 and for 3; twelve is 3 × 4; fifteen is 3 × 5. 9,348 has digits adding to 24 and ends in 48, so it passes both halves of the 12 test, and 9,348 = 12 × 779.

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: take the last digit off, double it, and subtract it from the number that remains. Repeat until you can see the answer. 861 gives 86 − 2 = 84, and 84 gives 8 − 8 = 0, so 861 divides by 7 — it is 7 × 123.

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?". The lesson in the app draws the tests and then asks you to use them.

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

Practice it in the game

Math Challenge asks these as swipe questions mixed in with the rest of the number work, and brings back the ones you miss more often.