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

 Mutually friendly (Posted on 2002-07-01)
Prove that any group of six people contains either 3 mutual friends or 3 mutual strangers.

(For the purpose of this problem any pair of people must be either friends or strangers.)

 See The Solution Submitted by levik Rating: 3.0000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): IQ Test | Comment 4 of 11 |
(In reply to re: IQ Test by levik)

Actually I did not prove that a group of 5 people can avoid triples. all I proved was that a group of six cannot.

To prove that five can avoid triples I would need to examine four reationships that I never toched on in my previous analysis: Bill-Dave, Bill-Ed, Carl-Dave and Carl-Ed.

Bill cannot be a friend to both Dave and Ed, Neither can Carl. But Dave cannot be a stranger to both Bill and Carl. Neither can Ed

The following relatioships fulfill these conditions, and finally prove it is possible for five people to have no triples:

Bill and Dave are friends
Bill and Ed are strangers
Carl and Dave are strangers
Carl and Ed are friends.
 Posted by TomM on 2002-07-02 08:01:25

 Search: Search body:
Forums (1)
Random Problem
Site Statistics
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox: