Factorial problem again !!

For such questions divide the number continuously by powers of 5 less than the number itself and then sum all the quotients.

• 8000=5 * 1600 +0
• 8000=25 * 320 +0
• 8000=125 * 64 +0
• 8000=625 * 12 +500
• 8000=3125 * 2 +1750

Thus sum of quotients is 1998.

I have a two line answer without too much calculation.

$\large\ { v }_{ p }(n) = \frac { n - { S }_{ p }(n) }{ p - 1 }$ .

This formula states that highest power of prime $p$ in $n$ is given by above statement.

${ S }_{ p }(n)$ is sum of digits of $n$ in base $p$ .

In our case, $n = 8000, p = 5$ . So $\boxed{1998}$ is our answer, as simple as that.

$5 \underline{|8000}$

$5 \underline{|1600}$

$5 \underline{|320}$

$5 \underline{|64}$ *

$5 \underline{|12}$ *

$\textcolor{#FFFFFF}{ab}2$

*All the divisions above are integer divisions.

$1600 + 320 + 64 + 12 + 2 = \boxed{1998}$

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Note: This is no different from the process described in the GIF wiki except that it's faster, looks good, and takes less space.

$\lfloor \frac{8000}{5} \rfloor\ + \lfloor \frac{8000}{5²} \rfloor\ + \lfloor \frac{8000}{5³} \rfloor\ + \lfloor \frac{8000}{5⁴} \rfloor\ + \lfloor \frac{8000}{5^5} \rfloor$

$= 1600 + 320 + 64 + 12 + 2$

$=1998$

This wiki from contest maths has got more details.