I can't seem to figure out this problem. I can factor to reduce the number, but this is too time consuming. Isn't FLT suppose to help here?
Can someone provide clarification please?
Answer
Since $15^{48}$ is nearly $15^{52}$, we can write
\begin{align}
15^{48} &\equiv 15^{52} \cdot 15^{-4} &\pmod{53}\\
&= 15^{-4}, &\pmod{53}
\end{align}
using Fermat's little theorem.
With the extended Euclidean algorithm, one can compute $15^{-1} = -7$, and so
\begin{align}
15^{-4} &\equiv (-7)^4 &\pmod{53}\\
&\equiv 49^2 &\pmod{53}\\
&\equiv (-4)^2 &\pmod{53}\\
&\equiv 16. &\pmod{53}
\end{align}
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