Monday, December 16, 2019

I am stuck on Fermat's Little Theorem. I know how to apply it, but does it apply here $15^{48}$ mod $53$.



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?



FLT problem


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}



No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...