There are floor(n/5) + floor(n/25) + floor(n/125)+... zeroes at the end of n! Since 1/5 + 1/25 + 1/125 + ... = 1/4, we would expect (4n!) to be a very good lower bound for our answer. Also, the number of zeroes in successive values of n! only increase when n is a multiple of five. Using those two facts, we can create a starting value that requires very few checks even when n is very large.

def fact_zeros(n):
    count = 0
    while n:
        count += n//5
        n //= 5
    return count

def solve(n):
    test = ((4*n)//5) * 5
    while fact_zeros(test) < n:
        test += 5
    return test