9.9 Define $\left\{a_{n}\right\}_{n=1}^{\infty}$ as follows:
$a_{1}=1989^{1989}$, and $a_{n}(n>1)$ is equal to the sum of the digits of $a_{n-1}$, what is $a_{5}$?
(21st Canadian Mathematics Competition, 1989)
[Solution] Since $a_{1}=1989^{1989}<10000^{1989}$,
and $10000^{1989}$ has $4 \cdot 1989+1=7957$ digits.
Therefore, $a_{1}$ has no more than 7957 digits.
By the problem statement, $a_{2}$ is the sum of the digits of $a_{1}$, then
$$
a_{2}<10 \cdot 8000=80000 .
$$

So $a_{2}$ is at most a 5-digit number, hence
$$
\begin{array}{l}
a_{3} \leqslant 7+4 \cdot 9=43, \\
a_{4} \leqslant 4+9=13 .
\end{array}
$$

Thus, $a_{5}$ is a single-digit number.
Since any positive integer is divisible by 9 if and only if the sum of its digits is divisible by 9.
Since $9 \mid 1+9+8+9=27$, then
$$
9 \mid 1989^{1989} \text {. }
$$

Therefore,
$$
9 \mid a_{5} \text {. }
$$

That is,
$$
a_{5}=0 \text { or } 9 .
$$
$a_{5}=0$ is obviously impossible, so $a_{5}=9$.