<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] You can seperate and do it $\pmod{2}$ and $\pmod{25}$  and use chinese remainder:
They are both odd so their sum is even and thus $\equiv 0 \pmod{2}$.
The euler function of 25 gives 20 and thus $3^{333}\equiv 3^{13}\pmod{25}$ and $7^{777}\equiv 7^{17} \pmod{25}$.
Now, $7^2=49\equiv -1 \pmod{25}$. Thus $7^{17}=7^{16}\cdot 7 \equiv 7\pmod{25}$.
$3^3=27\equiv 2 \pmod{25}$. Thus $3^{13}=3^{12}\cdot 3\equiv 2^4\cdot 3 \equiv -2 \pmod{25}$.
Thus their sum is $5\pmod {25}$.
In conclusuion, $3^{333}+7^{777} \equiv 30 \pmod{50}$.
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