# Math Help - Mod 15

1. ## Mod 15

Q: If 5x=5y (mod 15), then could someone please explain to me why it's NOT true to say that x=y (mod 15) but it IS true to say x=y (mod 3)? Just a little confused as to why I can't divide both sides by 5 in the first instance to give x=y (mod 15) and yet I can for x=y (mod 3).

2. Would a counterexample be enough? Try x=4, y=7.

3. Hello,

$5x=5y \bmod 15 \implies 5x-5y=15k \quad k \in \mathbb{Z}$

Simplifying : $x-y=3k$, that is to say $x=y \bmod 3$

This is partly because if you assume that x-y=15k' (which would yield x=y mod 15), it means that k was a multiple of 5, and nothing lets you say that.