Last updated on September 26th, 2022 at 11:00 pm
No matter whether you liked or hated mathematics in your school, I can say you must have loved fancying about knowing the divisibility tests of 2, 3, 5 & 10. And if you were among the brainy students you also knew the divisibility tests of 4, 6, 8 & 11. But, what about the divisibility test of 7? Do you know about it? In today’s article, we’ll talk about the previous and newly discovered divisibility tests of 7.
The divisibility of a number by 7 is not as easy as the other divisibility test. In order to find it out, you need to follow a couple of recursive steps. Most importantly, there has been a divisibility test of 7 with well-defined steps. But the recent discovery by Chika Ofili about the same has been a hot topic in the headlines.
The Foregoing Test
Step 1: Separate the last digit of the number.
Step 2: Double the last digit and subtract from the remaining number.
Step 3: Repeat the steps unless you get a number within 0-70.
Step 4: If the result is divisible by 7, the number you started with is also divisible by 7.
About Chika’s discovery
A 12-year old Nigerian boy, Chika Ofili, made history this year after his new discovery in the field of mathematics. He was awarded at the TruLittle Hero Awards for discovering the new divisibility test of 7, popularly called as Chika’s Test.
Miss Mary Ellis, Ofili’s teacher and the head of the Mathematics Department of Westminster Under School, said that she gave all the students a book called First Steps for Problem Solvers (published by the UKMT) to study during the holidays. The book had several divisibility tests but it had no memorable test for 7. After the vacations, Ofili came to her saying that he had discovered a formula for the divisibility of 7 and also had the algebraic proof to back it up.
Chika’s Test
It is similar to the previously foregoing rule, but it is simpler and faster than it.
Step 1: Separate the last digit of the number.
Step 2: Multiply the last digit by 5 and add it to the remaining number.
Step 3: Repeat the steps unless you get a number within 0-70.
Step 4: If the result is divisible by 7, the number you started with is also divisible by 7.
Proof of Chika’s Test
Honestly, I am not sure about the proof Chika had to prove his discovery. This is just a supporting proof of the working rules and credibility of Chika’s Test.
Let us say, a number N is divisible by 7.
( N = 7K, where K is an integer )
N = 7K = 10nan + 10n-1an-1 + 10n-2an-2 + 10n-3an-3 + ….. + 100a2 + 10a1 + a0
= 10[ 10n-1an + 10n-2an-1 + 10n-3an-2 + ….. + 10a2 + a1] + a0
Adding and subtracting ‘50a0’ on the RHS .
N = 10[ 10n-1an + 10n-2an-1 + 10n-3an-2 + ….. + 10a2 + a1] + 50a0 -50a0 + a0
N = 10[ 10n-1an + 10n-2an-1 + 10n-3an-2 + ….. + 10a2 + a1 + 5a0 ] – 50a0 + a0
N = 7K = 10[ 10n-1an + 10n-2an-1 + 10n-3an-2 + ….. + 10a2 + a1 + 5a0 ] – 49a0
7K + 49a0 = 10[ 10n-1an + 10n-2an-1 + 10n-3an-2 + ….. + 10a2 + a1 + 5a0 ]
Let P = 10n-1an + 10n-2an-1 + 10n-3an-2 + ….. + 10a2 + a1 + 5a0
7K + 49a0 = 10P
Now, LHR : (7K + 49a0) ≡ 0 (mod 7) => RHS : 10P ≡ 0 (mod 7)
Also, 10 ≡ 3 (mod 7) => for RHS : 10P ≡ 0 (mod 7) , P ≡ 0 (mod 7)
Therefore, this proves Chika’s Test working rule and credibility.
Well, you may say how is this beneficial or faster? Yes, it is better than the foregoing test, reason being,
- multiplying by 5 helps to reach a number within 0-70 at a faster rate compared to multiplication by 2.
- adding two numbers is psychologically simpler than subtraction.
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Abhijeet is a 4th-year Undergraduate Student at IIT Kharagpur. His major inclination is towards exploring the science behind the things of our day-to-day life.
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Abhijeet Mahatohttps://www.scilynk.in/author/abhijeet-mahato/
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Abhijeet Mahatohttps://www.scilynk.in/author/abhijeet-mahato/
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Abhijeet Mahatohttps://www.scilynk.in/author/abhijeet-mahato/
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Abhijeet Mahatohttps://www.scilynk.in/author/abhijeet-mahato/
Did the child provide the proof of his algorithm as well? What I intend to know, is if he got the result by a trial-and-error approach, or had a well-informed hunch about his test.
Hello Chandrashekar!!
The Algorithm was completely based on Chika’s Observation and Trial-and-Error Approach. Prof. Simon Ellis (the younger brother of Mary Ellis), who also taught mathematics, tested and wrote an algebraic proof for the theory proposed by Chika.
The proof of this test seems to allow false positives… Ie numbers which aren’t divisible by 7 but still satisfy the test. Because it’s already assuming the number to be divisible by 7. Is there a complete proof?
Can we not consider it as positive osculator for 7, in opposition to the already present and well established concept of the negative osculator of 7 which is given as negative 2?
Hey Alex, nice observation, ideally the last digit is being multiplied by a positive osculator (+5) and added to the rest. But, I am not sure where Chika knew about the Osculator Method of checking divisibility.
I wrote a paper that was published in the Spring 1995 issue of the Pi Mu Epsilon Journal that shows a method to derive a divisibility test for any prime. It shows how to find both the positive and negative osculators. Note that the positive and negative osculators are related in that 5 + 2 = 7. Once you’ve found one osculator, you’ve found the other.
Perhaps you feel that removing only one digit at a time and adding or subtracting is still rather slow. My paper also shows how one can find the osculators used to remove more than one digit at a time.
I forgot to add that there are a couple typos in the originally published version of my paper:
a) The value in the table, column y=8 for prime= 89 should be 2, not 22.
b) The equation on top of page 97 has the signs wrong. Instead of +, +, +, and – they should be -, +, -, and +.
We know that for a number ab if the a – 2b is divisible by 7 then ab is divisible by 7 ⇒ It means a-2b=7q ⇒ a=7q+2b ⇒ a+5b=7q+7b=7(q+b), this is the simplest proof of why a+5b can be us for divisibility of 7 test.