Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.2 9/17/84; site mhuxt.UUCP
Path: utzoo!watmath!clyde!burl!ulysses!mhuxr!mhuxt!js2j
From: js2j@mhuxt.UUCP (sonntag)
Newsgroups: net.puzzle
Subject: Re: A different weighing problem
Message-ID: <788@mhuxt.UUCP>
Date: Mon, 15-Apr-85 13:16:33 EST
Article-I.D.: mhuxt.788
Posted: Mon Apr 15 13:16:33 1985
Date-Received: Tue, 16-Apr-85 01:05:53 EST
References: <13940@watmath.UUCP>
Organization: AT&T Bell Laboratories, Murray Hill
Lines: 24

> Here's a different "weighing problem", stolen from a "Columbo" from
> years gone by...
> 
> 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.
> 
> I'll post the solution in one week.
> 
> Jan Gray (jsgray@watmath.UUCP)   University of Waterloo   (519) 885-1211 x3870

    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.  
-- 
Jeff Sonntag
ihnp4!mhuxt!js2j
     "In the long run, we'll all be dead."-John Maynard Keynes