1. Divisibility Tests
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 5if 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 for2 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 4 if 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.
2. 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 or9.
Because 10=9+1, 100=99+1, 1000=999+1 and so on, we can see that every power
of 10is just 1 more than a multiple of 9, and so the method for divisibility by 3 actually
transfers to 9too: 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.
3. 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).
