$\begingroup$How to solve $$23^{{2020}^{2020}} \mod 37.$$ Below given is my understanding of trying to solve the problem.
From $$x^{p-1} = 1 \mod p$$ I deduce that $$23^{2020} \mod 37$$ would be $$23^{56.36+4} \mod 37$$ which is further simplified as $$23^{4} \mod 37$$ as $$23^{\alpha .36} = 1 \mod 37$$
Keeping the above in mind, I am wondering if there is anyway of solving $$23^{{2020}^{2020}} \mod 37.$$ I'm clueless about how to simplify the double exponent.
$\endgroup$52 Answers
$\begingroup$To explain in more detail what Ross Millikan suggested,
note that $2020^{2020}\equiv0\pmod4$ and $2020^{2020}\equiv 4^{336\times6+4}\equiv4^4\equiv4\pmod9$
($4^6\equiv1\pmod9$ by Euler's Theorem),
so $2020^{2020}\equiv4\pmod{36}$ by the Chinese Remainder Theorem.
$\endgroup$$\begingroup$From $x^{36} \equiv 1 \pmod {37}$ what you care about is the exponent $\bmod 36$. Now you need to evaluate (not solve) $2020^{2020} \pmod {36}$. The factors of $2$ are easy, as you quickly have two of them. Then you are only interested in evaluating it $\bmod 9$. Back to you.
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