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

Home > Logic > Weights and Scales
Pearls (Posted on 2002-05-23) Difficulty: 3 of 5
You have nine pearls, one of which is (as is usually the case in these problems) fake. You know that the fake pearl weighs less than the others, but it is (of course) impossible to distinguish from the others in any other way.

What is the minimum number of weighings that must be performed to find the fake pearl? How would you go about it?

See The Solution Submitted by levik    
Rating: 2.7143 (7 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution | Comment 5 of 9 |
OK, without looking at what anyone else has posted, I reckon it must be two weighings. Weigh two groups of three, this will tell you which group the light one is in (bearing in mind that if the two groups balance, then you automatically know the dodgy pearl is in the third group);  then weigh two of the suspect group against each other (similar story - should they balance, then the other one can be declared fake without having to weigh it).
  Posted by Jenny Turner on 2004-06-02 17:29:13
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 (8)
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