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

# The number of ways of factoring 91,000 into two factors  $m$ and $n$ such that  and  is

1. A

7

2. B

15

3. C

32

4. D

37

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### Solution:

We have 91,000 = ${2}^{3}×{5}^{3}×7×13$

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

For  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 is equal to the number of ways of partitioning A into 2 unordered non-empty subsets. W can partition A as follows:

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

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