elementary number theory - Calculating $12^{20} bmod(41)$ by hand

Hi I'm practicing the Pohlig-Helman algorthm right now and I was wondering if I could get an explanation on how to easily compute something like

$12^{20} \bmod(41)$ by hand

I can't think of a smart way to do it, I won't be allowed a calculator on an exam. So any help would be greatly appreciated.

Answer

$$\begin{align}12^{20} &\equiv 3^{20}2^{40} \pmod{41}\\ &\equiv 3^{20} \pmod{41} \text{, Fermat's}\\ &= (3^4)^5 \pmod{41} \\ &\equiv (-1)^5 \pmod{41} , \text{since } 81 = 2(41)-1\\ &\equiv -1 \pmod{41}\end{align}$$