11 Given a positive integer $n$, among its positive divisors, there is at least one positive integer ending in each of the digits $0,1,2, \cdots, 9$. Find the smallest $n$ that satisfies this condition.
11. The smallest $n$ that satisfies the condition is $270$.

In fact, from the condition, we know that $10 \mid n$, and we start the discussion from the factor of the last digit of $n$ being 9. If $9 \mid n$, then $90 \mid n$, and it can be directly verified that 90 and 180 are not multiples of any number ending in 7; if $19 \mid n$, then $190 \mid n$, and $n=190$ is not a multiple of any number ending in 7. However, 270 is a multiple of $10, 1, 2, 3, 54, 5, 6, 27, 18, 9$, which meets the conditions. Therefore, the smallest $n$ is 270.