11.3. How many points are there on the hyperbola $y=\frac{2013}{x}$ such that the tangent at these points intersects both coordinate axes at points with integer coordinates
Answer: 48 points. Solution. Let $k=2013$. The equation of the tangent to the hyperbola $y=\frac{k}{x}$ at the point $\left(x_{0}, y_{0}\right)$ is $y-y_{0}=-\frac{k}{x_{0}^{2}}\left(x-x_{0}\right)$, where $y_{0}=\frac{k}{x_{0}}$. From this, we find the coordinates of the points of intersection of the tangent with the axes $O x$ and $O y$, specifically $x_{1}=2 x_{0}, y_{1}=\frac{2 k}{x_{0}}=2 y_{0}$. Thus, $2 x_{0}$ is an integer; let us denote it by $n$. Then $y_{1}=\frac{2 k}{x_{0}}=\frac{4 k}{n}=\frac{4 \cdot 2013}{n}$. Therefore, $n$ can take any value of a divisor of the number $N=4 \cdot 2013=2^{2} \cdot 3^{1} \cdot 11^{1} \cdot 61^{1}$. The number of natural divisors of the number $N$ is $(2+1)(1+1)(1+1)(1+1)=24$ (since any divisor has the form $2^{\alpha_{1}} \cdot 3^{\alpha_{2}} \cdot 11^{\alpha_{3}} \cdot 61^{\alpha_{4}}$, where $0 \leq \alpha_{1} \leq 2$, $0 \leq \alpha_{2,3,4} \leq 1$). Taking into account the negative divisors (corresponding to tangents in the third quadrant), we get a total of 48 points.