## 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]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 $N$ 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.

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.