18. Let $m$ and $n$ be relatively prime positive integers. If $m^{3} n^{5}$ has 209 positive divisors, then how many positive divisors does $m^{5} n^{3}$ have?
18. 217
Let $d(N)$ denote the number of positive divisors of an integer $N$. Suppose that the prime factorizations of $m$ and $n$ are $\prod\left(p_{i}^{a_{i}}\right)$ and $\prod\left(q_{i}^{b_{i}}\right)$ respectively. Observe that 209 has four positive divisors: $1,11,19,209$. If $m=1$, then $n^{5}$ would have 209 divisors. Thus, $d\left(n^{5}\right)=209$. However, it is known that $d\left(n^{5}\right)=\prod\left(5 b_{i}+1\right)$, but the latter implies that the remainder of $d\left(n^{5}\right)$ when divided by 5 is 1 , a contradiction.
Likewise, if $n=1$, then it implies that the remainder when $d\left(m^{3}\right)=209$ is divided by 3 is 1 , also a contradiction.
Thus, $m, n>1$, so $d(m), d(n)>1$. As $m$ and $n$ are relatively prime, then $d\left(m^{3} n^{5}\right)=$ $d\left(m^{3}\right) d\left(n^{5}\right)$. As the only way to factor 209 as a product of 2 integers greater than 1 is as $11 \cdot 19$, then $d\left(m^{3}\right)$ and $d\left(n^{5}\right)$ are 11 and 19 in some order. As the remainder when $d\left(m^{3}\right)$ and $d\left(n^{5}\right)$ is divided by 3 and 5 respectively is 1 , then $d\left(m^{3}\right)=19=(3 \cdot 6+1)$, and $d\left(n^{5}\right)=11=(5 \cdot 2+1)$. Thus, $m$ and $n$ can be expressed as $p^{6}$ and $q^{2}$ respectively. Therefore, $d\left(m^{5} n^{3}\right)=d\left(p^{30} q^{6}\right)=(30+1)(6+1)=217$.