Take a gamble

For the product $A$ to be divisible by $21$ at least one of the 21 digits must be divisible by $3$ , i.e., must be one of $3,6,9$ , and at least one of the digits must be $7$ . It will be easier though to first count the number of elements of $S$ that are $not$ divisible by $21$ .

There are $5^{21}$ elements of $S$ that have no digits that are $7$ or are divisible by $3$ , (i.e., those 21-digit integers formed using the digits $1,2,4,5,8$ ).

Next, as there are $8^{21}$ elements that do not have $7$ as a digit, there are $8^{21} - 5^{21}$ elements that have at least one digit that is a multiple of $3$ but have no $7$ .

Finally, as there are $6^{21}$ elements that have none of $3,6,9$ as digits, there are $6^{21} - 5^{21}$ elements that have none of $3,6,9$ as digits but do have at least one $7$ .

Thus there are $5^{21} + (8^{21} - 5^{21}) + (6^{21} - 5^{21}) = 8^{21} + 6^{21} - 5^{21}$ elements of $S$ that are not divisible by $21$ . As $S$ has $9^{21}$ elements in total, the probability $P$ that the product $A$ associated with a randomly chosen elements $N$ of $S$ is divisible by $21$ is

$P = \dfrac{9^{21} - (8^{21} + 6^{21} - 5^{21})}{9^{21}} = 0.915509802...$ , and so $\lfloor 1000P \rfloor = \boxed{915}$ .