AMC12 2014 A

We can list the first few numbers in the form $8 \cdot (8....8)$ (Hard problem to do without the multiplication, but you can see the pattern early on) $8 \cdot 8 = 64$ $8 \cdot 88 = 704$ $8 \cdot 888 = 7104$ $8 \cdot 8888 = 71104$ $8 \cdot 88888 = 711104$ By now it's clear that the numbers will be in the form $7$, $k-2$ $1$'s, and $04$. We want to make the numbers sum to 1000, so $7+4+(k-2) = 1000$. Solving, we get $k = 991$, meaning the answer is $\boxed{\textbf{(D) } 991}$ Another way to proceed is that we know the difference between the sum of the digits of each product and $k$ is always $9$, so we just do $1000-9=\boxed{\textbf{(D) } 991}$.