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


Click to listen to the single ▶

Support album on Kickstarter
perplexus dot info

Home > Logic > Weights and Scales
Weights that check themselves. (Posted on 2010-11-23) Difficulty: 3 of 5

Assume an old-fashioned scale balance in which weights can be placed on either side. The associated set of weights (each of which is greater than zero) is 'complete' for some W if it is capable of measuring all integer weights from 1 to W.

Clearly it is possible for such sets to exist even if no combination of the weights themselves can be balanced against any combination of the remaining weights - the set {1,2,4,8..} is just one such example.

On the other hand, the set {1,1,2,4} is also 'self-measuring', because, assuming that one of the weights were unmarked, a stranger could neverthless establish its value by weighing it in the scales with the others.

Question: You are allowed 5 weights. They must form a set which is complete, and also self-measuring.

What is the largest possible value of W, given these constraints?

Bonus: What is the largest possible value of W, if 8 weights are allowed?

See The Solution Submitted by broll    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
solution | Comment 3 of 6 |
Like I said, I was tired.

Edited on November 24, 2010, 5:49 am
  Posted by Justin on 2010-11-24 05:21:23

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 (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2014 by Animus Pactum Consulting. All rights reserved. Privacy Information