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From: muffy@lll-crg.ARPA (Muffy Barkocy)
Newsgroups: net.puzzle
Subject: Re: Yet another weighing problem
Message-ID: <538@lll-crg.ARPA>
Date: Sun, 21-Apr-85 18:20:45 EST
Article-I.D.: lll-crg.538
Posted: Sun Apr 21 18:20:45 1985
Date-Received: Wed, 24-Apr-85 02:36:51 EST
References: <5755@duke.UUCP> <419@talcott.UUCP>
Reply-To: muffy@lll-crg.UUCP (Muffy Barkocy)
Organization: Lawrence Livermore Labs, CRG group
Lines: 28

In article <419@talcott.UUCP> jak@talcott.UUCP (Joe Konstan) writes:
>> 	A piece of gold weighing 40 pounds was dropped and broken into 4
>> 	pieces. It was broken in such a way that you can, using the 4
>> 	pieces, weigh anything weighing from 1 up to 40 pounds. For example
>>         something weighing 5 pounds could be weighed using one piece of
>> 	gold weighing 15 pounds and another weighing 10 pounds. 
>
>> 	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 ? 
>> 
>>         Note: I do not know (yet) the solution to the second one.
>
>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.
>
>Mithrandir
>jak@talcott


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.

                         Muffy