## 7. Natural Fraction

What is the sum of all natural numbers $n$ for which the fraction

$$
\frac{6 n+900}{2 n-3}
$$

is a natural number?

Result: $\quad 672$
## Solution.

It holds that

$$
\frac{6 n+900}{2 n-3}=\frac{3(2 n-3)+909}{2 n-3}=3+\frac{909}{2 n-3}
$$

The given fraction $\frac{6 n+900}{2 n-3}$ is a natural number exactly when the fraction $\frac{909}{2 n-3}$ is a natural number.

The fraction $\frac{909}{2 n-3}$ is a natural number if $(2 n-3)$ is a divisor of 909.

Since $909=3 \cdot 3 \cdot 101$, the divisors of 909 are: 1, 3, 9, 101, 303, 909.

Therefore, there are six possibilities:

$$
\begin{array}{ll}
2 n-3=1 & n=2 \\
2 n-3=3 & n=3 \\
2 n-3=9 & n=6 \\
2 n-3=101 & n=52 \\
2 n-3=303 & n=153 \\
2 n-3=909 & n=456 \\
2+3+6+52+153+456=672
\end{array}
$$

The sum of all the sought numbers is 672.