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From: jeff@hpcnoe.UUCP (jeff)
Newsgroups: net.puzzle
Subject: Re: Orphaned Response
Message-ID: <17500002@hpcnoe.UUCP>
Date: Wed, 15-May-85 19:52:00 EDT
Article-I.D.: hpcnoe.17500002
Posted: Wed May 15 19:52:00 1985
Date-Received: Sat, 11-May-85 23:49:20 EDT
References: <-73900@whuxlm.UUCP>
Organization: Hewlett-Packard - Fort Collins, CO
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Nf-ID: #R:whuxlm:-73900:hpcnoe:17500002:37777777600:1064
Nf-From: hpcnoe!jeff    May  3 15:52:00 1985

>>> 	How about generalizing it, lets assume that the gold originally 
>>> 	weighed K pounds and was broken into n pieces, what are the weights
>>> 	of the n pieces given that the above property still holds ? 

>>In general, we want to break it into blocks of 3^n for n going from
>>0 up as high as possible.  I think that leaving the leftover as just one
>>block will be good enough to weigh everything, by always using that
>>block for weights above it.

> Yes, but...suppose the problem is: break a 40-lb block into *5* pieces?
> Then, using 3^n, we get 1 3 9 27 *0*.  How would you do this?  I assume
> that any of them could just be broken in two, but your answer should have
> included this case.

Or the case 40-lb block into 3 pieces, which there is no solution.
Or the case 40-lb block into 41 pieces, which means that at least two
blocks would be non-integral lb weights.

There is an algorithmic way of calculating if there is no solution for
a given n, K.  Is there a mathmatical formula which does this?

Jeff Wu / CNO
Hewlett-Packard
Fort Collins CO