Mathematical game: how many nuggets did the jeweler use?

A jeweler at seven gold nuggets whose weights are all distinct, from 1 to 7 grams. He has three jewelry projects, for each of which he selects two nuggets. When made, they weigh 4, 9 and 13 grams. What nuggets will he use?

Answers:

All except the 2 gram one.

The interest of this enigma does not reside in this result but in the various methods of resolution that can be classified according to their elegance, a delicate concept to define elsewhere.

Here are three demonstrations that you will be able to judge under this criterion, but that we have not classified at random:

1. Case exhaustion method

It is a question of finding the pairs of two numbers between 1 and 7 whose sum is 4, 9 and 13. A courageous method is to try all the cases. Their number is not large, just equal to 21 and we try them all… which we leave to you.

2. Reasoning by absurd

Assume the 2 gram nugget is used. We want to make jewelry of 4, 9 and 13 grams. It can therefore only be associated with the 7 gram nugget. The jewel of 13 grams is then impossible to make since there are no more nuggets left whose sum of weights is 13 grams. The 2 gram nugget cannot therefore be used. So all the others are.

3. Invariance of weight

The total weight of the nuggets is 1 + 2 + 3 + 4 + 5 + 6 + 7, i.e. 28 grams. The total weight of the jewels is 4 + 9 + 13 = 26. We must therefore exclude the nugget of 2 grams.