$\ 111^{333}+333^{111} is \ divisible \ by \ 7 $

$\ 111^{333}+333^{111} $ = $\ 111^{111} (111^{222}+3^{111}) $ = We know $ 7 \nmid 111 and hence, -7 \nmid 111^{111} $ Therefore, we have to show that $ 7 \mid (111^{222}+3^{111}) $ $ \therefore 11 = (15 \times 7) + 6$ $\therefore$ 111 = 6(mod7) = -1 (mod 7) $\therefore 111^{222} = (-1)^{2^{111}} = 1 (mod 7)$ --------------1 on the other hand, $\ 3^3 = (3 \times 7) +6$ $\ 3^3 = 6(mod 7) = -1 (mod 7)$ -----------------2 $ \therefore 111 = 37 \times 3, we get from 2$ $ \therefore 3^{111} = [(3)^3]^{37} (mod 7)$ $\ = [-1]^{37} (mod 7)$ $ \ 3^{111} =-1 (mod 7)$ ------------3 $\ 111^{333}+333^{111} $ = (1-1) (mod 7) $\ 111^{333}+333^{111} $ = 0 (mod 7) **Hence, $\ 111^{333}+333^{111} $ is divisible by 7**