How do I show that for all $x, y \in \mathbb{Z}$, if $13x \equiv 13y\pmod{12}$, then $x \equiv y \pmod{12}$

Question

Essentially the title. I'm trying to wrap my head around modular arithmetic and I got stuck trying to understand this problem in particular. My first thought was to organize the problem like this:

$13x - 13y = 12k$ (for some integer $k$)

$13x - 13y = 13(x - y)$

$13(x - y) = 12k$

But I got stuck here and don't know what to do. Can anyone point me in the right direction? (But pplease don't tell me the answer).

If $~a.~b,~$ and $~c~$ are integers, and $~a ~| ~(b \times c),~$ and $~a,~b~$ are relatively prime, does this imply that $~a~| ~c~?$ — user2661923, Apr 05 '23 at 04:50
By the Congruence Product Rule in the linked dupe $,13\equiv 1\Rightarrow 13x\equiv 1x\equiv x\ \ $ — Bill Dubuque, Apr 05 '23 at 06:20

Prem · Answer 1 · 2023-04-05T06:06:50.893

-1

Here is a hint , which you can complete :

$13=12+1$

Here is a stronger hint :

$13x=13y \mod 12$
$12x+x=12y+y \mod 12$

You can complete it now.

The Same hints work with your attempt too :

$13(x-y)=12(x-y)+(x-y)$

edited Apr 05 '23 at 06:06

answered Apr 05 '23 at 05:47

Prem

9,669

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Apr 05 '23 at 06:21

How do I show that for all $x, y \in \mathbb{Z}$, if $13x \equiv 13y\pmod{12}$, then $x \equiv y \pmod{12}$

1 Answers1