The number of ways of factoring 91,000 into two factors  $m$ and $n$ such that $m > 1, n > 1$ and $gcd (m, n) = 1$ is 

We have 91,000 = $2^3\times 5^3\times 7\times 13$ 

Let A = $\left\{2^3,5^3,7,13\right\}$ be the set associated with the prime factorization of 91,000.

 For $m, n$ to be relatively prime, each element of A must appear either in the prime factorization of $m$ or in the prime factorization of $n$ but not in both. 

Moreover, the 2 prime factorizations must be composed exclusively from the elements of A.

 Therefore, the number of relatively prime pairs $m, n$is equal to the number of ways of partitioning A into 2 unordered non-empty subsets. W can partition A as follows:

and   $\begin{array}{l}\left\{2^3\right\}\cup \left\{5^3,7,13\right\},\left\{5^3\right\}\cup \left\{2^3,7,13\right\}\\ \left\{7\right\}\cup \left\{2^3,5^3,13\right\},\left\{13\right\}\cup \left\{2^3,5^3,7\right\}\\ \left\{2^3,5^3\right\}\cup \{7,13\},\left\{2^3,7\right\}\cup \left\{5^3,13\right\}\\ \left\{2^3,13\right\}\cup \left\{5^3,7\right\}\end{array}$

Therefore, the required number of ways = 4 + 3 = 7.