## Task B-1.2.

Determine all natural numbers $n$ for which the fraction $\frac{4 n}{4 n-2019}$ is an integer.
## Solution.

Let's write the fraction in the following form

$$
\frac{4 n}{4 n-2019}=\frac{4 n-2019+2019}{4 n-2019}=1+\frac{2019}{4 n-2019}
$$

The obtained expression will be an integer if the denominator $4 n - 2019$ is a divisor of the number 2019.

Since $2019=3 \cdot 673$ is the prime factorization of 2019, 1 point for the fraction to be an integer, the number $4 n - 2019$ must be one of the divisors of 2019:
$-1, 1, -3, 3, -673, 673, -2019, 2019$.

1 point

This means that $4 n$ must be one of the numbers: 2018, 2020, 2016, 2022, 1346, 2692, 0, 4038.

1 point

Four of these numbers are not divisible by 4, and the number $n=0$ is not a natural number, so only three solutions remain.

The sought numbers are $n=504, n=505$, and $n=673$.

1 point

Note: If the student does not discard the solution $n=0$, they lose 1 point.