29, expressed in decimal; the integer $a$ consists of 1985 eights, and the integer $b$ consists of 1985 fives. What is the sum of the digits in the decimal representation of the integer $9ab$?
(A) 15880,
(B) 17856;
(C) 17865;
(D) 17874;
(E) 19851.
29. Due to the last digit of $9 a b$ being zero, the sum of its digits differs from the sum of the digits of the number $N=\frac{9 a b}{10}$. Additionally, if the number $M$ is formed by $h$ digits $d$ (where $d$ is an Arabic numeral) arranged in a row, then
$$
\begin{array}{c}
M=d d d \cdots d=\frac{d}{9}(999 \cdots 99) \\
k \uparrow \\
=\frac{d}{9}\left(10^{k}-1\right) .
\end{array}
$$

Thus, $N=\frac{9 a b}{10}=\frac{9}{10}\left(\frac{8}{9}\left(10^{1985}-1\right)\right)$
$$
\begin{aligned}
& \cdot\left[\frac{5}{9}\left(10^{1985}-1\right)\right] \\
= & \frac{4}{9}\left[10^{2 \cdot 1985}-2 \cdot 10^{1985}+1\right] \\
= & \frac{4}{9}\left[\left(10^{2 \cdot 1985}-1\right)-2\left(10^{1985}-1\right)\right] \\
= & \frac{4}{9}\left(10^{2 \cdot 1985}-1\right)-\frac{8}{9}\left(10^{1985}-1\right) \\
= & P-Q .
\end{aligned}
$$

Here, $P$ consists of $2 \cdot 1985$ fours, and $Q$ consists of 1985 eights. Subtracting these, we see that $N$ is composed of 1984 fours, 1 three, 1984 fives, and 1 six. Therefore, the sum of the digits of $N$ is $1984(4+5)+3+6=1985 \times 9$ $=17865$.
(C) Ans.