Smallest number whose number of factors is prime.

Since n is a prime number, the only way a number can have n factors is that the number, in its prime factorization form, is in the form $x^{(n-1)}$ . The lowest possible number that can have n factors, where n is prime, is $2^{(n-1)}$ , because 2 is the lowest prime number. $log_2{1000000}$ is approximately 19.9. Therefore, the highest possible prime value of n is 19. There are 8 prime numbers less than or equal to 19: 2, 3, 5, 7, 11, 13, 17, and 19. Therefore, the answer is $\boxed{8}$ .

Note that if an integer $D(n)$ has a prime factorization of $D(n) = p_1^{q_1}p_2^{q_2}p_3^{q_3}...p_k^{q_k}$ , then the number of divisors of $D(n)$ is $n = (q_1 + 1)(q_2 + 1)(q_3 + 1)...(q_k + 1)$ .

Because $n$ is a prime, only one factor of $n$ is $n$ itself, namely $n = q_1 + 1$ . Meanwhile every other factor $(q_2 + 1), (q_3 + 1), ...(q_k + 1)$ is $1$ , giving $q_2 = q_3 = ... = q_k = 0$ . So $D(n)$ is of the form $D(n) = p_1^{n - 1}$ , where $p_1, n$ are prime numbers. And because $D(n)$ is minimally chosen, $p_1$ has to be $2$ . Therefore for a prime number $n$ , $D(n) = 2^{n - 1}$ .

Now we determine the largest number $N$ such that $D(N) = 2^{N - 1} < 1000000 = 10^6$ . You can find $N$ by direct computation, or estimate it as follow: Since $5^3 = 125 < 128 = 2^7$ , and $2^6 \sqrt{2} < 2^6 * 1.5 = 96 < 125$ , we get $2^6 \sqrt{2} < 5^3 < 2^7$ . This implies $(2^6 \sqrt{2})^2 < (5^3)^2 < (2^7)^2$ or $2^{13} < 5^6 < 2^{14}$ , leading to $2^{19} < 2^6*5^6 = 10^6 < 2^{20}$ .

Hence, $2^{N - 1} = 2^{19}$ or $N = 20$ , and every prime number that are less or equal to $N$ will satisfy the requirement. This leaves us with $8$ possibilities of $n$ , which are $2,3,5,7,11,13,17,19$ with the corresponding $D(n)$ are $2,4,16, 64, 2^{10}, 2^{12}, 2^{16}, 2^{18}$ .