Maths Olympiad Prep

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, 2010

Combinatorics Difficulty 6.4 National Olympiad Prove it Estonia

Let nn be a positive integer. Find the largest integer NN for which there exists a set of nn weights such that it is possible to determine the mass of all bodies with masses of 11, 22, \ldots, NN using a balance scale (i.e. to determine whether a body with unknown mass has a mass 11, 22, \ldots, NN, and which namely).

Solution

The possibility to determine mass mm means the possibility to place the weights on the two scalepans so that the difference of total masses on the two scalepans is exactly mm.

Every weight can be placed on either of the two pans or on neither of the pans. For nn weights this makes 3n3^n different placements. Note that the placement where none of the weights is on the scales does not determine any mass. Also, for each placement there is a symmetric placement with all the weights on the two pans swapped, which determines the same mass. Therefore with nn weights it is possible to determine at most 3n12\frac{3^n-1}{2} different masses.

We show by induction that it is possible to determine all masses from 11 to 3n12\frac{3^n-1}{2} using nn weights with masses 11, 33, \dots, 3n13^{n-1}.

For n=1n=1 it is obvious. Assume that the claim holds for n=kn=k. By the induction assumption we can determine all masses from 11 to 3k12\frac{3^k-1}{2} by weights 11, 33, \dots, 3k13^{k-1}. Using the weight with mass 3k3^k, we can determine the mass 3k3^k, and using it together with the other weights also the masses 3k+13^k+1, \dots, 3k+3k123^k+\frac{3^k-1}{2} and 3k13^k-1, \dots, 3k3k123^k-\frac{3^k-1}{2}.

Since 3k3k12=3k12+13^k - \frac{3^k-1}{2} = \frac{3^k-1}{2} + 1 and 3k+3k12=3k+1123^k + \frac{3^k-1}{2} = \frac{3^{k+1}-1}{2}, the claim is also true for n=k+1n=k+1.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.