The product $(8)(888\dots8)$, where the second factor has $k$ digits, is an integer whose digits have a sum of $1000$.  What is $k$?
$\textbf{(A)}\ 901\qquad\textbf{(B)}\ 911\qquad\textbf{(C)}\ 919\qquad\textbf{(D)}\ 991\qquad\textbf{(E)}\ 999$
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 $\fbox{(D)}$
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 $1000-9=\boxed{\textbf{(D) } 991}$.

Proof of Solution 1
Since this solution won't fly on a proof-based competition, here's a proof that it's valid:
We will call $x_k=8(888\dots8)$ with exactly $k$ $8$s. We then rewrite this more formally, as:
\[x_k=8\biggr(\sum_{n=0}^{k-1}8(10)^n\biggr)\]
\[=64\biggr(\sum_{n=0}^{k-1}(10)^n\biggr)\]
\[=64\frac{10^{k}-1}{9}\]
Then, finding a recursive formula, we get:
\[x_{k+1}=64\times 10^{k}+x_k\]
We will now use induction, Our base case will be $k=2$. It's easy to see that this becomes $x_2=704$. Then, the $k+1$ case: let $x_k=7111\dots104$ with $k-2$ $1$s. Then $x_{k+1}=64000\dots000+7111\dots104$. Adding these numbers, we get $x_{k+1}=71111\dots104$. 
Summing these digits, we have $4+7+(k-2)=1000$, giving us $k=991$.