<s>[INST] Find last two digits of $302^{46}$ 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 $2^{46}$.
 [/INST] $2^{46} \equiv 4^{23} \equiv 4 \cdot 4^{22} \equiv 4 \cdot 16^{11} \equiv 4 \cdot 16 \cdot 16^{10} \equiv 4 \cdot 16  \cdot 56^{5}  \equiv$
$\quad 4 \cdot 16  \cdot 56 \cdot 56^{4}  \equiv 4 \cdot 16  \cdot 56 \cdot 36^{2} \equiv$
$\quad  4 \cdot 16  \cdot 56 \cdot 96 \equiv^\text{algorithm complete / now applying discretionary techniques}$
$\quad 4 \cdot 16  \cdot 56 \cdot (-4) \equiv (-1) \cdot 4\cdot 4 \cdot 16 \cdot 56 \equiv (-1)  \cdot 16 \cdot 16 \cdot 56 \equiv$
$\quad (-1) \cdot 56 \cdot 56 \equiv -36 \equiv 64 \pmod{100}$
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