Divisibility by 3

The idea

Every multiple of 9 is a multiple of 3, but not conversely; a number is divisible by 3 if the sum of its digits is divisible by 3.

Nines and threes

9 = 3 × 3, so a group of 9 is three groups of 3. Take 18: it is 2 groups of 9, and it is also 6 groups of 3. Now take 15: it is 5 groups of 3.

Which statement is true?

• Every multiple of 9 is a multiple of 3, but not every multiple of 3 is a multiple of 9
• Every multiple of 3 is a multiple of 9
• No multiple of 9 is a multiple of 3

Group the digit sum by 3

Same picture as before, with a choice of group size: 9 or 3. The blue part is always a multiple of 9. Look at how the leftover cells group into 9s and into 3s.

Same trick, smaller groups

Every multiple of 9 is a multiple of 3, but not conversely; a number is divisible by 3 if the sum of its digits is divisible by 3.

Each place value is 1 more than a multiple of 9. And a multiple of 9 is a multiple of 3 too. So each place value is 1 more than a multiple of 3, and only the digit sum can leave something over.

It also works the other way: if the digit sum is not divisible by 3, the number is not divisible by 3.

Multiples of 3, but not of 9: 15 has digit sum 6 33 has digit sum 6 87 has digit sum 15

Each of these digit sums is divisible by 3 but not by 9.

Notes

Every multiple of 9 is a multiple of 3, but not conversely; a number is divisible by 3 if the sum of its digits is divisible by 3.

Check yourself

Which number is divisible by 3 but not by 9?

How many of 135, 284, 1203, 4562, 7041 are divisible by 3?

Answer: 3

The digit sums are 9, 14, 6, 17 and 12. Those divisible by 3 are 9, 6 and 12, so 135, 1203 and 7041 are divisible by 3. That is 3 numbers.

How many digits ★ from 0 to 9 make 58★1 divisible by 3?

Answer: 3

14 + ★ must be a multiple of 3: 15 (★ = 1), 18 (★ = 4) and 21 (★ = 7). So 3 digits work: 5811, 5841 and 5871.

Is every multiple of 3 also a multiple of 9?

• 2475. 2 + 4 + 7 + 5 = 18 is divisible by 9, so 2475 is divisible by 9 as well as by 3.
• 7309. 7 + 3 + 0 + 9 = 19 is not divisible by 3, so 7309 is not divisible by 3.
• 9000. The digit sum is 9, so 9000 is divisible by 9 as well as by 3.
• 5412 — correct. Yes! 5 + 4 + 1 + 2 = 12. 12 is divisible by 3 but not by 9, so 5412 = 3 × 1804 is divisible by 3 but not by 9.
• Yes, because 3 × 3 = 9. It is the other way round: 9 is made of threes, so every multiple of 9 is a multiple of 3.
• No, because 15 is divisible by 3 but not by 9 — correct. Yes! 15 = 3 × 5, but 15 = 9 + 6 leaves 6. Its digit sum 6 is divisible by 3 but not by 9.
• Yes, because they have the same digit sum rule. The rules differ: 3 needs the digit sum divisible by 3, and 9 needs it divisible by 9.