322. Find at least one natural number $n$ such that each of the numbers $n, n+1, n+2, \ldots, n+20$ has a common divisor with the number $30030=2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ that is greater than one.
322. Answer: for example, 9440. Let $m=2 \cdot 3 \cdot 5 \cdot 7 \cdot k$. By choosing $k$ such that $m-1$ is divisible by 11, and $m+1$ is divisible by 13, we get that the number $n=m-10$ satisfies the condition of the problem (P2).