I'm reading a solution of the following congruence: $x^{59} \equiv 604 \mod 2013$. It says that it is equivalent to the following system of congruences: $$\begin{cases} x^{59} \equiv 604 \mod 3 \\ x^{59} \equiv 604 \mod 11\\ x^{59} \equiv 604 \mod 61\end{cases}$$ Why?
EDIT:
I know that $2013=3*11*61|x^{59}-604$. But why is this information sufficient to say that $3,11,61$ all divide $x^{59}-604$ when considered separately?
Hint $\ $ If $\,m,n\,$ are coprime then $\,{\rm lcm}(m,n) = \color{#c00}{mn},\ $ therefore
$$ a\equiv b\!\!\!\pmod{\!m,n} \iff m,n\mid a-b\color{#0a0}\iff \color{#c00}{mn}\mid a-b\iff a\equiv b\!\!\!\pmod{mn}$$
Applied twice yields the claim. This is a special case of CRT = Chinese Remainder Theorem.
Remark $\ $ The $\,\rm\color{#0a0}{middle}\,$ equivalence employs the universal property of lcm, $ $ i.e.
$$ m,n\mid k\color{#0a0}\iff {\rm lcm}(m,n)\mid k$$
Hint
Have you tried to multiply $3$, $11$ and $61$? Just to see what happens.
More hints
If $ab$ divides $c$, then $c=(ab)d=a(bd)=b(ad)$ for some $d$, that is, both $a$ and $b$ divide $c$.
Conversely, if $a$ and $b$ are coprime and $a$ and $b$ divide $c$, then we can write $ax+by=1\:(*)$ and $c=aj=bk$. Multiply $(*)$ then by $c$ to obtain $$c=acx+bcy=abkx+abjy=ab(kx+jy)$$
