Let \(n\) be the least positive integer such that \(168\) is a factor of \(1134^n\) . If \(m\) is the least positive integer such that \(1134^n\) is a factor of \(168^m\) , then \(m+ n \) equals

Question

The number of positive integers less than 50, having exactly two distinct factors other than 1 and itself, is

Answer 1

The prime factorizations of 1134 and 168 are: \(168 = 2^3 × 3 × 7\) \(1134 = 2 × 3^4 × 7 \) The smallest positive integer n for which 168 is a factor of \(1134^n\) is 3. \(1134^3 = 2^3 × 3^{12} × 7^3\) The smallest positive integer m for which \(1134^3\) is a factor of \(168^m\) is 12. Therefore, \(m + n = 12 + 3 = 15\). The correct option is (A): 15

Let \(n\) be the least positive integer such that \(168\) is a factor of \(1134^n\) . If \(m\) is the least positive integer such that \(1134^n\) is a factor of \(168^m\) , then \(m+ n \) equals

The Correct Option is A

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