House of Maths School Workshops Primary & Secondary in Dorset & South - A STRANGE TEST FOR DIVISIBILITY BY 7

Seven is a difficult number!

It’s really easy to see whether a whole number is divisible by $2$ (it has to end in $0, 2, 4, 6$ or $8$), by $3$ (the sum of its digits has to be divisible by $3$) or by $5$ (it has to end in $0$ or $5$). But what about $7$?

Here’s one way we can do it. We’ll then see some examples, and then see why this bizarre test works!

To see if a whole number is divisible by 7: remove the last digit, double it and subtract that from what’s left. If this gives you a number that’s divisible by 7 then your starting number is also divisible by 7.

Wow – what a strange looking test! Let’s see it in action:

AN EXAMPLE:
I wonder if $119$ is a multiple of $7$? Let’s subtract double the last digit from what’s left:
$11-(2\times9) = 11-18 = -7$.
$-7$ is a multiple of $7$ (it’s $-1\times 7$). This means that $119$ is also divisible by $7$ (in fact $119 = 17\times 7$).

ANOTHER EXAMPLE:
Let’s see if $1352$ is divisible by $7$:
We subtract twice the final digit from what’s left:
$135-(2\times 2)=131$.
Hmm, I still don’t know if $131$ is a multiple of $7$, so let’s go round again:
$13-(2\times 1) = 11$.
$11$ is NOT a multiple of $7$, meaning that neither is $131$ or $1352$.

WHEN WOULD I ACTUALLY USE THIS?
If this all seems a bit complicated: well yes, I agree! BUT it’s particularly quick and easy for short numbers that end in $1, 2$ or $3$, such as $91$ or $133$.
At first glance, both of these numbers look prime: they certainly don’t divide by $2, 3, 5$ or (if you know your times tables up to $12 \times 11 = 132$) by $11$
Let’s check for divisibility by $7$:
$91→9-2=7$ so YES
$133→13-6=7$ so YES – both $91$ and $133$ are both divisible by $7$ ($91=7\times 13$, and $133=7\times 19$).

Before we see why this strange trick works, why not try your skills out by playing Is This Prime? The game shows you a number (they all seem to be less than $150$ so it’s not too hard), all you have to do is decide if you think it is a prime number then click YES or NO. It’s great fun, although even after months of trying I cannot score more than $32$ points in a minute!! How many can you score?

OPTIONAL: WHY DOES THIS WEIRD TRICK WORK?
To answer this we will need a little algebra, and an intermediate result called a lemma. Algebra is the language of patterns, so it’s usually the best way to see why number tricks like this work! A lemma is just a small, less important result that we need to prove a bigger more important result.

LEMMA: multiplying a number by 10 does not change whether or not it is divisible by 7
For instance, $91$ is divisible by $7$ (it’s $7\times13$), this means that $910$ is too. $198$ is not divisible by $7$,this means that $1980$ isn’t divisible by $7$ either.
One way to see why it works is to think about the number written as the product of its prime factors (e.g.$198 = 2\times3^2\times11$). Multiplying by ten will add a $5$ and a $2$ to the product ($1980= 2^2\times3^2\times11\times 2 \times 5$). This cannot affect whether or not $7$ appears in the list! At a more sophisticated level, multiplying by $10$ does not affect a number’s divisibility by $7$ because $7$ and $10$ are coprime (they have no prime factors in common). If you’re happy to accept that the lemma works, we can now look at our main result:

Suppose we want to test a number for divisibility by $7$: let’s call this our starting number. I’m going to call the final digit $b$, and what’s left after removing it I will call $a$. After subtracting twice $b$ from $a$, we reach our end number.
EXAMPLE: if our starting number is $161$ then $a=16$ and $b=1$. Our end number is $16-(2\times1) = 14$
Our starting number is a moved one place to the left (you can do this by multiplying by $10$) and then with $b$ tagged on the end. So
start number = $10a+b$
end number = $a-2b$ because you subtract twice the final digit $b$ from what’s left $a$.
Algebra can give us a clever relationship between the two:
$10a+b=10(a-2b) + 21b$ (you can expand the brackets to check that both sides are the same).
The term $21b$ on the right is ALWAYS a multiple of $7$. This means that our starting number $10a+b$ will be a multiple of $7$ exactly when $10(a-2b)$ is a multiple of $7$ (because then all three terms are multiples of $7$). And our lemma from earlier tells us that $10(a-2b)$ is a multiple of $7$ exactly when our end number $a-2b$ is a multiple of $7$.

Now we know why test works, try using it to decide which of these numbers are divisible by $7$:
$182, 203, 231, 371, 471$.

And please do have a go at the Is This Prime game!