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

Home > Numbers
190 couples (Posted on 2018-05-03) Difficulty: 3 of 5
The set S consists of 20 distinct positive integers, each below 70.

Create a set S' of 190 pairwise absolute values of differences between two distinct members of S.

Prove that in S' there must exist at least one value that appears more than 3 times.

See The Solution Submitted by Ady TZIDON    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution A much stronger proof (spoiler) Comment 2 of 2 |
Forget the 190 couples.  Order the elements of S in ascending order and consider just the 19 pairwise differences between adjacent elements.  Among just these 19 there must be one number that appears at least 4 times!!

PROOF:  Assume that no number appears 4 or more times.  Then the minimum value of the sum of the 19 differences between adjacent elements must be 3*1 + 3*2 + 3*3 + 3*4 + 3*5 + 3*6 + 7 = 3*21 + 7 = 70.  But this sum is just the difference between the highest and lowest number in S, and that can be at most 69.  So we have a contradiction, and our initial assumption is wrong.  Thus, at least one number appears at least 4 times among the 19 pairwise differences.  q.e.d

(ASIDE: I expect that there are many more than just one number that must appear more than 3 times among the full 190 elements, but I am moving on.)

 

  Posted by Steve Herman on 2018-05-06 18:19:44
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 (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information