6. Let $a$ be the sum of the digits of $4568^{777}$, $b$ be the sum of the digits of $a$, and $c$ be the sum of the digits of $b$. Find $c$.
6. Solution: $\because 4568^{7777}<10000^{777}=10^{4 \times 777}$ has a total of $4 \times 7777+1=31109$ digits,
$$
\therefore a<9 \times 31109=279981, b<2+5 \times 9=
$$
$47, c<4+9=13$.
$$
\begin{array}{l}
\because 4568^{7777} \equiv 5^{7777} \equiv 5^{3 \times 2592+1} \equiv 5(-1)^{2592} \equiv \\
5(\bmod 9), \\
\therefore a \equiv b \equiv c \equiv 5(\bmod 9), \therefore c=5 .
\end{array}
$$