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?

(In reply to solution by Justin)

unfortunately, I believe that your submissions fail to meat the self-measuring requirement.  For example, with {1,3,9,27,81} you are not able to weight the "1" weight using only the remaining weights, namely {3,9,27,81}, no possible selection of these remaining weights will allow you to weight 1.  However, your insight does provide us with a theoretical maximum, although it remains to be seen if it is possible to achieve it.

please see my next post for the initial results of my computer search.

Posted by Daniel on 2010-11-24 05:26:41