Posted by **Lucy** on Tuesday, April 9, 2013 at 10:37am.

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 ...

- Math ( Number Theory ) -
**Steve**, Tuesday, April 9, 2013 at 11:11am
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 brute-force your way through.

Maybe not a lot of help, but it does show that there are more than 997 of them.

- Math ( Number Theory ) -
**Arpan**, Wednesday, April 10, 2013 at 9:12am
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....

- Math ( Number Theory ) -
**Anonymous**, Wednesday, April 10, 2013 at 4:49pm
When I calculated out I got 990 as the correct answer.

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