8.1. Provide an example of six different natural numbers such that the product of any two of them is not divisible by the sum of all the numbers, while the product of any three of them is divisible.
# Solution.

Let $\mathrm{p}$ be a sufficiently large odd prime number. Represent the number $\mathrm{p}^{2}$ as the sum $\mathrm{a}_{1}+\ldots+\mathrm{a}_{6}$ of distinct natural numbers, none of which are divisible by $\mathrm{p}$. The numbers $\mathrm{pa}_{1}, \ldots, \mathrm{pa}_{6}$ will be the desired ones: the product of any two of them does not divide their sum, which is $\mathrm{p}^{3}$, while the product of any three does divide it.

An example is already obtained for $\mathrm{p}=5$: the decomposition $25=1+2+3+4+6+9$ yields the set of numbers: $5,10,15,20,30,45$.

Criteria. Correct example without justification: 5 points. Correct example with justification: 7 points. It is not necessary to explain how the example was found.