Date: Tue, 16 Apr 1996 12:53:59 -0700
From: Julie
Organization: Industry Canada
Subject: Solve the problem!
X-Url: http://fermat.math.uregina.ca/math_central/

This problem came to me from a student, Simon, who is in Industrial engineering from Laval University in Quebec city.

You have in front of you TEN piles of ten one cent coins each. One of these piles is composed exclusively of fraudulent coins but you don't know which pile. You know the weight of a one cent coin, lets say 1g. and you also know that each fraudulent coin weights 1g. more than an authentic coin. How many weighings would you need to do to find out which pile is fraudulent???

You can dissasemble the piles as needed if you wish. The type of balance used has no bearing on this.

Julie Hebert, francophone projects SchoolNet's Virtual Products Industry Canada - http://rescol2.carleton.ca/francais/ hebert.julie@ic.gc.ca

Julie and Simon
This is an old problem with a nice solution. The same problem, in a different guise, is in Martin Gardner, aha!, Scientific American, Inc./W.H. Freeman and Company, New York, 1978, page 23.

Take 1 coins from pile 1, 2 coins from pile 2, etc., that is take i coins from pile i, and weigh this collection. If they were all real 1 cent coins then this collection would weigh 1+2+...+10 = 10(11)/2 = 55 grams. If the first pile contained the fraudulent coins then your collection would contain 1 fraudulent coin and weigh 55 + 1 = 56 grams, if the second pile contained the fraudulent coins then your collection would weigh 55 + 2 = 57 grams, so in general your collection will weigh 55 + x grams where x is the number of the pile with the fraudulent coins.

You only need one weighing.

Cheers
Penny Nom

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