We have 91,000 =
Let A = 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 or in the prime factorization of 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:
Therefore, the required number of ways = 4 + 3 = 7.