Divisibility Tests

Stage: 3, 4 and 5

Article by Tim Rowland

Published January 1997,October 2010,February 2011.

Multiples of 2 and 5

The easiest divisibility tests are for 2 and 5. A number is divisible by 2 if its last digit is even, by 5 if its last digit is 0 or 5.

(In this article 'number' will always mean 'positive whole number')

These tests refer to 'digits' in the (usual) base 10 representation of the number, so that (for example) 2645 represents the number (5×1)+(4×10)+(6×100)+(2×1000). The tests for 2 and 5 work because the rest of the number (apart from the last digit) is a multiple of 10, and so is always divisible by 2 and 5. If the last digit is a multiple of 2 (or 5), then the whole number must be.

Multiples of 4 and 8

Since 100, 1000 and so on are multiples of 4, it follows (as for 2) that a number is divisible by 4if the number represented by its last two digits is a multiple of 4.
Example: 3728 is divisible by 4 because 28 is.

Powers of 10, from 1000 on, are divisible by 8, therefore it follows that a number is divisible by 8 if the number represented by its last three digits is a multiple of 8.
Example: 3728 is divisible by 8 because 728 is.

Note: if you think that you need a calculator to decide whether (for example) 728 is divisible by 8, then it will help you to learn the 8 times table and to practice some divisions which you can check on your calculator until you are confident that you can divide by 8 and don't need the calculator.

Multiples of 3 and 9

A slightly more complicated version of such reasoning gives rise to a test for divisibility by 3.

Now 10 is (3×3)+1, so (for example) 50 is (15×3)+5. To decide whether 57 is divisible by 3, we can take out the 15 lots of 3 in 57 and just check whether the remaining 5+7 is divisibly by 3: which it is, since 5+7=12.

Put slightly differently, we reason that 57=(a multiple of 3)+(5+7).

Therefore 57 is a multiple of 3 if and only if 12 is.

For 257, we note that 100 is (33×3)+1, so 200=(66×3)+2. We looked at 57 above.

Therefore 257=(a multiple of 3)+(2+5+7).

Once again, 257 is a multiple of 3 if and only if the sum of its digits is a multiple of 3.

Actually, that sum is 14, which is a multiple of 3 if and only if 1+4 is.

Since 5 is not a multiple of 3, neither is 257.

In general, 10=9+1, 100=99+1, 1000=999+1 and so on:

every 'power' of 10 (like 10, 100, 1000, 10000 and so on) is just 1 more than a multiple of 3, and so the method for divisibility can be applied to a number with any number of digits.

Example: is 1997 divisible by 3?

Now 1+9+9+7=26, and 2+6=8 which is not divisible by 3.

Therefore 1997 is not divisible by 3.

Note: in this example, we added the digits of 1997, then we added the digits of the answer, and so on, until we arrived at an answer with just one digit, sometimes called the 'digital root' of the original number. So we can say that a number is divisible by 3 if and only if its digital root is 3, 6 or 9.

Because 10=9+1, 100=99+1, 1000=999+1 and so on, we can see that every power of 10 is just 1 more than a multiple of 9, and so the method for divisibility by 3 actually transfers to 9 too: a number is divisible by 9 if and only if its digital root is 9.

Multiples of 6 and 12

A number is divisible by 6 if and only if it is divisible by both 2 and 3.

This is not at all obvious: it is true because 2×3=6 and because 2 and 3 are 'coprime' - i.e. they have no common factor (apart from 1).

Example: 1638 is even and its digital root is 9. Therefore it is a multiple of 6.

Similarly, a number is divisible by 12 if and only if it is divisible by both 3 and 4 - because 3×4=12, and 3 and 4 are coprime.

Multiples of 11

The test for 11 is a modified version of that for 3 and 9.

Whereas every power of 10 is 1 more than a multiple of 3 (or 9), an alternating pattern emerges for multiples of 11. That is to say, 10 is 1 less than 11, 100 is 1 more than 9×11, 1000 is 1 less than 91×11, 10000 is 1 more than 909×11, and so on. If we write `m11' as shorthand for 'a multiple of 11', we see that odd powers of 10 are m11−1, and even powers of 10 are m11+1.

Example:

Is 54637 divisible by 11?

Solution: Start with the units digit and work 'left':

54637=7+3×(m11−1)+6×(m11+1)+4×(m11−1)+5×(m11+1),

which equals m11+(7−3+6−4+5) or m11+11. Therefore 54637 must be a multiple of 11.

This is only slightly more complicated than finding the digital root of a number, because we alternately add and subtract the digits, starting from the right. (We could call the answer the 'alternating digital root').

Example:

What is the remainder when 710−7 is divided by 11?

Solution:

710−7=282475242 whose alternating digital root is 2−4+2−5+7−4+2−8+2=−6. Our number is 6 less than a multiple of 11, so if we divide it by 11, the remainder will be 5.

A number is divisible by 11 if its alternating digital root is 0 or 11 (or any other multiple of 11).