# Unbalanced weight of boxes

You are given a traditional balance scale with two pans (no reading) and 6 boxes with the following properties:
1) the weight of each box is a positive integer number
2) if you put any number of boxes on the left pan of the scale and any number of boxes on the right pan of the scale, the scale is never in balance. Note it is not required to use all boxes for weighing, but at least one box is on a pan.

What is the minimum weight of the heaviest box?

Example: weight 1,2,4,8,16,32 is a possibility with weight 32 being the heaviest box. Find a lower weight than 32 for the heaviest box!

I have found a set where the weight of the heaviest box is

$$24$$

Here are the weights of the boxes

$$11, 17, 20, 22, 23, 24$$

General Strategy

It seems like a good approach here is to use an inductive strategy on the number of weights $$n$$.
For example, for $$n=2$$, it is easily seen that the best solution just has weights $$1, 2$$. To get the best solution for $$n=3$$, we could express the $$n=2$$ solution as $$0,1,2$$ and add some value to each weight to generate a set $$b, 1+b, 2+b$$. The smallest value of $$b$$ which will work for $$n=3$$ is $$b=2$$ and we get the set $$2,3,4$$.
Applying this strategy again generates the sets
$$3,5,6,7$$ for $$n=4$$ and
$$6,9,11,12,13$$ for $$n=5$$.
Applying it one more time gives the resulting set for $$n=6$$. Obviously this is not a proof but just a good approach to the problem.

Minimality confirmed by Oray using computation.

• I confirmed with a code. 24 is min. actually this is unique solution with 24. – Oray Feb 3 at 17:24
• @Oray cool, thanks – hexomino Feb 3 at 17:30