Math ( Number Theory )
posted by Lucy .
How many numbers
from 1 to 1000
inclusive can be
expressed as the sum
of k >= 2 consecutive
positive integers for
some value of k ? Sorry to post question from Brilliant, I got 997 integers, but it's wrong. If you think you can answer ...

Hmmm.
For k=2, we want all n<=1000 such that n=m+m+1 = 2m+1
That is, all odd numbers 3<=n<=999
There are 996/2+1 = 499 of those
For k=3, we want n=3m+3
so, that's all multiples of 3 with 6<=n<=999
There are 993/3+1 = 332 of those
For k=4, we want 4m+6
so, that's all multiples of 4 with 10<=n<=998
There are 988/4+1 = 248 of those
There's a pattern here, so you should be able to come up with some
∑
k=2
expression, or just bruteforce your way through.
Maybe not a lot of help, but it does show that there are more than 997 of them. 
The ans Steve gave is absolutely wrong...the ans is less than 823 and greater than 665...steve u have counted the same numbers more than one time...for example u have counted 15 twice....

When I calculated out I got 990 as the correct answer.