- Simon Ellis

# 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**