If \(k\) is a positive integer, what is the maximum value of \(k\) such that \(\dfrac{65!}{6^k}\) is an integer?

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by AAPL » Wed Oct 04, 2023 7:39 am

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If \(k\) is a positive integer, what is the maximum value of \(k\) such that \(\dfrac{65!}{6^k}\) is an integer?

A. \(21\)
B. \(28\)
C. \(30\)
D. \(36\)
E. \(63\)

OA C

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If \(k\) is a positive integer, what is the maximum value of \(k\) such that \(\dfrac{65!}{6^k}\) is an integer?

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If \(k\) is a positive integer, what is the maximum value of \(k\) such that \(\dfrac{65!}{6^k}\) is an integer?

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