I framed a new question just now. What is the Remainder when the number $20^{20}$ is divided by $2020$
My try:
$$\frac{20^{20}}{2020}=\frac{20^{19}}{101}$$
Now Consider: $$20^{18}=(400)^9=(404-4)^9=101k-2^{18}$$
Now i was trying to find Remainder without calculator or by manual division.
On
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
$$20^{20}\equiv\boxed{600}\pmod{2020}.$$
$20^{19}=100^9\cdot4^9\cdot20=100^9\cdot4^{10}\cdot5=100^9\cdot1024^2\cdot5 \equiv - 14^2\cdot5=-980 \equiv 30 \; (\mod 101)$