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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.

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    That is what we call Euclid's Lemma. If $a|bc$ and $\gcd(a,b)=1$ then $a|c$. – 2017-02-11
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    Don' say that $60|r(x-y)$ is equal to $60|(x-y)$. It should be $60|r(x-y)$ implies $60|(x-y)$ – 2017-02-11
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    Yes, I will keep in mind. Or maybe like the book say "we know that..."? I'm googling for Euclid's Lemma, see if I can make some more of it. – 2017-02-11
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    The x -y is a red herring. Let $x-y =b$. Then $\gcd(60,r) = 1$ means $60$ and $r$ have no factors in common. If $60|rb$ then all the factors of 60 are factors of $r$ and $b$. But $60$ and $r$ have no factors in common. So the factors are only factors of $b$ and $60|b$. – 2017-02-11
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    @fleablood Yes, it is really clear now! Thanks :-) – 2017-02-11
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    @ΘΣΦGenSan Thanks for heading me in the right direction! – 2017-02-11

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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 $60 \mid (x - y)\,\,$.

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    Thank you, this cleared it up really, really greatly :-) ! – 2017-02-11