## Task 2 - 340842

Determine all natural numbers $n$ that satisfy the following conditions (1) and (2)!

(1) The number $n$ is the product of exactly three prime numbers; any two of these prime numbers are different from each other; each of these prime numbers is greater than 10.

(2) The number $n$ can be represented as the product of two natural numbers whose sum is 600. The number $n$ can also be represented as the product of two natural numbers whose sum is 240.
I. If a natural number $n$ satisfies conditions (1) and (2), then:

According to (1), there exist prime numbers $p, q, r$ such that $p>10, q>10, r>10, p \neq q, p \neq r, q \neq r$, and $n=p \cdot q \cdot r$. All representations of $n$ as a product of two natural numbers are therefore

$$
n=1 \cdot p q r=p \cdot q r=q \cdot p r=r \cdot p q
$$

According to (2), one of the numbers $1+p q r, p+q r, q+p r, r+p q$ is equal to 600, and another is equal to 240.

Both 599 and 239 are prime numbers. This follows from $25^{2}>599$ and $16^{2}>239$, respectively, because 599 is not divisible by any of the prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, and 239 is not divisible by any of the prime numbers $2,3,5,7,11,13$.

Thus, the cases $1+p q r=600$ and $1+p q r=240$ are excluded.

Therefore, the labels $p, q, r$ can be chosen such that the equations

$$
p+q r=600 \quad, \quad q+p r=240
$$

hold.

Hence, $p r<240$, and due to the condition $p \geq 11$, we have $r<\frac{240}{11}<22$ and, due to the condition $r \geq 11$, also $p<22$. Therefore, only the prime numbers 11, 13, 17, 19 are possible for $p$ and $r$.

Thus, the number $q r=600-p$ is one of the numbers 589, 587, 583, 581. Of these, only $589=19 \cdot 31$ and $583=11 \cdot 53$ are the product of two prime numbers greater than 10, while this is not the case for the prime number 587 and for $581=7 \cdot 83$.

Therefore, the only possibilities are:
either $r=19, q=31, p=600-589=11, n=11 \cdot 31 \cdot 19=6479$

or $r=11, q=53, p=600-583=17, n=17 \cdot 53 \cdot 11=9911$

II. For each of these two numbers $n$, the given factorization shows that (1) is satisfied. Furthermore, the checks show that (2) is also satisfied.

With I. and II., it is shown that exactly the two numbers $n=6479$ and $n=9911$ satisfy conditions (1) and (2).