solve congruence $x^{59} \equiv 604 \pmod{2013}$ This is an exercise from my previous exam; how should I approach this?

Solve congruence $\;x^{59} \equiv 604 \pmod{2013}$

Thanks in advance :)

Hint We have that $3 \cdot  11\cdot 61=2013$. Break up your congruence into three.
By $x^2\equiv 1\mod 3$, the first one turns into $x\equiv 1\mod 3$, for example, since we can deduce $3\not\mid x$. Glue back using CRT.
ADD Just in case you want the solution. First we may write $x^{59}\equiv 604\mod 3$ as $ x^{2\cdot 27}x\equiv 1\mod 3$. The last equation reveals$3\not\mid x$, so  $x^2\equiv 1\mod 3$ and $x\equiv 1\mod 3$. The second one can be reduced to $x^{59}\equiv 10\mod 11$ which again reveals $11\not\mid x$. Thus $x^{10}\equiv 1\mod 11$ and then $x^{-1}\equiv 10\mod 11$ which gives $x\equiv 10\mod 11$. Finally we have $x^{59}\equiv 55\mod 61$. Again $61\not\mid x$ so $x^{60}\equiv 1\mod 61$ and we get $x^{-1}\equiv 55\mod 61$. Using the Euclidean algorithm, we find $55\cdot 10-61\cdot 9=1$ so $x\equiv 10\mod 61$. Thus we have that $$\begin{cases}x\equiv 1\mod 3\\x\equiv 10\mod 11\\x\equiv 10\mod 61\end{cases}$$
One may apply now the Chinese Remainder Theorem, or note $x=10$ is a solution of the above. 
