A natural number ‘n’ such that n! ends in exactly 1000 zeros. Then $n=$

Number of zero’s in ends of any number = highest power of 10 (i.e highest power of 5 in it)
i.e $10=2\times 5$
Number n! will have highest power of 5
$n!=2^{\alpha }{.2}^{\beta }{.5}^{{ }^{\gamma }}\mathrm{....}$
$\gamma =\left[\frac{{\displaystyle n}}{{\displaystyle 5}}\right]+\left[\frac{{\displaystyle n}}{{\displaystyle 5^2}}\right]+\mathrm{......}+\left[\frac{{\displaystyle n}}{{\displaystyle 5^k}}\right]$
Let $n=4009$
$\gamma =\left[\frac{{\displaystyle 4009}}{{\displaystyle 5}}\right]+\left[\frac{{\displaystyle 4009}}{{\displaystyle 5^2}}\right]+\mathrm{....}+\left[\frac{{\displaystyle 4009}}{{\displaystyle 5^5}}\right]$
$\gamma =801+160+32+6+1=1000$
$\therefore n=4009$