This week’s *Le Monde* mathematical problem is:

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

and we want to minimize .

Write , with and integers and as large as possible. Then it is clear that both and are multiples of . Write and . Suppose that we have a solution to the problem, with ; then the couple would give another solution. Hence , and , so .

The problem is now to find the smallest such that no perfect square divides (otherwise would not be as large as possible). and don’t work, so we move to , which does work.

The solution is then that in year , young will meet and they will live happily ever after.

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This entry was posted on 04/03/2010 at 14:24 and is filed under Mathematical games. You can follow any responses to this entry through the RSS 2.0 feed.
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05/03/2010 at 09:34 |

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