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 :)
 A: 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. 
A: As $(604,2013)=1, (x,2013)=1$
Using Carmichael function ( 1,2  )   $\lambda(2013)=$lcm$(\lambda(61),\lambda(11),\lambda(3)=$lcm$(60,10,2)=60$
So, $x^{60}\equiv1\pmod{2013}$ again $x^{59}\equiv604\pmod{2013}\implies x^{60}\equiv604x\pmod{2013}$
So, we have $604x\equiv1\pmod{2013}$
Method $1:$ By observation, $1\pmod{2013}\equiv 3\cdot2013+1=6040$
$\implies 604x\equiv 6040\pmod{2013}$
$\implies x\equiv10\pmod{2013}$ as  $(604,2013)=1$
Method $2:$ Alternatively, expressing as continued fraction, $$\frac{2013}{604}=3+\frac{201}{604}=3+\frac1{\frac{604}{201}}=3+\frac1{3+\frac1{201}}$$
So, the previous convergent of $\frac{2013}{604}$ is $3+\frac13=\frac{10}3$
Using Theorem 3 of this, $604\cdot10-2013\cdot3=1$
$\implies  604\cdot10\equiv1\pmod{2013}\implies  x\equiv10\pmod{2013}$
