A natural number n such that n! ends in exact | vedclass

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(C) The number of trailing zeros in $n!$ is given by Legendre's formula: $E_5(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{5^k} \rfloor$. We want $E_5(n

Vedclass · Accepted Answer

$4009$

Vedclass · Answer

(C) The number of trailing zeros in $n!$ is given by Legendre's formula: $E_5(n!) = \sum_{k=1}^{\infty} \lfloor \frac{n}{5^k} \rfloor$. We want $E_5(n!) = 1000$. Testing option $C$ $(n=4009)$: $E_5(4009!) = \lfloor \frac{4009}{5} \rfloor + \lfloor \frac{4009}{25} \rfloor + \lfloor \frac{4009}{125} \rfloor + \lfloor \frac{4009}{625} \rfloor + \lfloor \frac{4009}{3125} \rfloor$ $= 801 + 160 + 32 + 6 + 1 = 1000$. Thus,$n=4009$ is the correct value.

$A$ natural number $n$ such that $n!$ ends in exactly $1000$ zeros is

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