6. (USS) ${ }^{1 \mathrm{MO4}}$ Let $A$ be the sum of the digits of the number $16^{16}$ and $B$ the sum of the digits of the number $A$. Find the sum of the digits of the number $B$ without calculating $16^{16}$.
6. Let us denote by $C$ the sum of digits of $B$. We know that $16^{16} \equiv A \equiv$ $B \equiv C(\bmod 9)$. Since $16^{16}=2^{64}=2^{6 \cdot 10+4} \equiv 2^{4} \equiv 7$, we get $C \equiv 7(\bmod$ 9). Moreover, $16^{16}<100^{16}=10^{32}$, hence $A$ cannot exceed $9-32=288$; consequently, $B$ cannot exceed 19 and $C$ is at most 10 . Therefore $C=7$.