How many two-digit numbers have exactly 4 positive factors? (Here 1 and the number $n$ are also considered as factors of $n$.)

A positive integer $n$ has exactly 4 positive factors if and only if its prime factorization is of the form:

1. $n=p^3$ for a prime $p$, or

2. $n=p\cdot q$ for distinct primes $p<q$.

We are looking for two-digit numbers, which means $10\le n\le 99$.

Case 1: $n=p^3$

• If $p=2$: $2^3=8$ (not a two-digit number).

• If $p=3$: $3^3=27$ (valid).

• If $p=5$: $5^3=125>99$ (invalid).

This case yields exactly $1$ solution: $27$.

Case 2: $n=p\cdot q$ with $p<q$

• If $p=2$: we need $10\le 2q\le 99⟹5\le q\le 49$. The primes $q$ in this range are $5,7,11,13,17,19,23,29,31,37,41,43,47$ (13 primes).

• If $p=3$: we need $10\le 3q\le 99⟹3.33\le q\le 33$. Since $p<q$, $q\ge 5$. The primes $q$ in this range are $5,7,11,13,17,19,23,29,31$ (9 primes).

• If $p=5$: we need $10\le 5q\le 99⟹2\le q\le 19.8$. Since $p<q$, $q\ge 7$. The primes $q$ in this range are $7,11,13,17,19$ (5 primes).

• If $p=7$: we need $10\le 7q\le 99⟹1.42\le q\le 14.1$. Since $p<q$, $q\ge 11$. The primes $q$ in this range are $11,13$ (2 primes).

• If $p\ge 11$: the smallest product is $11\times 13=143>99$, so no solutions.

Summing Case 2: $13+9+5+2=29$ solutions.

Total: The total number of valid two-digit numbers is $1+29=30$.