For example, what is the last digit of 17^(3241)? It would help a ton if someone could show how to solve this using Fermat's Little Theorem. I have finals coming up and this was something I never quite understood
For example, what is the last digit of 17^(3241)? It would help a ton if someone could show how to solve this using Fermat's Little Theorem. I have finals coming up and this was something I never quite understood
Here is a solution via Fermat's Last Theorem (although it's not necessarily the simplest):
By Fermat's Last Theorem, $\displaystyle 17^4 \equiv 1 \mod 5$,
so since
$\displaystyle 17^{3241} = (17^4)^{810} \cdot 17$,
$\displaystyle 17^{3241} \equiv 1 \cdot 17 \equiv 2 \mod 5$.
We also have $\displaystyle 17 \equiv 1 \mod 2$, so $\displaystyle 17 ^ {3241} \equiv 1 \mod 2$.
Since $\displaystyle 17^{3241} \equiv 2 \mod 5$ and $\displaystyle 17 ^ {3241} \equiv 1 \mod 2$,
$\displaystyle 17^{3241} \equiv 7 \mod 10$.
We could use the Chinese Remainder Theorem at the last step, but it seems like overkill here.