Weights

From the Tournament of the Towns:

A balance and a set of metal weights are given, with no two the same. If any pair of these weights is placed in the left pan of the balance, then it is always possible to counterbalance them with one or several of the remaining weights placed in the right pan. What is the smallest possible number of weights in the set?

Let’s work our way up. One could argue that with a set containing just a single weight, the condition vacuously holds; but this clearly wasn’t the point of the question…

With two weights, there is nothing to counterbalance with.

With three weights, there is no way to counterbalance the two heaviest.

With four weights, again there is no way to counterbalance the two heaviest.

Five weights is the first interesting case. If five weights is possible, then the heaviest two must be counterbalanced by the three lightest weights. It must also be the case that the pair consisting of the heaviest and the middle weight must be counterbalance by the three remaining weights (2nd, 4th and 5th heaviest). But this is a problem – in moving from the first balancing act to the second, the left pan became lighter but the right pan became heavier, contradiction.

So six weights is the best we could hope for. And in fact this is possible; after a little experimenting, you might stumble upon the set {3,4,5,6,7,8}, in which every pair can be counterbalanced:

3+4=7

3+5=8

3+6=4+5

3+7=4+6

3+8=4+7=5+6

4+8=5+7

5+8=6+7

6+8=3+4+7

7+8=4+5+6.

Thus the smallest possible number of weights is six.

Thought-provoking and elegant: yet another beautiful puzzle from Russia.

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