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.
 A: As $143=11\cdot13,$ $$x^{11}\equiv12\pmod{143}$$
$\implies(i)x^{11}\equiv12\pmod{11}\equiv1$
But by Fermat's little theorem, $x^{11-1}\equiv1\pmod{11}\implies x\equiv1\pmod{11}\ \ \ \ (1)$
and $(ii)x^{11}\equiv12\pmod{13}\equiv-1,$
Now $12x\equiv x^{12}\equiv1\pmod{13}$ by Fermat's little theorem
$\implies 1\equiv12x\equiv-x\iff \equiv-1\pmod{13}\ \ \ \ (2)$
Using Chinese remainder theorem/by observation, $$x\equiv12\pmod{\text{lcm}(11,13)}$$
A: $x^{11}\equiv12\pmod{143}\implies x^{22}\equiv1\pmod{143}$, 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.
A: Hint:
$$
x^{11\cdot11}=x^{\varphi(143)+1}\equiv x\pmod{143}
$$
Furthermore, note that $12^2\equiv1\pmod{143}$.
