## Task A-3.4.

For how many natural numbers $n$ is the value of the fraction

$$
\frac{n+2022}{n-3}
$$

an integer?
## Solution.

Let's rewrite the expression in a different way:

$$
\frac{n+2022}{n-3}=\frac{2025}{n-3}+1
$$

If $n-3$ is positive, it must be a divisor of the number 2025. The number 2025 has 15 positive divisors:

1 point

$$
1,3,5,9,15,25,27,45,75,81,135,225,405,675,2025
$$

1 point

If the expression $n-3$ is negative, it can take the values -1 or -2 (since $n$ is a natural number). 1 point

Since 2 is not a divisor of 2025, the only possibility is $n-3=-1$ (i.e., $n=2$).

1 point

We conclude that there are a total of 16 natural numbers $n$ with the property described in the problem.

Note: The third point in the scoring scheme is given for the argument that 2025 has exactly 15 positive divisors. One alternative way to prove this claim is by using the factorization $2025=3^{4} 5^{2}$ and the formula for the number of positive divisors $\tau(m)$ of a number $m$ (where $m=p_{1}^{a_{1}} \cdots p_{k}^{a_{k}}$ is its prime factorization):

$$
\tau(m)=\left(a_{1}+1\right) \cdot \ldots \cdot\left(a_{k}+1\right)
$$