$X!$ Has 1000 Trailing Zeros

This problem was inspired by one of the daily problems .

From the previous problem, we observe that the number of trailing zeros approaches n/4, so a good starting point would be 4000 (although we would expect that to be slightly under 1000).

4000! has 800 multiples of 5; 160 multiples of 25; 32 multiples of 125; 6 multiples of 625; and 3125. As suspected, that's 999 trailing zeros.

4001, 4002, 4003, and 4004 don't add any more 5s.

4005! has 801 multiples of 5; 160 multiples of 25; 32 multiples of 125; 6 multiples of 625; and 3125. That's exactly 1000 trailing zeros.

$\sum\limits_{i=1} \lfloor \frac x{5^i} \rfloor = 1000 \\ \therefore \frac x4 \approx 1000 \Rightarrow x \approx 4000$

but $\sum\limits_{i=1} \lfloor \frac {4000}{5^i} \rfloor = 800 + 160 + 32 + 6 + 1 = 999$

So should add one 5 more, i.e. $x = 4005$

I used the Legendre equation. If 1000 traling zeros are needed then ${\huge \left\lfloor \frac{X}{5} \right\rfloor + \left\lfloor \frac{X}{5^2}\right\rfloor + \left\lfloor \frac{X}{5^3}\right\rfloor + \left\lfloor \frac{X}{5^4}\right\rfloor + \left\lfloor \frac{X}{5^5}\right\rfloor = 1000}$

${\huge X=\frac{1000 \cdot 3125}{781} \cong 4001}$

It's near the result because the next 5 multiple 4005 is the correct answer.