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Divisibility Tests

Stage:3, 4 and 5

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 $(5\times 1)+(4\times 10)+(6\times 100)+(2\times 1000)$. 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 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 need a calculator to decide whether (for example) 728 is divisible by 8, then this test is obviously of limited practical use!

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 $(3\times 3)+1$, so (for example) 50 is $(15\times 3)+5$. 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 $(33\times 3)+1$, so $200=(66\times 3)+2$. 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 $2\times 3=6$ and because 2 and 3 are `coprime' - i.e. they have no common factor (apart from 1).\par 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 $3\times 4=12$, 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 $9\times 11$, 1000 is 1 less than $91\times 11$, 10000 is 1 more than $909\times 11$, 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 $m11+1$.

Is 54637 divisible by 11?

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

$54637=7+3\times(m11-1)+6\times(m11+1)+4\times(m11-1)+5\times(m11+1)$,\par 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').

Solution:

$7^{10}-7=282475242$ 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).

INTERLUDE: Remainder arithmetic

What is the final digit of $34\times 57$?

Without doing the full multiplication, we know it must be 8, because 8 is the final digit of $4\times 7$.

But how do we know?

Because $34\times 57=(30\times 57)+(4\times 57)=m10+(4\times 50)+(4\times 7) =m10+200+8=m10+8$.

In this example `final digit' means `remainder when divided by 10'. To find the remainder in the product (of 34 and 57), we need only find the product of the remainders (4 and 7). This rule works, for essentially the same reason, for remainders when we divide by numbers other than 10.

Once upon a time, schoolchildren were taught a special case of this, called `casting out nines'. Suppose, for example, I work out $256\times 77$ by long multiplication, and I get the answer 19612. The remainders when I divide 256 and 77 by 9 are their digital roots: 4 and 5 respectively. The product of 4 and 5 is 20, with digital root (remainder) 2. My answer 19612 should also have a digital root of 2; in fact it has a digital root of 1, so I must have made a mistake in my long multiplication!

Beware: casting out nines may detect a wrong answer (like that above) but it cannot guarantee a correct one. For example, 19721 has digital root 2, but the `final digit test' (which could be called `casting out tens') shows that it cannot be the answer to $256\times 77$.

[Attending to remainders is the essence of 'modular arithmetic'. The genius C F Gauss gave the first formal account of this in his 1801 book Disquitiones Arithmeticae, which he published at the age of 24].

Multiples of 7

A test for divisibility by 7 (or any number, in principle) can be devised using remainder arithmetic, as follows. The remainder when 10 is divided by 7 is 3, so the remainder when 100 ($=10\times 10$) is divided by 7 is $3\times 3=9$ (which is $7+2$, so the actual remainder is 2).

[Editor's note:

$10 = m7 + 3$ so $10^2 = (m7+3)^2= (m7)^2 + 6\times m7 + 3^2 = m7 + 9$]

Using our shorthand: $10=m7+3$ so $100=m7+9=m7+2$.

Hence $1000=m7+2\times 3=m7+6$, or $m7-1$.

Special Tests

With some ingenuity, particular tests can be contrived for some integers, such as:

• 7: "Double the units and subtract from the tens", e.g. $1365\rightarrow 136-(2\times 5)=126\rightarrow 12-(2\times 6)=0$. If the chain ends in zero or a multiple of 7, then the original number is divisible by 7. Some help with this can be found if you look at an old AskNRICH discussion thread .
• 11: "Subtract the units from the tens", e.g. $1364\rightarrow 136-4$ etc.
  If the chain ends in zero, then the original number is divisible by 11.
• 13: "Add the tens to 4 times the units", e.g. $1365\rightarrow 136+20$ etc.
  If the chain ends in a multiple of 13, then the original number is divisible by 13.
• 19: "Add the hundreds to 4 times the rest", e.g. $1311\rightarrow 13+44$ etc.
  If the chain ends in a multiple of 19, then the original number is divisible by 19.

Can you explain why each of these four tests works?

By a happy 'coincidence', 1001 is the product of 7, 11 and 13. This fact underlies another test for divisibility by these three prime numbers: see the entry for '1001' in David Wells' fascinating book 'The Penguin Dictionary of Curious and Interesting Numbers'.

Composite Numbers

Suppose $m$ is composite (i.e. not prime) and $m=a b$ where $a$, $b$ are coprime.\par As remarked above (for 6, 12), $N$ is divisible by $m$ if $N$ is divisible by both $a$ and $b$. This can be extended to writing $N$ as a product of more than two factors, as in the following:

Applying each of the tests for 4, 9 and 11 shows that 351 648 792 is divisible by each of these three factors, and therefore by 396.

Final Problem: If the digits 5, 6, 7 and 8 are inserted at random in 3_1_4_0_92 (one in each space), what is the probability that the number created will be a multiple of 396?