Consider a string of n 1 's. We wish to place some + signs in between so that the sum is 1000 . For instance, if n =190 , one may put + signs so as to get 11 ninety times and 1 ten times, and get the sum 1000. If a is the number of positive integers n for which it is possible to place + signs so as to get the sum 1000 , then find the sum of the digits of a .

Step-by-step solution

1000=1 a₁+11 a₂+11 a₃+ where a₁, a₂, a₃, . . are non-negative integers for all a_i , when i>3, a_i=0 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 . _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _