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. 

$$7^2\equiv-1\pmod{50}$$
$$\implies7^{777}\equiv(7^{2})^{388}\cdot7\equiv(-1)^{388}7\equiv7$$
and
$$3^{333}=3(10-1)^{166}$$
Now $$(10-1)^{166}=(1-10)^{166}\equiv1-\binom{166}110\pmod{100}\equiv1-60\equiv41$$
