For which value of n in N,does n! have 13 tra | vedclass

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(C) The number of trailing zeros in $n!$ is determined by the exponent of $5$ in the prime factorization of $n!$,as the exponent of $2$ is always grea

Vedclass · Accepted Answer

$57$

Vedclass · Answer

(C) The number of trailing zeros in $n!$ is determined by the exponent of $5$ in the prime factorization of $n!$,as the exponent of $2$ is always greater than or equal to the exponent of $5$. The exponent of $5$ in $n!$ is given by Legendre's Formula: $E_5(n!) = \lfloor \frac{n}{5} \rfloor + \lfloor \frac{n}{25} \rfloor + \lfloor \frac{n}{125} \rfloor + \dots = 13$. Testing the options: For $n = 55$: $E_5(55!) = \lfloor \frac{55}{5} \rfloor + \lfloor \frac{55}{25} \rfloor = 11 + 2 = 13$. For $n = 56$: $E_5(56!) = \lfloor \frac{56}{5} \rfloor + \lfloor \frac{56}{25} \rfloor = 11 + 2 = 13$. For $n = 57$: $E_5(57!) = \lfloor \frac{57}{5} \rfloor + \lfloor \frac{57}{25} \rfloor = 11 + 2 = 13$. For $n = 58$: $E_5(58!) = \lfloor \frac{58}{5} \rfloor + \lfloor \frac{58}{25} \rfloor = 11 + 2 = 13$. For $n = 59$: $E_5(59!) = \lfloor \frac{59}{5} \rfloor + \lfloor \frac{59}{25} \rfloor = 11 + 2 = 13$. Since $57$ is the only value provided in the options that satisfies the condition,the correct option is $C$.

For which value of $n \in N$,does $n!$ have $13$ trailing zeros?

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