## Task B-4.1.

Determine all natural numbers $n$ for which the fraction $\frac{2 n^{2}-40 n+175}{n-10}$ is a natural number.
## Solution.

Notice that the following holds:

$$
\begin{array}{rlr}
\frac{2 n^{2}-40 n+175}{n-10}= & \frac{2 n^{2}-40 n+200-25}{n-10}=\frac{2(n-10)^{2}-25}{n-10} & 2 \text { points } \\
& =2 n-20-\frac{25}{n-10}
\end{array}
$$

The obtained expression will be a natural number if the fraction $\frac{25}{n-10}$ is an integer, which is only possible if the denominator $n-10$ is a divisor of the number 25. Therefore,

$$
n-10 \in\{ \pm 1, \pm 5, \pm 25\}
$$

From the condition that $n \in \mathbb{N}$, it follows that:

$$
n \in\{5,9,11,15,35\}
$$

For each of the obtained numbers $n$, it is still necessary to check whether the original fraction, or the expression $2 n-20-\frac{25}{n-10}$, is a natural number. By direct verification, we obtain that the final solution is

$$
n \in\{9,15,35\}
$$

Note: The student can transform the initial fraction in some other way, for example:

$$
\begin{gathered}
\frac{2 n^{2}-40 n+175}{n-10}=\frac{2 n^{2}-20 n-20 n+200-25}{n-10}=\frac{2 n(n-10)-20(n-10)-25}{n-10} \\
=2 n-20-\frac{25}{n-10}
\end{gathered}
$$