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Weights that check themselves. (Posted on 2010-11-23) Difficulty: 3 of 5

Assume an old-fashioned scale balance in which weights can be placed on either side. The associated set of weights (each of which is greater than zero) is 'complete' for some W if it is capable of measuring all integer weights from 1 to W.

Clearly it is possible for such sets to exist even if no combination of the weights themselves can be balanced against any combination of the remaining weights - the set {1,2,4,8..} is just one such example.

On the other hand, the set {1,1,2,4} is also 'self-measuring', because, assuming that one of the weights were unmarked, a stranger could neverthless establish its value by weighing it in the scales with the others.

Question: You are allowed 5 weights. They must form a set which is complete, and also self-measuring.

What is the largest possible value of W, given these constraints?

Bonus: What is the largest possible value of W, if 8 weights are allowed?

  Submitted by broll    
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Solution: (Hide)
I agree with Justin. The 'self-measuring' requirement is satisfied if there is any combination of all the weights that can be balanced against each other. So it makes sense to use the most effective length-covering set {3^0,3^1,3^2,3^3...} with the last weight weighing the same as the combined total of all the others. So, for 5 weights we have {1,3,9,27,40} and can weigh any amount up to 80, and for 8 weights we have {1,3,9,27,481,243,729,1093} giving a maximum W of 2186.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re(2): solutionJustin2010-11-24 06:54:57
5 weights initial searchDaniel2010-11-24 05:30:22
re: solutionDaniel2010-11-24 05:26:41
solutionJustin2010-11-24 05:21:23
re: Not quite there yetbroll2010-11-24 00:53:41
No Fibbing, No solution.ed bottemiller2010-11-23 22:34:49
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