All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
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.)
re(2): solution Comment 6 of 6 |
(In reply to re: solution by Daniel)

I kept thinking about that self-measuring part, kept thinking about it, kept thinking about it, completely forgot to account for it in my solution. So, unless your Mathematica code can find better, here are my new sequences, based on my already erased previous message:

As {1, 3, 9, 27, 81} is optimized for the sums and differences only, and NOT self-measuring, let us remove the last term. This leaves {1, 3, 9, 27}, which is the optimal set of size 4, for forming sums and differences. Now, the fifth number to be added will be the sum of those numbers (1 + 3 + 9 + 27 = 40). That way, we can use the first 4 numbers to represent anything from 1 to 40, and add 40 to those values for anything from 41 to 80. In fact, any set {x, 3, 9, 27, x+39} with 5 distinct positive integers, and x not being a multiple of 3 seems to result in W=80.


Likewise, for 8 terms, {1, 3, 9, 27, 81, 243, 729} will be optimized for the first 7 terms in creating the largest quantity of sums and differences possible, while adding the 8th term (1093 = 1 + 3 + 9 + 27 + 81 + 243 + 729) will ensure any unlabeled weight can be identified. So, W=2186 seems to hold for all sets {x, 3, 9, 27, 81, 243, 729, x+1092} of 8 distinct positive integers, again with x not a multiple of 3.

So, before I do anything else that's just downright stupid, I will submit these 2 answers as my new bests:

5 weights:
W = 80{x, 3, 9, 27, x + 39}, x mod 3 > 0

8 weights:
W = 2186{x, 3, 9, 27, 81, 243, 729, x + 1092}, x mod 3 > 0

  Posted by Justin on 2010-11-24 06:54:57
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