Divisible by 2, 3 or 5?
Spot it in seconds
You don't need to do the whole division. There are quick tricks that tell you straight away whether a number splits up evenly. Here are the three easiest ones.
Dividing by 2
This is the one you already know: even numbers. You only have to look at the last digit. If it ends in 0, 2, 4, 6 or 8, the whole number divides by 2.
| Number | Last digit | Divisible by 2? |
|---|---|---|
| 34 | 4 | Yes ✓ |
| 57 | 7 | No ✗ |
| 130 | 0 | Yes ✓ |
| 9 999 | 9 | No ✗ |
Why only the last digit? Because 10, 100, 1000… are all divisible by 2. So the tens, hundreds and thousands always split evenly on their own — only the units digit can leave anything behind.
Dividing by 5
Same idea, just as easy: look at the last digit. The multiples of 5 are 5, 10, 15, 20, 25… — notice they always end in 5 or 0. So if a number ends in 0 or 5, it divides by 5.
| Number | Last digit | Divisible by 5? |
|---|---|---|
| 45 | 5 | Yes ✓ |
| 90 | 0 | Yes ✓ |
| 72 | 2 | No ✗ |
| 1 005 | 5 | Yes ✓ |
Dividing by 3 — the digit trick
Three is sneakier: the last digit won't help you. Instead, add up all the digits. If that total divides by 3, then so does the whole number.
| Number | Add the digits | Total ÷ 3? | Divisible by 3? |
|---|---|---|---|
| 27 | 2 + 7 = 9 | Yes | Yes ✓ |
| 41 | 4 + 1 = 5 | No | No ✗ |
| 123 | 1 + 2 + 3 = 6 | Yes | Yes ✓ |
| 5 281 | 5 + 2 + 8 + 1 = 16 | No | No ✗ |
If the total is still big and you're not sure, just do the trick again on that total:
Quick recap
| Divides by | What to look at | It works when… |
|---|---|---|
| 2 | the last digit | it's 0, 2, 4, 6 or 8 |
| 5 | the last digit | it's 0 or 5 |
| 3 | the sum of all digits | that sum divides by 3 |
For the curious: why does the digit trick for 3 work?
This part is for older kids (or anyone who likes to know why). The trick isn't magic — it falls straight out of how our number system is built.
Every place value — 10, 100, 1000 — is just one more than a string of 9s:
And 9, 99, 999 … are all divisible by 3. Now let's break a number apart using that. Take 471:
= 4×(99+1) + 7×(9+1) + 1
= (4×99 + 7×9) + (4 + 7 + 1)
= a multiple of 3 + the digit sum (12)
The first bracket is always a multiple of 3 (because 99 and 9 are). So the only part that decides whether 471 divides by 3 is the digit sum, 4 + 7 + 1 = 12. Since 12 divides by 3, so does 471.
This works for any number: split each digit into "(a bunch of 9s) + 1", and the bunches of 9s always vanish into a multiple of 3, leaving just the sum of the digits. That's the whole secret.