Determine the remainder when $7^{7^{2019}}$ is divided by 47.

Question

Determine the remainder when $7^{7^{2019}}$ is divided by 47.

47 is prime, perhaps we can do something with that? I'm not sure how to approach this question, any and all help is appreciated.

Thanks!

Answer 1

By Fermat's little theorem $$7^{7^{2019}}\equiv 7^{(7^{2019} \mod{46})} \mod{47}$$ To calculate $$7^{2019} \mod{46}$$ We have $$7^{2019}\equiv1^{2019}\equiv1 \mod{2}$$ $$7^{2019}\equiv7^{(2019 \mod{22})}\equiv7^{17}\equiv (7)(49)^8\equiv(7)(3)^8\equiv(7)(27)^2(9)\equiv(7)(9)(4)^2\equiv19\mod{23}$$ $$\therefore 7^{2019} \equiv 19 \mod{46}$$ So, $$7^{7^{2019}}\equiv 7^{19}\equiv(7)(49)^9\equiv(7)(2)^9\equiv(7)(512)\equiv(7)(-5)\equiv-35\equiv12 \mod{47}$$