<s>[INST] Solve a congruence $1978^{20}\equiv x\pmod{125}$ I have checked the solution ($x=26$).
Solving modulo $5$ gives 
$$1978^{20}\equiv 1978^{2\cdot 10}\equiv 1\pmod{5}$$
Solving modulo $25$ also gives
$$1978^{20}\equiv 1\pmod{5}$$
How to evaluate the remainder $x$?
 [/INST] Note that, modulo $125$, $11^2=121\equiv-4$ and $2^7=128\equiv3$.
Therefore, $1978^{20}\equiv(-22)^{20}\equiv 2^{20}11^{20}\equiv 2^{20}(-4)^{10}\equiv 2^{40}\equiv2^{35}2^5\equiv3^52^5\equiv6^5\equiv26$.
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