1. **Problem Statement:** Find the smallest positive integer $n$ such that $n!$ (the factorial of $n$) ends with exactly six zeroes. 2. **Understanding trailing zeroes in factorials:** The number of trailing zeroes in $n!$ is determined by the number of factors of 10 in the product. Since 10 = 2 × 5, and factorials have more factors of 2 than 5, the count of 5s determines the number of trailing zeroes. 3. **Formula for counting trailing zeroes:** The number of trailing zeroes in $n!$ is given by: $$Z(n) = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{5^2} \right\rfloor + \left\lfloor \frac{n}{5^3} \right\rfloor + \cdots$$ where $\lfloor x \rfloor$ is the floor function (greatest integer less than or equal to $x$). 4. **Goal:** Find the smallest $n$ such that $Z(n) = 6$. 5. **Calculate $Z(n)$ for values around where $Z(n)$ might be 6:** - For $n=25$: $$Z(25) = \left\lfloor \frac{25}{5} \right\rfloor + \left\lfloor \frac{25}{25} \right\rfloor = 5 + 1 = 6$$ - For $n=24$: $$Z(24) = \left\lfloor \frac{24}{5} \right\rfloor + \left\lfloor \frac{24}{25} \right\rfloor = 4 + 0 = 4$$ 6. Since $Z(24) = 4 < 6$ and $Z(25) = 6$, the smallest $n$ with exactly 6 trailing zeroes is $n=25$. **Final answer:** $$\boxed{25}$$