Question Number 210719 by lmcp1203 last updated on 17/Aug/24
$${find}\:\mathrm{63}!^{\mathrm{36}!} {mod}\mathrm{97}\:\:\:{thanks} \\ $$
Answered by A5T last updated on 19/Aug/24
$$\phi\left(\mathrm{97}\right)=\mathrm{96} \\ $$$$\mathrm{36}!\equiv\mathrm{0}\left({mod}\:\mathrm{96}\right)\Rightarrow\mathrm{36}!=\mathrm{96}{k} \\ $$$$\left(\mathrm{63}!,\mathrm{97}\right)=\mathrm{1}\Rightarrow\mathrm{63}!^{\mathrm{36}!} =\left(\mathrm{63}!\right)^{\mathrm{96}{k}} \equiv\mathrm{1}\left({mod}\:\mathrm{97}\right) \\ $$
Answered by lmcp1203 last updated on 17/Aug/24
$${thank}\:{you} \\ $$