# Count On

## Explorer

Further Quick Divisibility Tests

**Test for multiples of 5**

- If the last digit is either 5 or 0, then the whole number is a multiple of 5.

Here is an interactive panel for you to practice on:

**Test for multiples of 10**- This is easy - if the last digit is 0, the whole number is a multiple of 10, otherwise it is not!

For example, 3295 ends in 5 so is*not*divisible by 10. But 3290 ends with 0 and so is a multiple of 10.

Combining Divisibility Tests

Now we know quick methods for divisibility by 2, 3, 4, 5, 8, 9, and 10. With these we can easily test for divisibility by

*any number that is a product of 2, 3, 4, 5, 8, 9 and 10*!

So, for example, let's take 6, the smallest number missing from the list above.

Quick tests for Divisibility by 6

Since 6 = 2 times 3, numbers divisible by 6 must be divisible by both 2 AND by 3.

For example:

- 432?
- This is a multiple of 2 because it ends in 2 which is even;

It is also a multiple of 3 since its digits sum to 4+3+2=9 which is a multiple of 3.

So 432*is*a multiple of 6. - 514?
- This is a multiple of 2 because it ends with 4 which is even;

Its digit sum is 5+1+4=10 which is not a multiple of 3, so 514 is not a multiple of 3.

Since 514 has failed one of the tests for divisibility by 2 AND by 3, it is NOT a multiple of 6. - 513?
- This time, 513 has a digit sum of 5+1+3=9, and so 513 is exactly divisible by 3;

However, it ends with 3 which is not even, so 513 is not a multiple of 2.

Again, one of the two tests has been failed, so 513 is NOT a multiple of 6.

Quick test for Divisibility by 12

To be divisible by 12, a number must also be divisible by 2 and 3 and 4 and 6, that is, by all the

**factors**of 12. So a multiple of 12 must pass

*all*the divisibility test for 2, 3, 4 and 6. But we have just seen that, if it is divisible by both 2 and 3, it is automatically divisible by 6, so we do not need to test for 6 explicitly.

Similarly, if a number is divisible by 4 it must automatically be a multiple of 2 also. So testing for divisibility by 4 is enough and we can forget about testing for 2 as well.

So we need to test just for divisibility by 3 and by 4 only - that's sufficient!

For example:

- 432?
- Divisible by 4? The last 2 digits are 32 and 32 is a multiple of 4, so, yes, 432 is divisible by 4.

Divisible by 3? The digit sum is 4+3+2=9 which is a multiple of 3, so, yes, 432 is divisible by 3.

Therefore 432 is also divisible by 12. - 428?
- Divisible by 4? Yes.

Divisible by 3? 4+2+8=14, not a multiple of 3, so No!

So 428 is*not*a multiple of 12. - 435?
- Divisible by 4? No because 35 is not a multiple of 4.

So even if it was divisible by 3 (which it is) it won't be a multiple of 12.

You can make up more divisibility tests of your own for 15, 18, 20, 24 and so on.

Here is an interactive panel for you to practice on: back toPrimes, Factors and Divisibility |