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.

Sincerely,
Jhonen

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
   http://mathforum.org/dr.math/problems/macdougall12.7.97.html   

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

Enjoy,

- Doctor Schwa, The Math Forum
  http://mathforum.org/dr.math/