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 :)

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, $604\cdot10-2013\cdot3=1$
$\implies  604\cdot10\equiv1\pmod{2013}\implies  x\equiv10\pmod{2013}$
