I am working on the following exercise: For which values of $a \in \mathbb{Z}$ is the following system of congruences
$$ \begin{alignat}{2} x^{27} & \equiv x^2 && \mod 144 \\ 10x & \equiv a && \mod 25 \\ 2^x - 1 & \equiv 4 && \mod 11 \end{alignat} $$
resolvable? Find the solutions.
From the last equation, I understand that $2^x \equiv 5 \mod 11 \Rightarrow x = 10n + 4, n \in \mathbb{Z}_{\ge 0}$ by Fermat’s little theorem. What to do with the other equations?