To find the number of zeros at the end of \( 100! \), we need to determine how many times \( 10 \) divides into \( 100! \). Since \( 10 = 2 \times 5 \), every pair of factors of 2 and 5 contributes a factor of 10, resulting in one zero at the end of the number.
The number of factors of 5 in \( 100! \) will determine the number of zeros at the end, because there are always more factors of 2 than factors of 5 in any factorial. To find the number of factors of 5, we use the following formula: \[ \left\lfloor \frac{100}{5} \right\rfloor + \left\lfloor \frac{100}{25} \right\rfloor \] The first term counts the multiples of 5, and the second term counts the multiples of 25 (which contribute an extra factor of 5). Let's calculate it: \[ \left\lfloor \frac{100}{5} \right\rfloor = 20 \] \[ \left\lfloor \frac{100}{25} \right\rfloor = 4 \] Thus, the total number of factors of 5 in \( 100! \) is: \[ 20 + 4 = 24 \] Therefore, the number of zeros at the end of \( 100! \) is \( 24 \).
\[ \boxed{24} \]
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: