Questions
The number of ways of factoring 91,000 into two factors and such that and is
detailed solution
Correct option is A
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.Similar Questions
The number of divisors of 240 which are of the form , is
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