Minimum Set of Weights PuzzleDate: 10/18/2001 at 23:58:12 From: Jhonen Subject: Minimum Set of Weights Puzzle Dear Dr. Math, What is the minimum number of weights needed for a scale that will be able to weigh objects from the weight of one pound to 100 pounds, inclusive, at one-pound increments? It seems as if the answer should be four, because of Bachet's Conjecture, and the fact that a set of four weights exists satisfying the conditions of the problem for other maximum numbers of similar size, such as 40. I'm a bit stumped as to how to proceed to find a conclusive proof for the number of weights needed, or the actual minimum set of weights in order to solve this problem. Sincerely, Jhonen Date: 10/21/2001 at 00:41:45 From: Doctor Schwa Subject: Re: Minimum Set of Weights Puzzle Hi Jhonen, For each weight, there are three things you can do: put it on the left pan, the right pan, or not on the balance at all. So, if you have n weights, there are 3^n things you can do with them. One of those things is not putting any weights on the scale, which is good if you want to weigh a 0-pound object, so really there are only 3^n - 1 arrangements. Then, for each arrangement there's also its mirror image (where all the weights are switched to the opposite pan of the scale), so there are at MOST (3^n - 1)/2 arrangements of n weights. That's enough to prove that 4 weights can weigh at most 40 different things ... 40 is really the upper limit for 4 weights. With a fifth weight, you should be able to get up to (3^5 - 1)/2 = 121 pounds. For some clues in the other direction (that is, how to figure out what the weights should actually be), see the Dr. Math archives at Three Weights http://mathforum.org/dr.math/problems/macdougall12.7.97.html or, if that doesn't help you, feel free to write us back! Enjoy, - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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