## Le Monde puzzle [#808]

**T**he current puzzle is as follows:

All integers are associated with one of two colours (gold and silver, say!), with the rule that for any triplet (with possible replications) with the same colour, the sum of this triplet is also of the same colour. If 58 is golden and a lot of integers are silver, what is the colour of 21? and of 40?

**H**ere is my resolution ** (spoiler alert!)**. The above rule tells us that for any x of a given colour, all multiples of the form 3

^{p}x are also of that colour. Also, given that 3=1+1+1, 5=3+1+1, &tc., all

*odd*numbers are of the same colour. Including 21. Starting with 2, a similar reasoning tells us that all integers of the form 2+4n are of the same colour as 2, which includes 58. Hence 2 is golden. Now, were 1 golden as well, 4=2+1+1 would be golden and all integers would be golden. Therefore, 1 is silver, therefore

*21 is silver*. Next, considering 4, if it was silver, 6=4+1+1 would be silver as well: impossible! Therefore, 4 is golden, which implies that all integers of the form 4+4n are also golden, hence

*40 is golden*. (Actually, all even integers are golden and all odd integers are silver.) The R resolution of this puzzle is not so obvious: either one sorts out that only 1,2, and 4 matter, in which case the issue is resolved, or else one tries “at random” but this does not mean much in this case: should one check 2⁵⁸ colourings? build a tree algorithm that severs impossible branches?

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