You have 4 weights weighing 2,3,5 and 7 pounds. The problem is none of them are marked. What is the fewest number of weighings you need using a balance scale figure out which weights are which?
Name the weights A, B, C and D
First weighing: A+B versus C+D
Second weighing: A+C against B+D
The scale that has 2 times the highest weight must contain the 7 pound weight (Why are we in pounds / not in Newton?), because all combinations with 7 are higher then any two other weights. Suppose weighing 1 gives (A+B) > (C+D) and (A+C) > ((B+D), then this gives A = 7 pounds.
Third weighing sets A against (B + D). Three possibilities. A = (B+D) then A being 7 results in B and D being 2 or 5, this is then solved by comparing them in a 4th weighing. If A > (B+D), then A being 7 results in B and D being 2 or 3, this is then solved by comparing them in a 4th weighing. If A < (B+D), then A being 7 results in B and D being 3 or 5, this is then solved by comparing them in a 4th weighing.
So my answer is 4 weighings.
I'm trying to solve in in 3 weighings using the 2+3=5 and 5+2=7 balance situations. Up till now I didn't find a solution.
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Posted by Hugo
on 2004-07-25 13:26:55 |