How many positive two-digit integers have exactly 8 positive factors?

Question

I solved this problem by listing all two-digit integers and going through each one. Is there an easy way to solve the problem?

How many positive two-digit integers have exactly 8 positive factors?

Answer 1

If $n$ has prime power factorization $n=p^aq^br^c$ then the number of factors for $n$ is $(a+1)(b+1)(c+1)$ ( A number with 4 or more distinct prime factors will exceed 100). For numbers of the type $p^aq^b$ the count is $(a+1)(b+1)$.

Now see how to get $8$ as product of the above type $(a+1)(b+1)(c+1)$, or $(a+1)(b+1)$ then choose primes $p,q,r$ to ensure the number is $<100$.