Zeros at the end of a factorial

Since a zero digit means a factor of ten, and since there are many more twos available than there are fives, counting zeroes really means counting fives.

Every fifth number gives a factor of five, every twenty-fifth gives an extra five, every one hundred and twenty-fifth gives and extra extra five, and so forth. So the number of factors of five we find before an integer x is simply the sum of $\frac{x}{5}+\frac{x}{5^{2}}+\frac{x}{5^{3}}+...$ , truncating any non-integer term.

This sum is slightly below the sum of the infinite geometric series, which will be $\frac{\frac{x}{5}}{1-\frac{1}{5}}=\frac{x}{4}$ . So our number must be just over $4,000,000,000$ .

Specifically, repeatedly dividing $4,000,000,000$ by $5$ and summing shows that we're three fives short (there is probably a better way here, but I'm only seeing the brute force right now).

So the answer must be between $4,000,000,015$ and $4,000,000,019$ , inclusive. However, $4,000,000,015$ is a multiple of five, $4,000,000,017$ is a multiple of three, and the evens are obviously not prime, so the answer must be $\boxed{4,000,000,019}$ .