$9 \cdot 7$ 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).
（17th International Mathematical Olympiad, 1975）
[Solution] First estimate the number of digits in $4444^{4444}$.
Since
$$
4444^{4444}<10000^{4444}
$$

and $10000^{4444}$ has $4 \cdot 4444+1=17777$ digits, and each digit can be at most 9,
therefore, $4444^{4444}$ has no more than 17777 digits, and
$$
A<17777 \cdot 9=159993.
$$

Thus, $A$ has no more than six digits.
$$
\text{Let } A=a_{5} \cdot 10^{5}+a_{4} \cdot 10^{4}+a_{3} \cdot 10^{3}+a_{2} \cdot 10^{2}+a_{1} \cdot 10+a_{0},
$$

where $a_{5} \leqslant 1, a_{4}, a_{3}, a_{2}, a_{1}, a_{0}$ are no more than 9.
Therefore,
$$
B \leqslant 1+5 \cdot 9=46.
$$

Thus, $B$ has no more than two digits.
Let $B=b_{1} \cdot 10+b_{0}$, where $0 \leqslant b_{0} \leqslant 9, 0 \leqslant b_{1} \leqslant 4$.
Let the sum of the digits of $B$ be $C$, then
$$
C<4+9=13.
$$

Since if an integer is not a multiple of 9, then the remainder when it is divided by 9 is the same as the remainder when the sum of its digits is divided by 9.
Therefore, we need to find the remainder when $4444^{4444}$ is divided by 9.
Since
$$
4444=493 \cdot 9+7,
$$

we have
$$
\begin{aligned}
4444^{4444} & \equiv 7^{4444} \quad(\bmod 9) \\
7^{4444} & =7^{4443} \cdot 7 \\
& =\left(7^{3}\right)^{1481} \cdot 7 \\
& =(342+1)^{1481} \cdot 7 .
\end{aligned}
$$

Thus,
$$
7^{4444} \equiv 1^{1481} \cdot 7 \quad(\bmod 9).
$$

Therefore,
$$
4444^{4444} \equiv 7 \quad(\bmod 9).
$$

Hence,
$$
4444^{4444} \equiv A \equiv B \equiv C \quad(\bmod 9).
$$

So,
$$
C \equiv 7 \quad(\bmod 9).
$$

Since
$$
C<13.
$$

Therefore,
$$
C=7.
$$

That is, the sum of the digits of $B$ is 7.