Solution. Suppose this is not the case. Let these numbers be denoted by $a_{1}, a_{2}, \ldots, a_{15}$, and their smallest prime divisors by $p_{1}, p_{2}, \ldots, p_{15}$, respectively. Since all the numbers are pairwise coprime, these prime divisors are distinct. The first 15 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, $37,41,43,47$. This means that the largest of the prime divisors $p_{i} \geq 47$. Then the corresponding number $a_{i} \geq 47^{2}=2209>2022$, which contradicts the condition. Therefore, one of the numbers $a_{1}, a_{2}, \ldots, a_{15}$ is prime.