Task B-2.5. (20 points) Find all solutions to the equation $\lfloor x\rfloor\{x\}=2009 x$, where $\lfloor x\rfloor$ is the greatest integer not greater than $x$, and $\{x\}=x-\lfloor x\rfloor$.
Solution. Obviously $x=0$ satisfies the equation.

For $x>0$, we have $0 \leq\lfloor x\rfloor \leq x, 0 \leq\{x\}<1$, so $\lfloor x\rfloor\{x\}<x<2009 x$. In this case, there is no solution.

For $x<0$, we have $x=-p+\alpha$, where $p=-\lfloor x\rfloor$ is a natural number, and $\alpha=\{x\}$. The equation now reads

$$
(-p) \cdot \alpha=(-p+\alpha) \cdot 2009
$$

whose solution is $\alpha=\frac{2009 p}{2009+p}$.

Since $\alpha<1$, we have $\frac{2009 p}{2009+p}<1$ or $p<\frac{2009}{2008}$. Therefore, $p=1, \alpha=\frac{2009}{2010}$, which gives $x=-\frac{1}{2010}$.

Thus, the equation has two solutions: $x=0, x=-\frac{1}{2010}$.

## NATIONAL MATHEMATICS COMPETITION

## 3rd grade - high school - B category

Pula, March 30, 2009.