aff@duke.UUCP (Amr F. Fahmy) (04/18/85)
All right guys here is another weighing problem. If nobody posts the solution
within a week I'll post mine. Enjoy it :
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.
What are the weights of the 4 pieces ?
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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.
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Amr Fawzy Fahmy
CSNet : aff@dukejak@talcott.UUCP (Joe Konstan) (04/20/85)
> 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. > > What are the weights of the 4 pieces ? Easy one: 1, 3, 9, and 27 pounds: to way 2, put 1 on the side with the stuff, 3 on other side similarly for 5, 6, 7, ... > 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
wws@whuxlm.UUCP (Stoll W William) (04/20/85)
> > All right guys here is another weighing problem. If nobody posts the solution > within a week I'll post mine. Enjoy it : > > 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. > > What are the weights of the 4 pieces ? > 1, 3, 9, 27 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. > Neither do I, but my gut instinct is that it can't be done. > > Amr Fawzy Fahmy > > CSNet : aff@duke Bill Stoll, ..!whuxlm!wws
muffy@lll-crg.ARPA (Muffy Barkocy) (04/22/85)
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