Chika’s Test

Something very exciting happened last Friday when one of my pupils, Chika Ofili, popped into the classroom and asked if he could tell me something he had thought of over the summer holidays. I was intrigued.

I had given him a book called First Steps for Problem Solvers (published by the UKMT) to look at over the holidays and inside the book was a list of the divisibility tests, which are used to quickly work out whether a number is exactly divisible by either 2, 3, 4, 5, 6, 7, 8 or 9 before you actually start dividing. Except that there was no test listed for checking divisibility by 7. The reason why it was missing is because there is no easy or memorable test for dividing by 7, or so I thought!

In a bored moment, Chika had turned his mind to the problem and this is what he came up with. He realised that if you take the last digit of any whole number, multiply it by 5 and then add this to the remaining part of the number, you will get a new number. And it turns out that if this new number is divisible by 7, then the original number is divisible by 7. What an easy test!

For example, take the number 532

53 + 2 x 5 = 63
63 is a multiple of 7, so 532 is a multiple of 7 (and therefore divisible by 7)

Or take the number 987

98 + 7 x 5 = 133
13 + 3 x 5 = 28
28 is a multiple of 7, so both 133 and 987 are multiples of 7

In fact if you actually keep going, you will always end up with either 7 or 49, if the original number is divisible by 7.

For example, take the number 2996

299 + 6 x 5 = 329
32 + 9 x 5 = 77
7 + 7 x 5 = 42
4 + 2 x 5 = 14
1 + 4 x 5 = 21
2 + 1 x 5 = 7

7 is a multiple of 7 and so is 21, 14, 42, 77, 329 and the original number 2996.

The opposite is also true in that if you don’t end up with a multiple of 7, then the original number is not divisible by 7.

For example, take the number 114

11 + 4 x 5 = 31
3 + 1 x 5 = 8

And since 8 is clearly not divisible by 7, neither is 31 nor our original number 114.

Chika demonstrated his test to 8E before lunch on Friday and no one could find a counter-example to disprove it. But mathematically we needed to find a proof. Once a theorem is proved mathematically it will always be true. Pythagoras proved his famous theorem over 2500 years and when he died, he knew it would never be disproved. It is important to realise that a result cannot be proved by finding thousands or even millions of results which support it.

I woke up early on Saturday morning still thinking about Chika’s test and rang my younger brother, Simon Ellis, who also happens to teach Maths, and asked him if he had ever come across the test. He hadn’t and immediately set about trying to write a proof. An hour later he sent me a watertight algebraic proof! He also discovered that the test works if you start by multiplying the last digit by 12, 19, 26, 33 … and then add it to the remaining part of the number. And furthermore, it works if you double the last digit and then subtract it from the remaining part of the number, of if you multiply the last digit by 9, 16, 23, 30 … and subtract. And actually the doubling and subtracting test can be easily found on the internet.

But we both agreed multiplying the last digit by 5 and adding it to the remaining part of the number is much more appealing! Mr O’Donnell urged me to tell the school about Chika’s discovery, so an impromptu demonstration followed in Monday’s assembly. We are now trying to figure out how to make this test more widely known, as it is so simple that it really should be taught alongside the other divisibility tests.

For all those interested, the proof can be currently found on my brother’s website: www.simonellismaths.com/post/new-maths

Also, to be entirely accurate, the test for divisibility by 6 was also missing from the book but this is because you only have to apply both the divisibility by 3 test and divisibility by 2 test (as 2 and 3 are coprime) to a number to determine if it is divisible by 6.

And finally thank you to Stephan for telling me on Tuesday about a Russian test, which involves multiplying the truncated number (without its last digit) by 3 and then adding the last digit. It does work but it is harder on the whole. So for example, take 154: 15 x 3 + 4 = 49, so since 49 is a multiple of 7, so is 154.

Miss Mary Ellis
HEAD OF MATHEMATICS
Author of The Aliens Have Landed and 174 Other Problems