HAPPY NEW YEAR $2020$ Remainder Problem

$20^{19}=100^9\cdot4^9\cdot20=100^9\cdot4^{10}\cdot5=100^9\cdot1024^2\cdot5 \equiv - 14^2\cdot5=-980 \equiv 30 \; (\mod 101)$

Let’s start with a simpler problem:

What is the remainder of $20^{19}$ when divided by $101$?

We can solve this by Exponentiation by Squares, at each step, just squaring the previous result. This is easy enough to do by hand.

$$20^1\equiv20\pmod{101},$$ $$20^2\equiv97\pmod{101},$$ $$20^4\equiv16\pmod{101},$$ $$20^8\equiv54\pmod{101},$$ $$20^{16}\equiv88\pmod{101}.$$

Since $19=16+2+1$, our desired remainder will be

$$20^{19}=20^{16}\times20^2\times20^1\equiv30\pmod{101}.$$

Finally, using that $a\equiv b\pmod{c}$ iff $ak\equiv bk\pmod{ck}$, for any non-zero $k$, we can deduce

$$20^{20}\equiv\boxed{600}\pmod{2020}.$$