A number theory problem by Rajasree Basu

$1990 = 2 * 5 * 199$

By Euler's Theorem / Fermat's Little Theorem:

For any prime $p$ and any positive integer $a$ , $a^{p-1} \cong 1 \ (mod \ p)$ .

This means that $a^n \cong a^{n-1} \ (mod \ 2)$ ;

$a^n \cong a^{n-4} \ (mod \ 5)$ ;

And $a^n \cong a^{n-198} \ (mod \ 199)$

The Least Common Multiple of $1$ , $4$ and $198$ is $396$

As $396|1980$ , $2^{1990} \cong 2^{10} \cong 1024 \ (mod \ 1990)$