modular exponentiation large power mod composite number $$x^{11} \equiv 12 \pmod{143}$$
How to solve this without any sophisticated methods?
$$x^{11} \equiv 12 \pmod{143}$$    
$${x^{120}}\equiv 12^{10}\pmod{143} $$
Is this correct knowing that x could divide 143?
Thanks in advance.

$x^{11}\equiv12\pmod{143}\implies x^{22}\equiv1\pmod{143}$, as $12^2=1\pmod{143}$
Fermat's little theorem gives:
$$x^{\varphi(143)}=x^{120}=1\pmod{143}\implies 1^6=(x^{22})^6=x^{132}\equiv x^{12}\pmod{143}$$
So, $x^{11}=12$ and $x^{12}=1$.Noting that $x^{12}$ is an invertible element of $\mathbb{Z}_{143}$, we know that $x$ must also be invertible. We thus divide the two equations and see that $x$ must be the inverse of $12\pmod{143}$ - which is $12$.
We thus have that $x=12$ is our only solution to this equation.
