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

Home > Logic > Weights and Scales
Special Similar Set (Posted on 2005-06-27) Difficulty: 3 of 5
A man offered me a set of eleven weights, not all them equal, each an integer number of pounds, which he said had the following property: if you removed any of the eleven weights, the other ten could form two five weights sets that balanced each other. Is this possible?

And if the weights didn't weigh an integer number of pounds each?

No Solution Yet Submitted by Old Original Oskar!    
Rating: 3.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Part one solution (and part 2) | Comment 1 of 11

If all the weights are even, then a different solution can be obtained by dividing each by two. This process can be continued until at least one is odd.  But if one is odd, all must be odd, so that what's left is an even number regardless of whether an odd or an even was taken.

A set that works is 1,1,1,3,3,5,5,7,7,9,9.

These add up to 51.

If a 1 is removed, then 25 is half the remaining 50, and 25 can be formed from 9+9+7.

If a 3 is removed, then 24 is half the remaining 48, and 24 can be formed from 5+5+7+7.

If a 5 is removed, then 23 is half the remaining 46, and 23 can be formed from 9+9+5.

If a 7 is removed, then 22 is half the remaining 44, and 22 can be formed from 9+9+3+1.

If a 9 is removed, then 21 is half the remaining 42, and 21 can be formed from 9+7+5.

For part 2 just divide each weight by 2.

  Posted by Charlie on 2005-06-27 18:46:19
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information