Divisibility by 3 and 9

A number is divisible by 3 if the sum of its digits is divisible by 3.
A number is divisible by 9 if the sum of its digits is divisible by 9.

Examples

Big digit sum? Add again

987,654 → 9 + 8 + 7 + 6 + 5 + 4 = 39 → 3 + 9 = 12 → divisible by 3, not by 9
99,999 → 9 + 9 + 9 + 9 + 9 = 45 → 4 + 5 = 9 → divisible by 9 and by 3

Why it works

Every place value is a number of nines plus 1:

10 = 9 + 1,   100 = 99 + 1,   1,000 = 999 + 1, …

So a number splits into a "nines part" and its digit sum:

324 = 3 × 100 + 2 × 10 + 4 = 3 × 99 + 2 × 9 + (3 + 2 + 4)

The nines part is always divisible by 9 (and by 3), so 324 is divisible by 9 or 3 exactly when 3 + 2 + 4 is. This uses the rule for the divisibility of a sum.

Careful: every number divisible by 9 is also divisible by 3, but not the other way around: 12, 21 and 4,512 are divisible by 3, but not by 9.
The last digit tells you nothing here: 13 and 23 end in 3 but are not divisible by 3, while 27 ends in 7 and is.

Try it yourself

Click a question to see the answer.

1. Which of these numbers are divisible by 3: 111, 205, 7,002, 9,871, 12,345?

111, 7,002 and 12,345 – their digit sums are 3, 9 and 15. The digit sums of 205 and 9,871 are 7 and 25.

2. Which of these numbers are divisible by 9: 7,002, 12,345, 4,536, 70,120?

7,002 and 4,536 – their digit sums are 9 and 18. The digit sums of 12,345 and 70,120 are 15 and 10.

3. Which digit can go in the box so that 47 is divisible by 9?

4 + 7 = 11, and the digit sum must reach 18, so the answer is 7: 477.

4. Which digits can go in the box so that 52 is divisible by 3?

5 + 2 = 7, and the digit sum must be 9, 12 or 15, so the answer is 2, 5 or 8: 522, 552 and 582.

5. Is 111,111,111 divisible by 9?

It has nine 1s, so its digit sum is 9. Yes

6. True or false: a number that ends in 3 is divisible by 3.

False. 13 and 23 end in 3, but their digit sums are 4 and 5.