In my math book it says:
"This means that $60 \mid r(x - y)$. But, since $\gcd(r, 60) = 1$, we know that $60 \mid (x - y)$."
I would like to know why when gcd(r, 60) = 1, we can say that $60 \mid r(x - y)$ is equal to $60 \mid (x - y)$. It must be simple, but I'm not seeing it.
To provide a very simple and intuitive explanation of the Euclid's lemma in this case: if $60 \mid r(x - y) \,\,$, then all prime factors of $60$ must be included in those of $r(x - y)\,$. On the other hand, if the gcd of $60$ and $r $ is $1$, they have no common prime factors. So all prime factors of $60$ must be included in those of $x - y \,$, and then $60 \mid (x - y)\,\,$.