Task B-1.2. How many integer solutions does the inequality $\frac{x^{2}-2015 x+2014}{2013 x^{2}-2013 x} \leq 0$ have?
## Solution.

Let's write the numerator and denominator of the fraction on the left side of the given inequality as a product,

$$
\begin{aligned}
& \frac{x^{2}-2014 x-x+2014}{2013 x(x-1)} \leq 0 \\
& \frac{x(x-2014)-(x-2014)}{2013 x(x-1)} \leq 0 \\
& \frac{(x-2014)(x-1)}{2013 x(x-1)} \leq 0
\end{aligned}
$$

(2 points)

For $x \neq 0$ and $x \neq 1$, the last inequality is equivalent to the following:

$$
\frac{x-2014}{x} \leq 0
$$

This implies

$$
\left\{\begin{array} { l } 
{ x - 2 0 1 4 \leq 0 } \\
{ x > 0 }
\end{array} \quad \text { or } \left\{\begin{array}{l}
x-2014 \geq 0 \\
x<0
\end{array}\right.\right.
$$

Thus, the solution to the inequality is the interval $\langle 0,2014] \backslash\{1\}$.

(2 points)

Within this interval, there are 2013 integer solutions to the given inequality.