1-17 Let $A$ be the sum of the digits of the decimal number $4444^{4444}$, and let $B$ be the sum of the digits of $A$. Find the sum of the digits of $B$ (all numbers here are in decimal).
[Solution] First, we estimate the number of digits in $10000^{4444}$, which has $4 \times 4444 + 1 = 17777$ digits. Since each digit can be at most 9, and $4444^{4444} < 10000^{4444}$, we have
$$A < 17777 \times 9 = 159993.$$

Now, let's represent $A$ as
$$A = a_{5} \cdot 10^{5} + a_{4} \cdot 10^{4} + a_{3} \cdot 10^{3} + a_{2} \cdot 10^{2} + a_{1} \cdot 10 + a_{0} \cdot 10^{0},$$

where $0 \leqslant a_{i} \leqslant 9 \quad (i=0,1,2,3,4)$, and $a_{5} \leqslant 1$. Therefore, we have
$$B \leqslant 1 + 5 \times 9 = 46.$$

Let $B = b_{1} \times 10 + b_{0}$, where $0 \leqslant b_{0} \leqslant 9, 0 \leqslant b_{1} \leqslant 4$. Similarly, for the sum of the digits of $B$, denoted as $C$, we have
$$C < 4 + 9 = 13.$$

On the other hand, we have
$$\begin{aligned}
4444^{4444} & \equiv 7^{4444} \quad(\bmod 9) \\
7^{4444} & \equiv 7^{4443} \times 7 \\
& \equiv (7^{3})^{1481} \times 7 \\
& \equiv (342 + 1)^{1481} \times 7 \quad(\bmod 9) \\
7^{4444} & \equiv 1^{1481} \times 7 \equiv 7 \quad(\bmod 9)
\end{aligned}$$

Therefore, $4444^{4444} \equiv 7 (\bmod 9) \quad (1)$
Since the sum of the digits of an integer is congruent to the integer itself modulo 9, we have
$$4444^{4444} \equiv A \equiv B \equiv C \quad(\bmod 9)$$

From (1) and (2), we get $C \equiv 7 (\bmod 9)$
Since $C < 13$, the sum of the digits of $B$ is 7.