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

Home > Logic > Weights and Scales
Ninety nine coins plus one catch (Posted on 2014-06-01) Difficulty: 3 of 5
You are given a task of locating one faulty coin out of 99 identical in appearance coins i.e. 98 of equal weight and one of a lesser weight; and to achieve it within seven weighings, using a balance scale.

Sounds familiar and easy (37>99)?

Familiar? Yes.

Easy? Not quite!

How can you accomplish it if no coin may be weighed more than twice ?

Bonus task: After devising the requested procedure please generalize:
How many coins (n-1 normal and one lighter) can be resolved within k weighings?

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Any thoughts on this one? | Comment 4 of 7 |
(In reply to re: Any thoughts on this one? by Ady TZIDON)

I guess I'd prefer just a small hint.  I'd like to be able to solve it myself but I can't seem to get past the condition that no coin may be weighed more than twice. 

As you mention, this is a familiar problem and the traditional solution might require weighing a coin more than twice.  I don't see how adding that additional restriction produces a different solution.  This puzzle is not in "Tricks" so I assume there's a legitimate solution but I'm having a hard time seeing what it could be. 


  Posted by tomarken on 2014-07-02 09:46:41
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 (0)
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