Problem 2. Let $a, b$ and $c$ be integers such that $a b=-6, a c=-10$ and $b c=15$. Find the product $a b c$ and the numbers $a, b$ and $c$.
Solution. Since $a$ is an integer that is a divisor of the numbers -6 and -10, we get that $a \in\{-1,1,-2,2\}$. In the cases where $a=1$ or $a=-1$, we get that $bc=60$, which is not correct. If $a=2$, then $b=-3$ and $c=-5$, so $bc=15$ and therefore $abc=30$. If $a=-2$, then $b=3$ and $c=5$, so $bc=15$ and therefore $abc=-30$.

Thus, the only solutions to the problem are:

$$
a=2, b=-3, c=-5, abc=30 \text { and } a=-2, b=3, c=5, abc=-30 \text {. }
$$