How to tell if a number is divisible by 3

There is no way to see a multiple of 3 by looking at the end of a number: 12, 13 and 15 all end in different digits and only two of them work. The digits have to be added up instead, and that turns a hard question into one a child can do in their head.

The rule

A number is divisible by 3 when its digits add up to a multiple of 3.

How to check it

  1. Add all the digits of the number together.
  2. If the total is still too big to judge, add its digits together again.
  3. If the final total is 3, 6 or 9, or any multiple of 3, then the original number divides by 3.

Worked examples

Is 4620 divisible by 3?

4 + 6 + 2 + 0 = 12

12 is 3 × 4, so yes. 4620 = 3 × 1540.

Is 1147 divisible by 3?

1 + 1 + 4 + 7 = 13

13 is not a multiple of 3. 12 and 15 are the nearest. So no.

Is 987 654 divisible by 3?

9 + 8 + 7 + 6 + 5 + 4 = 39 → 3 + 9 = 12

12 is a multiple of 3, so yes. Adding twice is allowed and changes nothing.

Why adding the digits works

Every ten is 9 + 1. Every hundred is 99 + 1. Every thousand is 999 + 1. All those 9s, 99s and 999s divide by 3 perfectly, so they can never be the reason a number fails.

What is left over from each place is just the digit itself: one for each ten, one for each hundred, and so on. Adding the digits adds up exactly the leftovers, which is why their total decides the whole question.

Written out for 4620: it is 4 × (999 + 1) + 6 × (99 + 1) + 2 × (9 + 1) + 0. Everything in brackets is a pile of 3s; what remains is 4 + 6 + 2 + 0.

Where it goes wrong

Questions people ask

Does the order of the digits matter?

No. 132, 321 and 213 all have digits adding to 6, and all three divide by 3. Shuffling the digits of a multiple of 3 always gives another multiple of 3.

Why does my teacher call this casting out nines?

Casting out nines is the same idea used for 9, and the 3 rule is its smaller sibling: both come from every power of ten being one more than a run of 9s.

Is there a quicker way for very long numbers?

Yes. Cross out any digits that already add to 3, 6 or 9 as you go, and add only what is left. 3, 6 and 9 themselves can always be ignored.

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