9. [6] Five guys each have a positive integer (the integers are not necessarily distinct). The greatest common divisor of any two guys' numbers is always more than 1 , but the greatest common divisor of all the numbers is 1 . What is the minimum possible value of the product of the numbers?
Answer: 32400 Let $\omega(n)$ be the number of distinct prime divisors of a number. Each of the guys' numbers must have $\omega(n) \geq 2$, since no prime divides all the numbers. Therefore, if the answer has prime factorization $p_{1}^{e_{1}} p_{2}^{e_{2}} \ldots p_{k}^{e_{k}}$, then $e_{1}+e_{2}+\ldots+e_{k} \geq 10$. If $p^{2}$ divided any of the guys' numbers, we could divide their number by $p$ to reduce the product. Therefore we may assume $e_{i} \leq 4$ for each $i$, so the smallest possible product is $2^{4} 3^{4} 5^{2}$. This bound is achievable: give the guys the numbers $10,6,6,6$, and 15 .