I need to find the last two digits of $302^{46}$ without resorting to Euler's theorem or Chinese remainder theorem (they have not been introduce so far in the course; I can user Fermat's little theorem though). This is what I tried:
We have to work $\pmod{100}$ and it is easy to see that:
$302 = 2 \pmod{100}$
So I can write
$302^{46} = 2^{46} \pmod{100}$
I'm stuck here I don't know know to further reduce $2^{46}$.