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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?

Nεo Pλατo
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    $7, 11, $ and $13$ are factors of $1001$ – J. W. Tanner Mar 24 '20 at 19:02
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    See for example this post. – Dietrich Burde Mar 24 '20 at 19:05
  • https://math.stackexchange.com/questions/328562/divisibility-criteria-for-7-11-13-17-19 – lab bhattacharjee Mar 24 '20 at 19:10
  • @ J. W. Tanner Wouldn't that make them have the same divisibility tests? – Nεo Pλατo Mar 24 '20 at 19:15
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    Partition $N$ into 3 digit numbers from the right ($d_3d_2d_1,d_6d_5d_4,\dots$). The alternating sum ($d_3d_2d_1 - d_6d_5d_4 + d_9d_8d_7 - \dots$) is divisible by 7, 11, or 13 if and only if $N$ is divisible by 7, 11, or 13, respectively – J. W. Tanner Mar 24 '20 at 19:25
  • And $10\equiv 1\pmod 9$ and that's the $9$ test for you. And $10\equiv -1 \pmod {11}$ so $k10^{2k}\equiv k\pmod{11}$ while $k10^{2k+1}\equiv -k\pmod {11}$ so the $11$ divisibility test of adding the digits in even positions and subtracting the digits of the odds. I can never remember the $7$ test. But it's pobably provable. Every number divides some $111....111\times 10^n$ so you can make up some rules on that but they may not be practical. – fleablood Mar 24 '20 at 19:56
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    @ J.W. Tanner Got it, thanks – Nεo Pλατo Mar 24 '20 at 20:01

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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

Matteo
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$7\times11\times13=1001,$ 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

J. W. Tanner
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