Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » Math Topics » geometry.puzzles.independent

Topic: coins problem
Replies: 7   Last Post: Jun 16, 2009 10:34 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Helena Verrill

Posts: 8
Registered: 12/6/04
Re: coins problem
Posted: Feb 1, 1999 5:09 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

John Conway wrote:
>
> Here's the solution to the n-weighing version of this problem that


...

> Neat, isn't it?

Yes, it's very nice, thank you.

By the way, you said that an information count would make you
think you could do (3^n-1)/2 --- I suppose you mean that there are 3
possibilities for weighing, so n weighing gives 3^n outcomes,
and k coins means 2k possibilitis, (anyone light or heavy),
so, max k has 2k=3^n, and since that k is not an integer, you get
(3^n-1)/2.... so how do you prove you can't do this many?
And why is it, that if you just have the extra information
that you know the odd coin is light, then the 'information
count' does work, and you can do 3^n in n weightings?
what's the difference?





Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.