## Task $5 / 72$

Determine all natural numbers $n$ with the following properties:

1. The number $n$ is the product of exactly three different prime numbers, each greater than 10.
2. The number $n$ can be expressed as the product of two natural numbers in such a way that their sum is 600 once and 240 another time.
The three prime factors of $n$ are $p_{1}, p_{2}$, and $p_{3}$, where $n=p_{1} p_{2} p_{3}$ and $p_{1} \neq p_{2}, p_{2} \neq p_{3}, p_{1} \neq p_{3}$, $p_{1} ; p_{2} ; p_{3}>10$. Then $n$ can be represented as the product of two natural numbers in three different ways:

$$
n=\left(p_{1} p_{2}\right) p_{3}=\left(p_{2} p_{3}\right) p_{1}=\left(p_{1} p_{3}\right) p_{2}
$$

Without loss of generality, we set $\left(p_{1} p_{2}\right)+p_{3}=600$ and $\left(p_{2} p_{3}\right)+p_{1}=240$.

Then $p_{1}=240-p_{2} p-3$, from which it immediately follows that $p_{2} ; p_{3}<23$ because $p_{1} ; p_{2} ; p_{3}>10$. Therefore, the only primes that $p_{2}$ and $p_{3}$ can be are 11, 13, 17, and 19. Since $p_{1} \geq 11$, it follows that $p_{2} p_{3} \leq 229$ or $p_{2} \leq \frac{229}{p-3}$.

For $p_{3}=19$, this gives $p_{2} \leq 12$, so $p_{2}=11$ and thus $p_{1}=31$; however, these values do not satisfy the second given equation: $\left(p_{1} p_{2}\right)+p_{3}=360 \neq 600$.

From $p_{3}=17$, it follows similarly that $p_{2} \leq 13$, so $p_{2}=13$ or $p_{2}=11$ with $p_{1}=19$ or $p_{1}=53$. By substituting into the equation $\left(p_{1} p_{2}\right)+p-3=600$, it is found that $p_{1}=53 ; p_{2}=11 ; p_{3}=17$ is a solution.

Similarly, $p_{3}=13$ (yields no solution) and $p_{3}=11$ are checked, where a second solution $p_{1}=31 ; p_{2}=$ $19 ; p_{3}=11$ is found. Since the solution method excludes further solutions, these are the only ones. The sought numbers are thus $n_{1}=53 \cdot 11 \cdot 17=9911$ and $n_{2}=31 \cdot 19 \cdot 11=6479$.