Today I was marking my year 9 classes books and came across some work on prime factor decomposition and tests of divisibility. Yesterday I had been arguing with Colin Beveridge (@icecolbeveridge) about the use of formula triangles, (see the comment section on this post) and any other method of anything that was presented in the fashion “do it this way and don’t worry about the why.” I felt a wave of hypocrisy flow over me. I had never explained the tests of divisibility to the class, and furthermore I didn’t even know myself why they worked! I thought I’d explore them and see what I came up with.
If the last digit is divisible by 2, the number is divisible by 2.
This test of divisibility is obvious, the definition of an even number is that it has 2 as a prime factor, and all even numbers have a last number divisible by 2. No further exploration needed. I know why this one works and I’m certain the class do too.
If the sum if the digits is a multiple of 3, then the number is divisible by 3.
This is a fact I’ve known since I was at primary school back in the 1980s, but I can honestly say that I’ve never thought about why it works. I wrote a couple if equations out, realised I should be working in modulo 3 and came up with this:
Which basically boils down to the fact any number can be split into the digits multiplied by a powerful of ten.
A= (x0)10^0 + (x1)10^1 +… + (xn)10^n
As (10^n) mod 3 = 1 for all natural numbers (and 0) then it follows that:
(A)mod 3 = (x0 +x1+…+xn) mod 3
Which implies our test of divisibility. This also implies the test if divisibility for 9 (ie is the sum of digits is divisible by 9 then so is the number) as (10^n)mod9 = 1 for all natural numbers and 0. To prove you just follow the exact working but in mod 9.
A number is divisible by 4 if it’s last two digits are divisible by 4.
This one uses the fact that if two numbers are both divisible by a number then so is their sum.
We know any number bigger than 100 can be expressed 100a + b where a and b are natural numberso and b<100. We also know that as 4 is a factor of 100, 100a is divisible by 4 if a is a natural number. Hence the whole number is divisible by 4 iff the best is divisible by 4, and b is the last two digits.
From this we can deduce the test of divisibility for 8 (ie a number larger than 1000 is divisible by 8 iff the last three digits are divisible by 8) as 1000 is divisible by 8. The proof is the same, but you split the number into 1000a + b where a and b are natural numbers and b is less than 1000.
Testing for 5 and 6
Testing for 5 (the last number is 5 or 0) is another obvious one that needs no further exploration. And the test for 6 is simple, if 2 and 3 are prime factors 6 must be a factor, as all numbers can be expressed as a product of their prime factors and 6 is the product of 2 and 3.
Testing divisibility for 10
Again, this is obvious as we are talking about a base ten system! That leaves just one more test.
To test for divisibility by 7, take the last digit, double it. Take this away from the number you are left with if you subtract the last number and divide by ten. If the answer is 0 or divisible by 7 (ie if the answer mod 7 = 0) then the original number us divisible by 7.
What a ridiculously complicated divisibility test. It looks much nicer algebraically:
(10a + b) mod 7 = 0 (a and b are natural numbers, b < 10)
iff
(a -2b)mod7 = 0
Once I expressed it algebraically I had two simultaneous equations and solved like this:
I enjoyed working through them and now I’m confident I know why these tests work, but I’m fairly sure that the ones that aren’t obvious would be too complicated to explain to year 9. This made me think about their use, and I certainly think it’s fine to use them even without understanding. Which brought me back to the formula triangles debate. This is one that will rage, but as I explained in my post on it, I don’t have a great issue with people who understand the algebra using them. I also don’t have a great issue with people who struggle at maths and aren’t going to pursue it past GCSE using them. My issue is with higher attainers who could and should understand the algebra being taught then with no deeper conceptual understanding. I guess it’s a topic I will need to think more on.
cavmaths 
Tests of divisibility
Today I was marking my year 9 classes books and came across some work on prime factor decomposition and tests of divisibility. Yesterday I had been arguing with Colin Beveridge (@icecolbeveridge) about the use of formula triangles, (see the comment section on this post) and any other method of anything that was presented in the fashion “do it this way and don’t worry about the why.” I felt a wave of hypocrisy flow over me. I had never explained the tests of divisibility to the class, and furthermore I didn’t even know myself why they worked! I thought I’d explore them and see what I came up with.
If the last digit is divisible by 2, the number is divisible by 2.
This test of divisibility is obvious, the definition of an even number is that it has 2 as a prime factor, and all even numbers have a last number divisible by 2. No further exploration needed. I know why this one works and I’m certain the class do too.
If the sum if the digits is a multiple of 3, then the number is divisible by 3.
This is a fact I’ve known since I was at primary school back in the 1980s, but I can honestly say that I’ve never thought about why it works. I wrote a couple if equations out, realised I should be working in modulo 3 and came up with this:
Which basically boils down to the fact any number can be split into the digits multiplied by a powerful of ten.
A= (x0)10^0 + (x1)10^1 +… + (xn)10^n
As (10^n) mod 3 = 1 for all natural numbers (and 0) then it follows that:
(A)mod 3 = (x0 +x1+…+xn) mod 3
Which implies our test of divisibility. This also implies the test if divisibility for 9 (ie is the sum of digits is divisible by 9 then so is the number) as (10^n)mod9 = 1 for all natural numbers and 0. To prove you just follow the exact working but in mod 9.
A number is divisible by 4 if it’s last two digits are divisible by 4.
This one uses the fact that if two numbers are both divisible by a number then so is their sum.
We know any number bigger than 100 can be expressed 100a + b where a and b are natural numberso and b<100. We also know that as 4 is a factor of 100, 100a is divisible by 4 if a is a natural number. Hence the whole number is divisible by 4 iff the best is divisible by 4, and b is the last two digits.
From this we can deduce the test of divisibility for 8 (ie a number larger than 1000 is divisible by 8 iff the last three digits are divisible by 8) as 1000 is divisible by 8. The proof is the same, but you split the number into 1000a + b where a and b are natural numbers and b is less than 1000.
Testing for 5 and 6
Testing for 5 (the last number is 5 or 0) is another obvious one that needs no further exploration. And the test for 6 is simple, if 2 and 3 are prime factors 6 must be a factor, as all numbers can be expressed as a product of their prime factors and 6 is the product of 2 and 3.
Testing divisibility for 10
Again, this is obvious as we are talking about a base ten system! That leaves just one more test.
To test for divisibility by 7, take the last digit, double it. Take this away from the number you are left with if you subtract the last number and divide by ten. If the answer is 0 or divisible by 7 (ie if the answer mod 7 = 0) then the original number us divisible by 7.
What a ridiculously complicated divisibility test. It looks much nicer algebraically:
(10a + b) mod 7 = 0 (a and b are natural numbers, b < 10)
iff
(a -2b)mod7 = 0
Once I expressed it algebraically I had two simultaneous equations and solved like this:
I enjoyed working through them and now I’m confident I know why these tests work, but I’m fairly sure that the ones that aren’t obvious would be too complicated to explain to year 9. This made me think about their use, and I certainly think it’s fine to use them even without understanding. Which brought me back to the formula triangles debate. This is one that will rage, but as I explained in my post on it, I don’t have a great issue with people who understand the algebra using them. I also don’t have a great issue with people who struggle at maths and aren’t going to pursue it past GCSE using them. My issue is with higher attainers who could and should understand the algebra being taught then with no deeper conceptual understanding. I guess it’s a topic I will need to think more on.
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