333. Find the remainder of the division of the number $\left(85^{70}+19^{32}\right)^{16}$ by 21.
333. $85^{70}=(84+1)^{70}=21 n+1$; since $\varphi(21)=12$, then $19^{12}-1$ is divisible by 21; therefore, $19^{32}=$ $=19^{8}\left(19^{24}-1\right)+19^{8}=21 m+19^{8} . \quad$ Therefore $\quad\left(85^{70}+\right.$
$.\left.+19^{32}\right)^{16}=\left(21 n+1+21 m+19^{8}\right)^{16}=[21 k+1+(21-$

$.\left.-2)^{8}\right]^{16}=\left(21 q+1+2^{8}\right)^{16}=(21 r+5)^{16}=21 t+5^{16}=$

$=21 t+5^{4}\left(5^{12}-1\right)+5^{4}=21 t+21 r+625=21 u+16$.

The required remainder is 16.