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

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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

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The following are the prime factorizations of 1134 and 168:  $168 = 2^3 × 3 × 7$ $1134 = 2 × 3^4 × 7 $ Clearly, 3 is the least positive integral number of n that allows 168 to be a factor of  $1134^n$ .  $1134^3 = 2^3 × 3^{12} × 7^3 = 1134^n $ It is evident that 12 is the least positive integral value of m that allows  $1134^3$  to be a factor of  $168^m.$   It follows that  $m + n = 12 + 3 = 15 $ The correct option is (A): 15

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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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