smc

We can list the first few numbers in the form $8\cdot (\mathrm{8....8})$ (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 $\boxed{\text{(D) }991}$ 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 $1000-9=\boxed{991}$.