How to compute: $(89^{3} \bmod 79)^4\bmod 26$?

Question

How to compute: $(89^{3} \bmod 79)^4\bmod 26$??

It's easy to calculate it by evaluating $89^{3}$ first and then mod $79$, but it seems stupid to do it this way. Do we have a faster way to evaluate it?

Answer 1

I’m assuming that the version of the problem originally posted is correct, and that what is wanted is $(89^{3} \bmod 79)^4\bmod 26$, where $\bmod$ is the binary operator. First, $89\bmod79=10$, so we want $10^3\bmod79$. $100\bmod79=21$, so $10^3\bmod79=210\bmod79=52$. Now $52\bmod26=0$, so $52^4\bmod26=0^4\bmod26=0$.