2. Provide an example of six different natural numbers such that the product of any two of them does not divide the sum of all the numbers, while the product of any three of them does divide. (S. Volchenkov)
Solution. Let $p-$ be a sufficiently large odd prime number. Represent the number $p^{2}$ as the sum $a_{1}+\ldots+a_{6}$ of distinct natural numbers, none of which are divisible by $p$. The numbers $p a_{1}, \ldots, p a_{6}$ will be the desired ones: the product of any two of them does not divide their sum, equal to $p^{3}$, while the product of any three does. An example is already obtained at $p=5$: the decomposition $25=1+2+3+4+6+9$ yields the set of numbers $5,10,15,20,30,45$.