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

Home > General
Non-arithmetic Triplets (Posted on 2013-04-08) Difficulty: 3 of 5
Find nine different integers from 1 to 20 inclusive such that no combination of any three of the nine integers form an arithmetic sequence.

(For example, if two of the integers chosen were 7 and 13, then that would preclude 1, 10 and 19 from being included.)

See The Solution Submitted by K Sengupta    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(4): Counting Up,,,,Why not up and down? | Comment 7 of 9 |
(In reply to re(3): Counting Up,,,,Why not up and down? by Ady TZIDON)

How would you know to start with a difference of 1 at one end and 2 at the other?

If for example the puzzle asked for 12 different integers from 1 to 30, you'd want

1, 3, 4, 8, 9, 11, 20, 22, 23, 27, 28, 30

with a difference of 2 at each end. In this case it's its own mirror structure.

Also, as another example, the following are the reversable kind, but still has a difference of 1 at each end, as the asymmetry is farther in:

Thirteen out of 32:

1, 2, 4, 8, 9, 11, 19, 22, 23, 26, 28, 31,32

1, 2, 5, 7, 10, 11, 14, 22, 24, 25, 29, 31, 32

Edited on April 9, 2013, 1:57 am
  Posted by Charlie on 2013-04-09 01:52:20

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 (5)
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