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New Maths?

Updated: Nov 20, 2019

Note from the Author:


No - it's not strictly "New" Maths but...


  • It was discovered independently by an 11 year-old boy who was very excited to share his discovery so it was NEW to him.

  • None of the teachers at his school - nor anyone they consulted - had seen it before so it was NEW to all of them, too.

  • I had only seen something like it as an exercise in Modular Arithmetic (and not as a divisibility test) so when I was asked to come up with a watertight proof that 11 year olds could understand (see below) that was NEW to me, too.

At the time, I did an extensive internet search and could only find one small reference tucked away inside a Wikipedia page - without the proof below...


...and, finally, I never actually said it was "NEW" Maths. There was always a question mark in the title (and in the first line of the article) and I am very grateful to those who have answered that question by pointing out other places where they have seen this divisibility test before - especially those who did it with good grace!


Let us all celebrate creativity in Mathematics and be grateful that there are 11 year olds out there who do this sort of thing for fun!


Simon Ellis

20th November 2019


A new test for divisibility by 7?


My sister, Mary Ellis, is the Head of Maths at Westminster Under School. She was very excited when one of her pupils, Chika Ofili, presented her with a test for divisibility by 7 which he had thought of while “bored” during the long summer holiday…


Chika's test goes like this:


Take any number (say 532), split off the last digit (2), multiply it by 5 and add it to the rest of the number (53). If the result is divisible by 7 then so is the original number. The process can be repeated until you get a number that you recognize as being divisible by 7 (or not).


To test 532, we find “Chika’s number”, in this case 53 + 2×5 = 63, and spot that it is divisible by 7. This means, according to the method, that 532 is divisible by 7, which it is!


To test 977, Chika’s number is 97 + 7×5 = 132. Now, we may not know if this is divisible by 7 so we repeat the process on 132. Chika’s number this time is 13 + 2×5 = 23 which is NOT divisible by 7 and so, according to the method, neither is 132 or 977.


It seems to work (play with some other numbers…) but a method such as this is just conjecture until it is PROVED to work…


…so, early the next morning, my sister rang me up and asked if I could help…

“Of course,” I said and one of the two proofs I came up with before breakfast follows.


Enjoy!


Proof of Chika’s Divisibility Test for 7


Let 𝑛 be the number we are testing and 𝑐 be “Chika’s number” (remember this!).


We write 𝑛 = 10𝑎 + 𝑏 (𝑏 is the final digit and 𝑎 is the rest of the number)


then 𝑐 = 𝑎 + 5𝑏 (the rest of the number plus 5 times the final digit)


First we need to show that if Chika’s number (𝑎 + 5𝑏) is divisible by 7 then the number we are testing (10𝑎 + 𝑏) is also divisible by 7.


If 𝑐 is divisible by 7


then 𝑐 = 7𝑘 for some 𝑘 ∈ ℤ where ℤ is the set of integers (whole numbers)


and 𝑎 + 5𝑏 = 7𝑘


so 𝑎 = 7𝑘 − 5𝑏


then 𝑛 = 10(7𝑘 − 5𝑏) + 𝑏


so 𝑛 = 70𝑘 − 50𝑏 + 𝑏


so 𝑛 = 70𝑘 − 49𝑏


so 𝑛 = 7(10𝑘 − 7𝑏)


which, since (10𝑘 − 7𝑏) ∈ ℤ, means 𝑛 is divisible by 7.


So far, so good.


But all we have shown is that IF 𝑐 is divisible by 7 THEN so is 𝑛.


Now we need to show that IF 𝑛 is divisible by 7 THEN so is 𝑐 (ie the method always works).


Here we go again...


If 𝑛 is divisible by 7


then 𝑛 = 7𝑙 for some 𝑙 ∈ ℤ


and 10𝑎 + 𝑏 = 7𝑙


so 𝑏 = 7𝑙 − 10𝑎


then 𝑐 = 𝑎 + 5(7𝑙 − 10𝑎)


so 𝑐 = 𝑎 + 35𝑙 − 50𝑎


so 𝑐 = 35𝑙 − 49𝑎


so 𝑐 = 7(5𝑙 − 7𝑎)


which, since (5𝑙 − 7𝑎) ∈ ℤ, means 𝑛 is divisible by 7.


So we have shown that:

IF 𝑐 is divisible by 7 THEN so is 𝑛

and IF 𝑛 is divisible by 7 THEN so is 𝑐


But, there is still one more question to consider:


What if 𝑐 IS divisible by 7 for some 𝑛 that is NOT?


Let’s check this, too...


If 𝑛 is NOT divisible by 7


then 𝑛 = 7𝑚 + 𝑞 for some 𝑚 ∈ ℤ and 𝑞 ∈ {1,2,3,4,5,6}


so 10𝑎 + 𝑏 = 7𝑚 + 𝑞


so 𝑏 = 7𝑚 + 𝑞 − 10𝑎


then 𝑐 = 𝑎 + 5(7𝑚 + 𝑞 − 10𝑎)


so 𝑐 = 𝑎 + 35𝑚 + 5𝑞 − 50𝑎


so 𝑐 = 35𝑚 − 49𝑎 + 5𝑞


so 𝑐 = 7(5𝑚 − 7𝑎) + 5𝑞


Since 5𝑞 cannot have a factor of 7 if 𝑞 ∈ {1,2,3,4,5,6}, 𝑐 does not have a factor 7 and therefore Chika’s number will not be divisible by 7 unless the original number 𝑛 is.


So we have now shown that the method will never give a “false positive” and our proof is complete!


Further thoughts – and another proof – to follow very soon…


Simon Ellis

18th September 2019

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