Consider a string of n 1 's. We wish to place some + signs in...

$\because 1000=1 \cdot a_1+11 \cdot a_2+11 \cdot a_3+\ldots \ldots$ where $a_1, a_2, a_3, \ldots .$. are non-negative integers for all $a_i$, when $i>3, a_i=0$ $\therefore \quad 1000=111 p+11 q+r$ If $p=0$, then there are 91 possibilities for $q, r$. For $p=1$, there are 81 possibilities for $q, r$. For $p=2$, there are 71 possibilities for $q, r$. For $p=3$, there are 61 possibilities for $q, r$. For $p=4$, there are 51 possibilities for $q, r$. $\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$ $\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$ $\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_$ For $p=9$, there are 1 possibility for $q, r$. Hence, total number of possibilities = 460 Sum of digits of 460 is 10.