$32 \cdot 9$ in decimal notation, the integer $a$ consists of 1985 eights, and the integer $b$ consists of 1985 fives. What is the sum of the digits of the integer $9ab$ in decimal notation?
(A) 15880.
(B) 17856.
(C) 17865.
(D) 17874.
(E) 19851.
(36th American High School Mathematics Examination, 1985)
[Solution] From the given, we know $9|9ab$, and $1+5+8+8=22, 1+9+8+5+1=24$, so $(A), (E)$ are not suitable.
$$
\begin{array}{l}
\text { Also, } 17856=1984 \cdot 9, \quad 17865=1985 \cdot 9, \quad 17874=1986 \cdot 9. \\
\text { Also, } 9ab=9 \cdot \underbrace{88 \cdots 8}_{1985 \uparrow} \cdot \underbrace{55 \cdots 5}_{1985 \uparrow} \\
=9 \cdot 8 \cdot 5 \cdot \underbrace{11 \cdots 1}_{1985 \uparrow} \cdot \underbrace{11 \cdots 1}_{1985 \uparrow}=\underbrace{44 \cdots 4}_{1985 \uparrow} \cdot \underbrace{99 \cdots 9}_{1985 \uparrow} \cdot 10 \\
=\underbrace{44 \cdots 4}_{1985 \uparrow} \cdot (10^{1985}-1) \cdot 10 \\
=(\underbrace{44 \cdots 4}_{1985 \uparrow} \cdot 10^{1985}-\underbrace{44 \cdots 4}_{1985 \uparrow}) \cdot 10 \\
=\underbrace{44 \cdots 4}_{1984 \uparrow} \underbrace{455 \cdots 560}_{1984 \uparrow}, \\
\end{array}
$$

Therefore, the sum of its digits is $4 \cdot 1984 + 5 \cdot 1984 + 3 + 6 = 9 \cdot 1985$. Hence, the answer is $(C)$.