<s>[INST] Find $3^{333} + 7^{777}\pmod{ 50}$ As title say, I need to find remainder of these to numbers. I know that here is plenty of similar questions, but non of these gives me right explanation. I always get stuck at some point (mostly right at the beginning) and don't have idea how to start. 
Thanks in advance. 
 [/INST] I always start by writing a couple of powers, in this case modulo $50$:
$$\begin{align}3^1\equiv 3&\mod 50\\
3^2\equiv 9&\mod 50\\
3^3\equiv 27&\mod 50\\
3^4=81\equiv 31&\mod 50\\
3^5\equiv 93\equiv 43&\mod 50\\
3^6\equiv 129\equiv 29&\mod 50\\
3^7\equiv 87\equiv 37&\mod 50\\
3^8\equiv 111\equiv 11&\mod 50\\
3^9\equiv 33&\mod 50\\
3^{10}\equiv 99\equiv-1&\mod 50\\
\end{align}$$
Oh look, a pattern! If $3^{10}\equiv -1\pmod {50}$, this means that $3^{20}\equiv 1\pmod {50}$,
so from here, I know that $3^{333}=(3^{20})^{16}\cdot 3^{10}\cdot 3^3\equiv 1\cdot(-1)\cdot 27=-27\equiv 23\pmod {50}$.

For the $7$, it's even easier, since $7^2\equiv -1\pmod {50}$, meaning that $7^{777}=(7^2)^{388}\cdot 7\equiv (-1)^{388}\cdot 7=7\pmod {50}$
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