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] 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.
Tags: Le Monde