Thread: Algebra sum problem

The sum of a certain number of consecutive positive integers is 2000. Find an example of these numbers. Is this solution unique? If not, how many solutions are there?

I got to the stage of n = start number m = number of consecutive numbers

then (m+1)n + m(m+1)/2 = 2000, given the sum of integers formula from 1.

Every odd divisor of 2000 admits a solution
(that might contain negative terms).
For example, divide 2000 by 5 to get 400, the middle of five consecutive integers with sum 2000.
Check this: 398 + 399 + 400 + 401 + 402 = 2000

On the other hand, 2000/125 = 16, so 16 is the middle of a list of 125 consecutive integers with sum 2000. But many of these are negative.