Computation and Sums Problem

Question

Compute $\sum_{k = 1}^{1000} ((1000 - k) \text{ div } 100)$.

I need help with this one. I tried to simplify it into two summations but got lost.

Answer 1

Using the information in the comments, I will assume $∑^{1000}_{k=1}((1000−k) div 100)= ∑^{1000}_{k=1}\lfloor {(1000−k)/100}\rfloor$ .

Then, with $100\ge k \ge 1$ , the $k$th term of the sum equals $9$. Then the sum equals $100 *9=900$. Doing this for other values of $k$ we get to the answer $∑^{1000}_{k=1}((1000−k) div 100)$= $100*\frac {9*10}{2}= 4500$