Question:
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
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
Updated On: Jan 13, 2026
- 15
- 12
- 24
- 9
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The Correct Option is A
Solution and Explanation
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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