This post is a bit mathematical, and talks about the divisibility test of 3 we all learnt in high school. As I’m not aiming for brevity but for completeness and ease of explanation let me remind you all of the theorem involved to test the divisibility of a number by 3 and then move onto its proof.
To test the divisibility of a number N by 3, recursively sum the digits that make up the number till you reach a manageable number whose divisibility of 3 can be inferred trivially.
For example if we’re testing the divisibility of 3423 by 3, we’ll add up 3+4+2+3 and check if that sum is divisible by 3. We know that 12 is divisible by 3, hence the number 3423 is divisible by 3 as well . If we did not know that 12 was divisible by 3 we would have summed up the digits of 12 and and try to see if that sum is divisible by 3 and so on.
Proof for theorem:
The main crux of the proof is representing the number as a specific set of additions and multiplications and your work is almost done. If we have a number with digits , then the number can also be represented as follows
With out proof in mind let’s separate the sum of all the digits of i.e, and see what we get.
Now if we can prove that the right glob we have here is divisible by 3 then our job is done (We’ll know that the if the sum of its digits is divisible by 3 then is divisible by 3).
A number with just 9 as its digits is divisble by 3
Proof for Corollary #1:
We know that any number which is divisible by 9 is divisible by 3 as well . So a number which has digits in it and all digits are 9 can be represented as
On further simplification we get
The above proves the corollary. With the corollary proven we can concur that the sum of the digits of a number should also be divisible by 3 if a number is divisible by 3 and vice versa.
Acknowledgement: I first saw this proof on the One Mathematical Cat website, but the instructor just proved it for a specific case of 3 digits. This was my effort in generalizing the proof for any arbitrary number of digits.