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  • The number of ways of factoring 91,000 into two factors, and <img alt="\mathrm{n}" src="https://entra

The number of ways of factoring 91,000 into two factors, \mathrm{m} and \mathrm{n},such that \mathrm{m>1, n>1} and \mathrm{\operatorname{gcd}(m, n)=1} is

Option: 1

7


Option: 2

15


Option: 3

32


Option: 4

37


Answers (1)

best_answer

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

Let \mathrm{A= \left \{ 2^{3},5^{3},7,13 \right \}} \mathrm{A=\left\{2^{3}, 5^{3}, 7,13\right\}}\mathrm{A=\left\{2^{3}, 5^{3}, 7,13\right\}}be the set associated with the prime factorization of 91,000 . For \mathrm{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 \mathrm{m,n} is equal to the number of ways of partitioning A into 2 unordered non-empty subsets. We can partition A as follows:

\mathrm{\left\{2^{3}\right\} \cup\left\{5^{3}, 7,13\right\},\left\{5^{3}\right\} \cup\left\{2^{3}, 7,13\right\}}
\mathrm{\{7\} \cup\left\{2^{3}, 5^{3}, 13\right\},\{13\} \cup\left\{2^{3}, 5^{3}, 7\right\}}
and
\mathrm{\left\{2^{3}, 5^{3}\right\} \cup\{7,13\},\left\{2^{3}, 7\right\} \cup\left\{5^{3}, 13\right\},}
\mathrm{ \left\{2^{3}, 13\right\} \cup\left\{5^{3}, 7\right\}}

Therefore, the required number of ways \mathrm{=4+3=7}.

Posted by

sudhir.kumar

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