2014 AMC 10A Problem 20

The $k$-digit number made entirely of $8$s is $8\frac{10^k-1}{9}$, so the product is $$\begin{aligned} 64\frac{10^k-1}{9} &=7\cdot10^k\\ &\quad+100\frac{10^{k-2}-1}{9}\\ &\quad+4. \end{aligned}$$ For $k\ge2$, this is the number whose digits are $7$, followed by $k-2$ ones, then $0,4$. This means that for any $k \geq 3,$ the sum of the digits in the product is $$7 + 4 + 0 + k - 2 = k + 9.$$ Finally, we get $$k + 9 = 1000$$$$k = 991.$$ Thus, D is the correct answer.