46.1.4 ** Define the sequence $\left\{a_{n}\right\}$ as follows: $a_{1}=1989^{1989}, a_{n}$ is the sum of the digits of $a_{n-1}$, what is $a_{5}$?
Because $a_{1}<10000^{1989}=b_{1}, b_{1}$ has a number of digits $4 \times 1989+1=7957$, so $a_{2}<10 \times 8000=80000$, hence $a_{2}$ has at most 5 digits, thus $a_{3} \leqslant 5 \times 9=45, a_{4} \leqslant 4+9=$ 13, therefore $a_{5}$ must be a single digit.
On the other hand, since $9 \mid 1989$, it follows that $9\left|a_{1}, 9\right| a_{2}, 9\left|a_{3}, 9\right| a_{4}, 9 \mid a_{5}$, so $a_{5}=9$.