Are the divisibility tests rooted in number theory?

Question

I had a good look at some mathematics I was doing at age 9. I remembered the divisibility tests we used to do and I thought that I could take a shot at proving them.

I managed to prove it for 3.

It is usually stated as follows:

If the sum of digits of a number is divisible by 3, then the number is divisible by 3.

This one is not that hard.

$10 \bmod 3=1 \implies (k \cdot 10^n) \bmod 3 \equiv k$ for $n \in \mathbb{N}$ and any one-digit number $k$

The divisibility of a number by 3 can therefore be contingent on the sum of it's digits.

My question is, how would you prove the divisibility tests for 7 and 11? And can you then create any divisibility test you want?

Answer 1

You can find a very good reference of the divisibility by $11$ at this page: https://artofproblemsolving.com/wiki/index.php/Divisibility_rules/Rule_for_11_proof While the divisibility by $7$, is explained here: https://artofproblemsolving.com/wiki/index.php/Divisibility_rules#Divisibility_Rule_for_7

Answer 2

$7\times11\times13=1001,$ so $1000\equiv-1\bmod 7,11,13$,

so the following test works for divisibility of $N$ by $7, 11, $ and $13:$

partition $N$ into 3 digit numbers from the right $ (d_3d_2d_1,d_6d_5d_4,…). $

The alternating sum $(d_3d_2d_1−d_6d_5d_4+d_9d_8d_7−…)$ is divisible by $7, 11,$ or $13$

if and only if $N$ is divisible by $7, 11,$ or $13,$ respectively