Le Monde problem: “The prediction”

This week’s Le Monde mathematical problem is:

The scene is set in 2010. Young N, an integer looking for his soul-mate, consults a fortune-teller, who says: “Yes, you shall encounter the beloved number [call it X], a number strictly greater than you, but later! That year [call it Y], you will be the only [my emphasis]  two positive integers whose sum is equal to the year, and whose product is a multiple of the year.” When (at the earliest) will N meet his soul -mate?

This is the kind of question which could be asked at the FFJM semi-finals in a couple of weeks (which my schedule doesn’t allow me to attend, unfortunately). The word only is key. (Christian solves the problem without that word.)

We have, for some integer k,

X>N\\Y>2010\\X+N=Y\\XN=kY

and we want to minimize Y.

Write Y=ab^2, with a and b integers and b as large as possible. Then it is clear that both X and N are multiples of ab. Write X=xab and N=nab. Suppose that we have a solution to the problem, with (x,n)\neq (1,2); then the couple (x+1,n-1) would give another solution. Hence X=ab, N=2ab and ab^2=Y=X+N=3ab, so b=3.

The problem is now to find the smallest Y=9a>2010 such that no perfect square divides a (otherwise b would not be as large as possible). 2016=9\times 4^2 \times 14 and 2025=9\times 15^2 don’t work, so we move to 2034=9\times 2\times 113, which does work.

The solution is then that in year Y=2034, young N=678 will meet X=1356 and they will live happily ever after.

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One Response to “Le Monde problem: “The prediction””

  1. robinryder Says:

    Actually, X and N need only be multiples of the least common multiple of a and b, rather than of ab. Luckily, that doesn’t matter in this case.

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