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 = 23×53×7×13 Let A = 23,53,7,13 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   23∪53,7,13,53∪23,7,13{7}∪23,53,13,{13}∪23,53,723,53∪{7,13},23,7∪53,1323,13∪53,7Therefore, the required number of ways = 4 + 3 = 7.