## Task 1 - 250921

Determine all triples $(a, b, c)$ of natural numbers $a, b, c$ for which $a \leq b \leq c$ and $a \cdot b \cdot c=19 \cdot 85$ hold!
The prime factorization of $19 \cdot 85=1615$ is $1615=5 \cdot 17 \cdot 19$. Therefore, for the triples to be determined, there are exactly the following possibilities.

1. Exactly the two numbers $a$ and $b$ are equal to 1. This leads to the triple $(1,1,1615)$.
2. Exactly one number $a$ is equal to 1. Then $b$ must contain at least one of the prime factors 5, 17, 19. If $b$ contained more than one of these factors, then $b \geq 5 \cdot 17=85$. On the other hand, $c$ would then contain at most one of these factors, so $c \leq 19$. This contradicts the condition $b \leq c$.

Therefore, $b$ contains exactly one of the prime factors 5, 17, 19, and $c$ contains the other two. This leads to the triples $(1,5,323),(1,17,95),(1,19,85)$.

3. None of the three numbers $a, b, c$ is equal to 1, leading to the triple $(5,17,19)$.

Thus, exactly the five triples listed in 1., 2., 3. are the desired ones.