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 m > 1, n > 1 and gcd (m, n) = 1 is 

  1. A

    7

  2. B

    15

  3. C

    32

  4. D

    37

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    Solution:

    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   2353,7,13,5323,7,13{7}23,53,13,{13}23,53,723,53{7,13},23,753,1323,1353,7

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

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