2006 AMC 12B Problem 22

Since factors of $2$ are more plentiful than factors of $5,$ $n$ equals the number of factors of $5$ in $a!\,b!\,c!,$ namely $$n = \sum_{k \ge 1}\left(\left\lfloor \tfrac{a}{5^k}\right\rfloor + \left\lfloor \tfrac{b}{5^k}\right\rfloor + \left\lfloor \tfrac{c}{5^k}\right\rfloor\right).$$ For each $k,$ $\lfloor \frac{a}{5^k} \rfloor + \lfloor \frac{b}{5^k} \rfloor$ $+ \lfloor \frac{c}{5^k} \rfloor$ $\ge \lfloor \frac{2006}{5^k} \rfloor - 2.$ Summing over $k = 1, 2, 3, 4$ (as $2006 \lt 5^5$) gives $$\begin{aligned} &n \ge (401 + 80 + 16 + 3) \\ &\quad {}- 4 \cdot 2 = 492. \end{aligned}$$ Equality is attainable, for example with $a = b = 624$ and $c = 758.$ So the minimum is $492.$ Thus, the correct answer is B.