Problem with factorials: find the least n for which (n + 16)!(n+20)! ends with a number of zeros divisible by 2016

Question

I have been working on this problem during the last few days but I can't find a good solution.

Answer 1

Number of zeros ended $n!$ gives formula

$$p = \left[\frac{n}{5^1}\right] + \left[\frac{n}{5^2}\right]+... + \left[\frac{n}{5^k}\right]$$ At first check $n=4040$:

$$\left[\frac{4040}{5^1}\right] + \left[\frac{4040}{5^2}\right]+ \left[\frac{4040}{5^3}\right]+ \left[\frac{4040}{5^4}\right]+ \left[\frac{4040}{5^5}\right]=808+161+32+6+1=1008$$ and $n+4=4044$:

$$\left[\frac{4044}{5^1}\right] + \left[\frac{4044}{5^2}\right]+ \left[\frac{4044}{5^3}\right]+ \left[\frac{4044}{5^4}\right]+ \left[\frac{4044}{5^5}\right]=808+161+32+6+1=1008.$$ For $n=4024$ product $(n+16)!\cdot (n+20)!=4040!\cdot 4044!$ ends with a 2016 zeros