, 2018

Solution:

We claim the answer is $72$, as it has $6$ primer factors: $6, 12, 24, 18, 36, 72$, and $6$ is a primer.

We now prove that there is no smaller primest number. Suppose there were a smaller primest number $r < 72$. We do casework on the number of distinct prime factors of $r$.

- $r$ has $\geq 4$ distinct prime factors. Then $r \geq 2 \cdot 3 \cdot 5 \cdot 7 = 210$, which is larger than $72$.

- $r$ has $3$ distinct prime factors. If each of these factors has multiplicity $1$, i.e. $r = p q s$ for distinct primes $p, q, s$, then $r$ has precisely $4$ primer factors: $p q, q s, s p, p q s$, and $4$ is not a primer. Thus, $r$ contains at least one factor of multiplicity at least $2$. If $r$ is $p^{2} q s$ for distinct primes $p, q, s$, then $r$ has $7$ distinct primer factors: $p q, q s, s p, p q s, p^{2} q, s p^{2}, p^{2} q s$, and $7$ is not a primer. Thus, if $r = p^{a} q^{b} s^{c}$, $a + b + c \geq 5$, and $r \geq 2^{3} \cdot 3 \cdot 5 = 120$, which is $\geq 72$.

- $r$ has $2$ distinct prime factors. If $r = p^{a} q^{b}$, for distinct primes $p, q$, then $r$'s primer factors are precisely its divisors of the form $p^{i} q^{j}$, where $1 \leq i \leq a$ and $1 \leq j \leq b$, meaning that it has $a b$ primer factors. Thus, $a b$ is a primer, meaning that $a b \geq 6$. Thus $r \geq 2^{3} \cdot 3^{2} = 72$, where the other possibilities can be ruled out through easy casework.

- $r$ has $1$ distinct prime factor. Then it doesn't have any primer factors, and thus cannot possibly have a primer number of them.

We conclude that $72$ is the smallest primest number.