The last two digits of $217^{382}$

Question

I am currently solving this problem but it's getting out of my head so I want to ask your help in order to resolve it.

Find the last two digits of $217^{382}$

I'm supposed to use Euler's theorem. Thank you in advance.

Answer 1

Euler's Theorem says that $a^{\phi(n)}\equiv 1\pmod n$. Since $100=2^2\cdot 5^2$, we have $\phi(100)=(4-2)(25-5)=40$. Now

\begin{align} 217^{382}&\equiv 17^{382}\\ &\equiv (17^{40})^9(17^{22})\\ &\equiv 17^{22}\pmod {100} \end{align}

Now $17^2=289\equiv 89 \pmod {100}$, so we have

\begin{align} 217^{382}&\equiv 89^{11}\\ &\equiv 89^5\cdot 89^5\cdot 89\\ &\equiv 49\cdot 49\cdot 89\\ &\equiv 89. \end{align}

I'll leave it to you to find a snappier way to end it if you want, but the problem becomes fairly manageable once you get to the second half.