+283 Let $m$ and $n$ be positive integers such that $m$ has exactly $8$ positive divisors, $n$ has exactly $6$ positive divisors, and $mn$ has exactly $12$ positive divisors. What is the smallest possible value of $mn?$
+9673 From my response to your other post, m is either of the form \(pq^3\) or pqr, where p, q, r are distinct primes.
By a similar approach, n is of the form \(pq^2\) where p, q are distinct primes.
Case 1: m is of the form pq^3.
Suppose n = rs^2, where r, s are distinct primes. r and s must be equal to p and q, regardless of order, since otherwise mn has more than 12 positive divisors. Then mn = p^2 q^5 or p^3 q^4. Taking the smallest values require q = 2 and p = 3. Then the minimum value of mn is min(3^2 * 2^5, 3^3 * 2^4) = 288.
Case 2: m is of the form pqr.
No matter what, mn will have more than 12 positive divisors.
Hence, the answer is 288.
