How to use modular arithmetic to find a rule for the divisibility of 13?

Question

I am failing, I managed to read a proof for the number 9, unfortunately I can't seem to have a clear idea of how to do it for 13.

I started by decomposing it in terms of 10's... but that's as far as I could get.

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

0

As $40-13\cdot3=1,$ we can write $4(10x+y)-13\cdot3x=x+4y$

So, $10x+y$ will be divisible by $13\iff x+4y$ is

Similarly we can use, $7\cdot13-9\cdot10=1$

Also as $10^3\equiv-1\pmod{13},$

$\sum_{r=0}^na_r10^r\equiv(a_0+10a_1+100a_2)-(a_3+10a_4+100a_5)+(a_6+10a_7+100a_8)-\cdots$

See also: Divisibility criteria for $7,11,13,17,19$

edited Apr 13 '17 at 12:21

Community

answered Jan 23 '15 at 06:20

lab bhattacharjee

13 will be divisible by 13 iff 1 + 10(3) is divisible by 13 according to the second formula? I am confused since 31 is not. – JOX Jan 23 '15 at 09:17
@DanielOrtizCosta, Which formula, you are referring to? Please share the steps – lab bhattacharjee Jan 23 '15 at 09:18
Your formula with the sum – JOX Jan 23 '15 at 09:19
Nevermind, I was looking at it inverted. – JOX Jan 23 '15 at 09:20

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