 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Consecutive Sums (Posted on 2016-07-12) In the list below I have tried to express each integer 1-9 as a sum of consecutive positive integers:

1 = ?
2 = ?
3 = 1+2
4 = ?
5 = 2+3
6 = 1+2+3
7 = 3+4
8 = ?
9 = 2+3+4

This suggest that there is a sum for every positive integer which is not a power of 2, and there is never a sum for powers of 2.

Prove or disprove my claims.

 No Solution Yet Submitted by Brian Smith No Rating Comments: ( Back to comment list | You must be logged in to post comments.) No Subject | Comment 1 of 2
If 2^m=n+n+1+...+n+a=(a+1)n+(1+2+...a)=
=((a+1)*(2n+a))/2

Then 2^(m+1)=(a+1)*(2n+a)
The only divisor of 2^(m+1) is 2 so both a+1 and 2n+a need to be powers of 2

But if a+1 is a power of 2 it must be even and then 2n+a should be odd.

So the claim of the puzzle is true.

Edited on July 12, 2016, 4:30 pm
 Posted by armando on 2016-07-12 16:27:30 Please log in:

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