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Mathematics 17 Online
OpenStudy (anonymous):

Is there a way to calculate\[17^{21}(\text{mod }23)\]without using a calculator?

OpenStudy (anonymous):

From Mathematica: \[\text{Mod}\left[17^{21},23\right]=19 \]

OpenStudy (anonymous):

I know... ... I was thinking perhaps an application of Euler's theorem,\[a^{\phi(n)}\equiv1(\text{mod }n),\]on these numbers:\[17^{\phi(23)}=17^{22}\equiv1(\text{mod }23).\]

OpenStudy (kinggeorge):

This is a little late, but you can solve using the Extended Euclidean Algorithm. The first thing to notice is that if \(p\) is prime, then by Fermat's little theorem \[a^{p-2} \equiv a^{-1}\;\; (\!\!\!\!\!\mod p) \]Since \(21=23-2\), you know that \(17^{21} \equiv 17^{-1} (\!\!\mod 23) \). And this is easy to find using the Extended Euclidean Algorithm.

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