The largest non-negative integer k such that | vedclass

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(B) To find the largest $k$ such that $24^k$ divides $13!$,we first find the prime factorization of $13!$.$13! = 2^{10} 	imes 3^5 	imes 5^2 	imes 7

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$3$

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(B) To find the largest $k$ such that $24^k$ divides $13!$,we first find the prime factorization of $13!$.$13! = 2^{10} 	imes 3^5 	imes 5^2 	imes 7 	imes 11 	imes 13$.We have $24^k = (2^3 	imes 3)^k = 2^{3k} 	imes 3^k$.For $24^k$ to divide $13!$,we must have $3k \le 10$ and $k \le 5$.From $3k \le 10$,we get $k \le \frac{10}{3} \approx 3.33$.From $k \le 5$,we get $k \le 5$.The largest integer $k$ satisfying both conditions is $k = 3$.

The largest non-negative integer $k$ such that $24^k$ divides $13!$ is

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