## Task 3 - 260913

Determine the number of all different triples of integers $(x, y, z)$ for which

(1) $x \leq y \leq z$ and

(2) $x y z=1986$ holds!

Hint: Two triples $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ are called different if and only if at least one of the inequalities $x_{1} \neq x_{2} ; y_{1} \neq y_{2} ; z_{1} \neq z_{2}$ holds.
Due to the prime factorization $1986=2 \cdot 3 \cdot 331$, there are exactly the following five different triples of natural numbers that satisfy (1) and (2):

$$
(1,1,1986), \quad(1,2,993), \quad(1,3,662), \quad(1,6,331), \quad(2,3,331)
$$

All other triples of integers that satisfy (1) and (2) are obtained by replacing exactly two numbers in each of the mentioned triples with their respective opposite numbers and choosing the order of the resulting numbers according to (1).

Starting from $(1,1,1986)$, this leads to exactly two additional triples

$$
(-1986,-1,1), \quad(-1,-1,1986)
$$

Starting from each triple $(x, y, z)$ with $0<x<y<z$, this leads, considering $-z<-y<-x<x<y<z$, to exactly the three additional triples

$$
(-z,-y, x), \quad(-z,-x, y), \quad(-y,-x, z)
$$

Therefore, the total number of triples is $5+2+4 \cdot 3=19$.