Question

Find n such that n! has 12 zeros

Answer 1

The values of n such that n! has 12 zeros are 50, 51, 52, 53, 54.

Answer 2

The number of trailing zeros in n! is determined by the number of factors of 5 in its prime factorization. This can be calculated using Legendre's formula: $Z(n) = \sum_{k=1}^{\infty} \lfloor n/5^k \rfloor$. We need $Z(n) = 12$. $Z(n) = \lfloor n/5 \rfloor + \lfloor n/25 \rfloor + \lfloor n/125 \rfloor + \dots$ Assuming $\lfloor n/125 \rfloor = 0$, we have $Z(n) = \lfloor n/5 \rfloor + \lfloor n/25 \rfloor = 12$. If $\lfloor n/25 \rfloor = 2$, then $50 \le n < 75$. This implies $\lfloor n/5 \rfloor = 10$, so $50 \le n < 55$. The intersection is $50 \le n < 55$. For $n = 50, 51, 52, 53, 54$, $Z(n) = 12$.