If 15^k divides 47! but 15^{k+1} does not div | vedclass

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Question

(C) Since $15 = 3 	imes 5$,the exponent of $15$ in $47!$ is determined by the exponent of the prime factor $5$ because $5$ is less frequent than $3$

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Since $15 = 3 	imes 5$,the exponent of $15$ in $47!$ is determined by the exponent of the prime factor $5$ because $5$ is less frequent than $3$ in the prime factorization of $47!$. Using Legendre's Formula,the exponent of a prime $p$ in $n!$ is given by $E_p(n!) = \sum_{i=1}^{\infty} \left[ \frac{n}{p^i} ight]$. For $p=5$ and $n=47$: $E_5(47!) = \left[ \frac{47}{5} ight] + \left[ \frac{47}{25} ight] = 9 + 1 = 10$. Thus,the highest power of $15$ that divides $47!$ is $15^{10}$. Therefore,$k = 10$. Hence,option $(C)$ is correct.

If $15^k$ divides $47!$ but $15^{k+1}$ does not divide it,then $k=$

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