Find the last $3$ digits of $13^{1201}$.

Question

I believe I am supposed to use a mod in this problem. I am unsure exactly how to do so. Help would be much appreciated and thanks in advance!

Answer 1

This corresponds to finding $13^{1201}\bmod 1000$ since the last three digits are in between $0$ and $999$.

Now, since $13$ and $1000$ are co-prime; Euler's theorem says $$13^{\varphi(1000)}\equiv 1\bmod 1000,$$ where $\varphi(x)$ is Euler's totient function evaluated at $x$.

Now $\varphi(1000)=400$ so $13^{400}\equiv 1$, that is, $13^{1200}\equiv 1$ and so $$13^{1201}\equiv 13.$$