1985 AMC 12 Problem 29

Write a string of $k$ identical digits using $R=\frac{10^k-1}{9}$

After removing the terminal zero, express the product as a string of $2k$ fours minus a string of $k$ eights

Let $k=1985$ and $R=\frac{10^k-1}{9}.$ Then $a=8R,$ $b=5R,$ and $9ab$ ends in $0.$ Removing that zero does not change the digit sum and leaves $$\begin{aligned} N=\frac{9ab}{10} &=36R^2\\ &=\frac49(10^{2k}-1)\\ &\quad{}-\frac89(10^k-1). \end{aligned}$$ Thus $N$ is a string of $2k$ fours minus a string of $k$ eights. The subtraction produces $k-1$ fours, then $3,$ then $k-1$ fives, then $6.$ Its digit sum is $$\begin{aligned} &4(k-1)+3+5(k-1)+6\\ &\quad=9k=17865. \end{aligned}$$ Therefore the correct answer is C.