# Problem 1.

A librarian at a physics and mathematics school noticed that if the number of geometry textbooks in the school library is increased by several (integer) times and the number of algebra textbooks is added to the resulting number, the total is 2015. If the number of algebra textbooks is increased by the same factor and the number of geometry textbooks is added to the resulting number, the total is 1580. How many algebra textbooks are in the library?
Solution. Let the number of geometry textbooks be $x$, and the number of algebra textbooks be $y$. We can set up the system

$$
\left\{\begin{array}{l}
x n+y=2015 \\
y n+x=1580
\end{array}\right.
$$

We can write an equivalent system, where the equations represent the sum and difference of the equations in the original system:

$\left\{\begin{array}{l}x(n+1)+y(n+1)=3595, \\ x(n-1)-y(n-1)=435\end{array} \Leftrightarrow\left\{\begin{array}{l}(x+y)(n+1)=5 \cdot 719, \\ (x-y)(n-1)=3 \cdot 5 \cdot 29 .\end{array}\right.\right.$

The factor $(n+1)$ of the number $5 \cdot 719$ must be 2 more than the factor $(n-1)$ of the number $3 \cdot 5 \cdot 29$. Clearly, this is only satisfied when $n=4$. Then

$$
\left\{\begin{array} { l } 
{ x + y = 7 1 9 } \\
{ x - y = 1 4 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=432 \\
y=287
\end{array}\right.\right.
$$

Answer: 287.