Solve the congruence relation $$1978^{20}\equiv x \pmod{125}$$
We have
$$125=5^3$$ $$1978^{2}\equiv -1 \pmod 5\Rightarrow 1978^{20}\equiv 1978^{2\cdot10+0}\equiv 1\pmod 5$$
The remainder $x$ should be $26$.
If the remainder $\bmod 5$ is $1$, how to evaluate remainder $\bmod 25$ and combine them to give $x=26$?
Here is an alternative solution:
$1978^1\equiv103\pmod{125}\implies$
$1978^2\equiv103^2\equiv109\pmod{125}\implies$
$1978^4\equiv109^2\equiv6\pmod{125}\implies$
$1978^8\equiv6^2\equiv36\pmod{125}\implies$
$1978^{16}\equiv36^2\equiv46\pmod{125}$
Hence:
$1978^{20}\equiv1978^{4+16}\equiv1978^4\cdot1978^{16}\equiv6\cdot46\equiv26\pmod{125}$
