Let $m=30030=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ and let $M$ be the set of its positive divisors which have exactly two prime factors. Determine the minimal integer $n$ with the following property: for any choice of numbers from $M$, there exist three numbers $a, b, c$ among them satisfying $a \cdot b \cdot c=m$.

Answer: $n=11$.
Taking the 10 divisors without the prime 13 shows that $n \geq 11$. Consider the following partition of the 15 divisors into five groups of three each with the property that the product of the numbers in every group equals $m$.

$$
\begin{array}{ll}
\{2 \cdot 3,5 \cdot 13,7 \cdot 11\}, & \{2 \cdot 5,3 \cdot 7,11 \cdot 13\}, \\
\{2 \cdot 11,3 \cdot 5,7 \cdot 13\}, & \{2 \cdot 13,3 \cdot 11,5 \cdot 7\} .
\end{array}
$$

If $n=11$, there is a group from which we take all three numbers, that is, their product equals $m$.