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 $2^{46}$.
Best Answer
So you wanna calculate $2^{46}$ modulo $100$. For that note that $$2^{46}=(2^{20}\times 2^{3})^2=((2^{10})^2\times 8)^2=(24^2\times 8)^2=(76\times 8)^2=(8)^2=64$$in $\mathbb Z/100\mathbb Z$. Thus, $2^{46}\equiv 64\pmod{100}$.