Talk:Puzzles/Decision puzzles/Weighings Once More

I would tentatively say 8, the number of divisors of 40? Dysprosia 04:38, 11 Oct 2003 (UTC)

You can see the weights as a kind of "physical bits", because it can be either in the collection or not.

With n bits you can represent no more than $$2^n$$ different states (including 0), which is proved trivially. Therefore, for a given 40 states, the lower bound for the number of weights is 6.

At first, we select 5 weights to represent the numbers 1 to 31:


 * 1 2 4 8 16

For the sixth weight, we can choose any weight $$a$$ with $$9\le a\le 32 $$

doopdoop


 * The solution is OK, however, it can be done with less weights. (hint: you can also place weights on the side of the object you want to weigh - so one weight can carry more information than just a bit:). Thomas

This looks like the same puzzle as Scales and Weights (also solved, with the better solution listed) Valhalla 12:45, 11 Jun 2005 (UTC)