There are several natural numbers 1, 2, 3. N, and there are 53 zeros at the end of the known product, so we can find the maximum value of n

There are several natural numbers 1, 2, 3. N, and there are 53 zeros at the end of the known product, so we can find the maximum value of n

Because the divisor of 2 in n! Is much more than that of 5, we only need to consider the divisor of 5. F (n) = [n / 5] + [n / 25] + [n / 125] +.. 31n / 125 ~ 53125x53 / 31 = 213f (213) = 42 + 8 + 1 = 51F (215) = 43 + 8 + 1 = 52F (220) = 44 + 8 + 1 = 53F (224) = 44 + 8 + 1 = 53F (225) = 45 + 8 + 1 = 54n, the maximum is 224