TITLE: Prove that $13$ divides the number ${}^910 + 23$.
QUESTION [5 upvotes]: Prove that the number $13$ divides the number $\large \left( 10^{10^{10^{10^{10^{10^{10^{10^{10}}}}}}}} + 23\right)$.

REPLY [9 votes]: $$K\equiv 10^{10^{\ldots}}\pmod{13}$$
By Fermat's Little Theorem we have:
$$K\equiv 10^{10^{\ldots}\pmod{12}}\equiv 10^{(-2)^{\ldots}\pmod{12}}\pmod{13}$$
But $(-2)^k\equiv 4\pmod{12}, \forall k\in\mathbb Z_{\ge 2}$.
Hence
$$K\equiv 10^{4}\equiv 3^4\equiv 27\cdot 3\equiv 3\equiv -23\pmod{13}\ \ \ \square$$