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From: riks@athena.UUCP (Rik Smoody)
Newsgroups: net.puzzle
Subject: Re: A different weighing problem
Message-ID: <236@athena.UUCP>
Date: Thu, 18-Apr-85 14:50:05 EST
Article-I.D.: athena.236
Posted: Thu Apr 18 14:50:05 1985
Date-Received: Sat, 20-Apr-85 08:07:49 EST
References: <13940@watmath.UUCP> <788@mhuxt.UUCP>
Organization: Tektronix, Beaverton OR
Lines: 24

> > You are given, say, 100 bags of 100 coins each; one of the bags contains
> > counterfeit coins, which weigh, say, 10% less than the real coins.
> > You must determine, in ONE weighing, which bag contains the counterfeits.
> 
>     Easy!  Just pile up 1 coin from bag #1, 2 coins from bag #2, ..., and
> 100 coins from bag #100.  Weigh the entire pile together.  Subtract this
> weight from the weight of (100 triangular (100+99+98+..+1)) real coins.
> Divide this difference by the difference in weight between real and
> counterfiet coins.  The result is the number of the bag containing the
> counterfiet coins.
> 
>     I'm assuming that you know pretty accurately just what a single real
> coin and a single counterfiet coin weigh.  

I have used this as an example of the difference between puzzle solving
and problem solving.
The mathematics is easy, but the pragmatics of counting out those coins
and keeping track of exactly which 37 coins came from bag 37, etc,
makes the clever solution rather impractical.

Do you really expect some bloke from the FBI to be able to count to 50?
And divide yet?
Cheers,
Rik Smoody