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No doubles (Posted on 2022-05-02) Difficulty: 2 of 5
Can you create a subset of (1, 2, 3, ..., 3k) such that none of its 2k-1 members is twice the value of another?

Either provide such a set or show none exists.

Inspired by: Austrian-Polish Math. Competition.

See The Solution Submitted by Ady TZIDON    
Rating: 4.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Idea, no proof | Comment 5 of 10 |
(In reply to re: Idea, no proof by broll)

I don't understand your variables. What do m and n have to do with k?

My method is sometimes one number short though:

n=128 is not possible as it is not a multiple of 3.
Suppose k=43 so 3k=129 and 2k-1=85
The included intervals are [1],[4,7],[16-30],[65,129]
1+4+15+65 = 85
This is fine.

Suppose k=42 so 3k=126 and 2k-1=83
The included intervals are [1],[4,7],[16-31],[64,126]
1+4+16+63 = 84 
So this method doesn't quite do it.

  Posted by Jer on 2022-05-02 13:35:07
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