All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Be the term-inator (Posted on 2019-06-16) Difficulty: 3 of 5
We have a series where the sum of any 7 consecutive terms is negative and the sum of any 11 consecutive terms is positive. What is the maximum number of terms in this series?

No Solution Yet Submitted by Danish Ahmed Khan    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Tightening the range | Comment 2 of 6 |
Assume the series has at least 17 terms.  For any subset of 4 consecutive terms there is either a subset of 7 terms on its left or its right.  The sum of the subset of 7 terms is positive and the sum of the union of the 4 and 7 term subsets is negative.  This implies the sum of the subset of four terms is negative.

Then every sum of four consecutive terms is negative.  This implies the sum of any eight consecutive terms is also negative.  However the sum of the first 7 or last 7 terms is positive, which implies the first and eighth terms are negative.

Then by going through each possible 8 term subset of the series, it is concluded that all the individual terms are negative.  This is a contradiction since the sum of any 7 consecutive terms is positive.  Therefore the series must have fewer than 17 terms.

By explicit construction the 14 term series {1, 1, 1, -5, 1, 1, 1, 1, 1, 1, -5, 1, 1, 1} satisfies the constraints.  So the maximum possible number of terms in a series satisfying the constraints is 14, 15, or 16.

Edited on June 18, 2019, 1:10 pm
  Posted by Brian Smith on 2019-06-18 13:09:23

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (8)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information