1. Given that $A$ is the sum of the digits of $4568^{7777}$ in decimal notation, $B$ is the sum of the digits of $A$, and $C$ is the sum of the digits of $B$, then $C=(\quad)$
A. 5
B. 32
C. 9
D. 14
1. A Hint: Knowing $A \leqslant \lg 4568^{7777} \times 9 < 7777 \times 9 \times 4 = 279972$
$$
\begin{array}{l}
B \leqslant 1 + 5 \times 9 = 46, \\
C \leqslant 3 + 9 = 12 .
\end{array}
$$

Also, $C \equiv 5(\bmod 9), C \in \mathbf{N}_{+}$, so $C=5$.