2. Natural numbers $x$ and $y$ are such that the following equality holds:

$$
x^{2}-3 x=25 y^{2}-15 y
$$

How many times greater is the number $x$ than the number $y$?
Answer. 5

Solution. The equality $x^{2}-3 x=25 y^{2}-15 y$ is equivalent to the equality $(5 y-x)(5 y+x-3)=$ 0. Since $x$ and $y$ are natural numbers, $x, y \geq 1$. Therefore, the second bracket $5 y+x-3 \geq 3$, in particular, is non-zero. Therefore, the first bracket $5 y-x$ is zero, meaning $x$ is 5 times $y$.

Criteria.

## $(-)$

$(-$.$) An example of specific numbers is considered.$

$(-/+)$ From the equality $x(x-3)=5 y(5 y-3)$, the conclusion $x=5 y$ is drawn without satisfactory explanations.

(+/2) A complete square is factored out on both sides, but an incorrect conclusion is made $(x-1.5)^{2}=$ $(5 y-1.5)^{2} \Rightarrow x-1.5=5 y-1.5$. The second case $(x-1.5)=-(5 y-1.5)$ is not considered.

(+/-) The equality $(x-5 y)(x+5 y-3) \neq 0$ is obtained, from which it is concluded that $x-5 y=0$, but it is not explained why $(x+5 y-3) \neq 0$.

(+.) Minor logical steps are omitted

$(+)$