can anyone help me with this? calculate 45^41 mod13 i know i'm probably meant to use fermat's little theorem but i can't see where.. thanks!
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Originally Posted by MariaJJ can anyone help me with this? calculate 45^41 mod13 i know i'm probably meant to use fermat's little theorem but i can't see where.. thanks! $\displaystyle 45\equiv 6\,(mod\,13)\Longrightarrow 45^{41}=6^{41}=\left(6^{13}\right)^3\cdot 6^2=6^3\cdot 6^2=\left(6^2\right)^2\cdot 6=10^2\cdot 6=2\,\,(mod\,\,13)$ Tonio
Originally Posted by tonio $\displaystyle 45\equiv 6\,(mod\,13)\Longrightarrow 45^{41}=6^{41}=\left(6^{13}\right)^3\cdot 6^2=6^3\cdot 6^2=\left(6^2\right)^2\cdot 6=10^2\cdot 6=2\,\,(mod\,\,13)$ Tonio @Tonio-If you can confirm (a yes/no would be sufficient plz) $\displaystyle \left(6^{13}\right)^3\cdot 6^2=6^3\cdot 6^2$ This came from Fermat't little thoerm = $\displaystyle a^p \equiv a mod p$ Correct?
Originally Posted by aman_cc @Tonio-If you can confirm (a yes/no would be sufficient plz) $\displaystyle \left(6^{13}\right)^3\cdot 6^2=6^3\cdot 6^2$ This came from Fermat't little thoerm = $\displaystyle a^p \equiv a mod p$ Correct? Yes ssssir! Tonio
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