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CTK Exchange
Alex
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Nov-13-06, 03:06 AM (EST) |
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"Weights and Counterbalances"
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There is a rock that weights an integer K pounds, from 1 to 40 inclusive. You also have a two-sided scale. What is the minimal number n of weights that allows you to measure the weight of the rock, if you are free to choose any set of weights, use an unlimited number of tries and both sides of the scales. I am having trouble figuring out a formula for the solution. I understand that each weight may be added to or subtracted from the others to form combinations of numbers. I also understand that the sum of the all weights must at least be 39. There are similiar problems around, but I am curious what the solution is to this particular example. |
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Alex
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Nov-16-06, 04:20 PM (EST) |
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2. "RE: Weights and Counterbalances"
In response to message #1
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Thank you, https://mathworld.wolfram.com/Weighing.html was helpful. I guess my question is whether these solutions account for strategies where weights can be determined by bounding. For example, we do not need to be able to add our weights up to 40 because we know that if something weighs more than 39lbs, it has to be 40. Similiarly we do not need to be able to add up to weight x if we can account for weights x-1 and x+1. I understand this might sound like a technicality, but I am interested in the practical solution to this particular example. |
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