smc

By the formula for the sum of a geometric series, $$\begin{array}{rl}a& =8\cdot 10^0+8\cdot 10^1+\cdots +8\cdot 10^{1984}\\ & =\frac{8\left(10^{1985}-1\right)}{10-1}\\ & =\frac{8\left(10^{1985}-1\right)}{9},\end{array}$$ and similarly $$b=\frac{5\left(10^{1985}-1\right)}{9},$$ so $$\begin{array}{rl}9ab& =9\cdot \frac{8\left(10^{1985}-1\right)}{9}\cdot \frac{5\left(10^{1985}-1\right)}{9}\\ & =\frac{40{\left(10^{1985}-1\right)}^2}{9}\\ & =\frac{40\left(10^{3970}-2\cdot 10^{1985}+1\right)}{9}\\ & =\frac{4\left(10^{3971}-2\cdot 10^{1986}+10\right)}{9}.\end{array}$$ We now compute the decimal expansion of this expression. Firstly, $10^{3971}=100\cdots 0$, with $1$ one and $3971$ zeroes, and $2\cdot 10^{1986}=200\cdots 0$, with $1$ two and $1986$ zeroes. Subtracting therefore gives $$10^{3971}-2\cdot 10^{1986}=99\cdots 9800\cdots 0,$$ where there are $3971-1986-1=1984$ nines followed by $1$ eight and then $1986$ zeroes. Adding $10$ transforms this to $99\cdots 9800\cdots 010$, now with $1984$ nines followed by $1$ eight, $1984$ zeroes, $1$ one, and a final zero. Using long division, and noting that $80=8\cdot 9+8$ and $81=9\cdot 9$, it follows that $$\frac{4\left(10^{3971}-2\cdot 10^{1986}+10\right)}{9}=11\cdots 1088\cdots 890,$$ with $1984$ ones, $1$ zero, then $1984$ eights, $1$ nine, and a final zero. Lastly, using long multiplication and noting that $9\cdot 4=36$, $8\cdot 4=32$, and $8\cdot 4+3=35$, we obtain $$\frac{4\left(10^{3971}-2\cdot 10^{1986}+10\right)}{9}=44\cdots 4355\cdots 560,$$ where there are $1984$ fours, $1$ three, $1984$ fives, $1$ six, and a final zero, so the sum of the digits is $$\begin{array}{rl}1984\cdot 4+3+1984\cdot 5+6+0& =1984\cdot 9+9\\ & =1985\cdot 9\\ & =\boxed{17865}.\end{array}$$