Pieces of Eight!

$8 * (8 * 10^n + 8 * 10^{n-1} + ... + 8 *10 + 8 * 1)$ $64 * 10^n + 64 * 10^{n-1} + ... + 64 *10 + 64 * 1)$

1
2
3
4
5
6
7
8
9

11100
------
   64  
  640
 6400
64000
  ...
------
71104

Here we can see that after the first two additions we'll always have a pair of 6 and 4 with 1 as a carry. This addition will continue until the last one where we'll have 6+1 = 7. The sum of the digits is: $7 + (k-2) * 1 + 4 = 9 + k$ $9 + k = 1000 => k = 991$

$8*8=64$

$8*88=704$

$8*888=7104$

$8*8888=71104$

$8*88888=711104$

so it appears for $k>2$ the pattern is $7$ , followed by $(k-2$ ) $1$ 's, then a $0$ and a $4$ .

So from this we get that for $k>2$ the sod is

$7+(k-2)*1+0+4=k+9$

so if we want

$k+9=1000$

$k=991$