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!
modular arithmetic
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!
Best Answer
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, $$7^{7^{2019}}\equiv 7^{19}\equiv(7)(49)^9\equiv(7)(2)^9\equiv(7)(512)\equiv(7)(-5)\equiv-35\equiv12 \mod{47}$$