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

Question

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

Infinity Learn · Accepted Answer

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 $$=$$ $$4$$ $$+$$ $$3$$ $$=$$ $$7$$.

The number of ways of factoring 91,000 into two factors $m$ and $n$ such that $m > 1, n > 1$ and $g c d (m, n) = 1$ is

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

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

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

The number of ways of factoring 91,000 into two factors $m$ and $n$ such that $m > 1, n > 1$ and $g c d (m, n) = 1$ is