I have this specific congruence relation, and although solving it with a calculator is trivial, I'd like to know what strategy I can use to solve this by hand.
The relation is
$79^7 \equiv x \mod 403$.
(I also have two other similar problems, but I assume they will be about the same in terms of solving.)
I've looked at Fermat's little theorem, but haven't been able to successfully apply it to this problem, though I suspect that would be the most likely candidate.
Thanks for the help.
In general (when fermat's little theorem is not applicable, or there is not some nice trick) we do the following. Since $79^2$ is larger than 403, reduce it mod 403. We know that $79^7=79^279^279^279$. You can replace every $79^2$ with its value reduced mod 403. Call this value $y$. You problem is now reducing $y^379$ modulo 403. See if $y^2$ is bigger than 403, if it is, perform the aforementioned process again. If not, work with $y^3$ and then finish the problem by reducing by that value times 79 modulo 403.