Imagine as 2015 fractions:

$$
\frac{2}{2016}, \frac{3}{2015}, \frac{4}{2014}, \ldots, \frac{2014}{4}, \frac{2015}{3}, \frac{2016}{2}
$$

Is it possible to choose three of these fractions with a product equal to 1?
Solution

Yes. See that each fraction is of the form $\frac{x}{2018-x}$. Thus, for $x=1009$, the fraction $\frac{1009}{1009}=1$ is part of the list. It then suffices to multiply the fractions:

$$
\frac{2}{2016} \cdot \frac{1009}{1009} \cdot \frac{2016}{2}=1
$$