15. How many ordered pairs $(x, y)$ of positive integers $x$ and $y$ satisfy the relation $x y + 5(x + y) = 2005$?
15. The answer is 10. By adding 25 to both sides, we obtain the equivalent relation $x y+5(x+y)+25=2030$ from which, by factoring the first member, we get $(x+5)(y+5)=2030=2 \cdot 5 \cdot 7 \cdot 29$. Since $x$ is an integer, then $x+5$ is also an integer, and the same applies to $y$. Therefore, it is sufficient to distribute the prime numbers of the factorization of 2030 between the two factors $x+5$ and $y+5$, in all possible ways.

Since $x$ must be positive, then $x+5$ must be at least 6, as must $y+5$. In particular, both factors must be positive. Furthermore, the equation is symmetric in $x$ and $y$, so we can assume that the only factor 2 divides $x+5$. For $x+5$, there are then 5 possibilities: $2 \cdot 5, 2 \cdot 7, 2 \cdot 29, 2 \cdot 5 \cdot 7, 2 \cdot 5 \cdot 29$. The other cases are $x+5=2$, which is to be excluded, $x+5=2 \cdot 7 \cdot 29$ and $x+5=2 \cdot 5 \cdot 7 \cdot 29$, which are to be excluded because they imply $y+5=5$ and $y+5=1$ respectively. The cases where 2 divides $y+5$ are just as many, and are all distinct from the previous ones because there is only one factor 2 in 2030. Therefore, the answer is $5 \cdot 2=10$.