
3. Find all integers greater than 1 by which a cancellation can be made in some fraction of the form

$$
\frac{3 p-q}{5 p+2 q}
$$

where $p$ and $q$ are mutually prime integers.


Solution. The fraction admits cancellation by an integer $d>1$ if and only if $d$ is a common divisor of its numerator and its denominator. Let us thus assume that $d \mid 3 p-q$ and at the same time $d \mid 5 p+2 q$, where $p$ and $q$ are mutually prime integers. Adding up the appropriate multiples of the two binomials $3 p-q$ and $5 p+2 q$, we get

$$
2(3 p-q)+(5 p+2 q)=11 p \quad \text { and } \quad 3(5 p+2 q)-5(3 p-q)=11 q
$$

Since both $3 p-q$ and $5 p+2 q$ are assumed to be divisible by $d$, the two numbers $11 p$ and $11 q$ must also be multiples of $d$. However, $p$ and $q$ are mutually prime by hypothesis and 11 is a prime, thus the numbers $11 p$ and $11 q$ have only one common divisor greater than 1 , namely the number 11. Thus $d=11$.

We now need to show that it is indeed possible to make a cancellation by 11 in some fraction of the given form. That is, we need to find a pair of mutually prime integers $p$ and $q$ so that $11 \mid 3 p-q$ and at the same time $11 \mid 5 p+2 q$. Solving the system of equations

$$
3 p-q=11 m \text { and } \quad 5 p+2 q=11 n
$$

we get $(p, q)=(2 m+n, 3 n-5 m)$, and $p, q$ will certainly be prime if we choose them so that $q=3 n-5 m=1$, thus e.g. for $n=2$ and $m=1$, when $(p, q)=(4,1)$ and the corresponding fraction is $11 / 22$.

Answer. The only integer greater than 1 , which can be cancelled in some fraction of the given form, is the number 11 .

## Second Round of the 58th Czech Mathematical Olympiad (January 20th, 2009)

![](https://cdn.mathpix.com/cropped/2024_04_17_5885ae9551045f8982ddg-12.jpg?height=163&width=384&top_left_y=322&top_left_x=575)
