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Minimum Set of Weights Puzzle

Date: 10/18/2001 at 23:58:12
From: Jhonen
Subject: Minimum Set of Weights Puzzle

Dear Dr. Math,

What is the minimum number of weights needed for a scale that will be 
able to weigh objects from the weight of one pound to 100 pounds, 
inclusive, at one-pound increments?  

It seems as if the answer should be four, because of Bachet's 
Conjecture, and the fact that a set of four weights exists satisfying 
the conditions of the problem for other maximum numbers of similar 
size, such as 40. I'm a bit stumped as to how to proceed to find a 
conclusive proof for the number of weights needed, or the actual 
minimum set of weights in order to solve this problem.


Date: 10/21/2001 at 00:41:45
From: Doctor Schwa
Subject: Re: Minimum Set of Weights Puzzle

Hi Jhonen,

For each weight, there are three things you can do: put it on the left 
pan, the right pan, or not on the balance at all.

So, if you have n weights, there are 3^n things you can do with them.

One of those things is not putting any weights on the scale, which is 
good if you want to weigh a 0-pound object, so really there are only 
3^n - 1 arrangements.

Then, for each arrangement there's also its mirror image (where all 
the weights are switched to the opposite pan of the scale), so there 
are at MOST (3^n - 1)/2 arrangements of n weights.

That's enough to prove that 4 weights can weigh at most 40 different
things ... 40 is really the upper limit for 4 weights.

With a fifth weight, you should be able to get up to (3^5 - 1)/2 = 
121 pounds.

For some clues in the other direction (that is, how to figure out
what the weights should actually be), see the Dr. Math archives at

   Three Weights

or, if that doesn't help you, feel free to write us back!


- Doctor Schwa, The Math Forum
Associated Topics:
High School Permutations and Combinations
High School Puzzles

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