Question

设 $a$ 、 $b$ 、 $c$ 为正整数，且 $a + b + c = 2006$ ，并且 $a!\,b!\,c! = m \cdot 10^n$ ，其中 $m$ 和 $n$ 为整数，且 $m$ 不能被 $10$ 整除。 $n$ 的最小可能值是多少？

Suppose $a,$ $b,$ and $c$ are positive integers with $a + b + c = 2006,$ and $a!\,b!\,c! = m \cdot 10^n,$ where $m$ and $n$ are integers and $m$ is not divisible by $10.$ What is the smallest possible value of $n?$

$489$

$492$

$495$

$498$

$501$

答案：B

知识点：勒让德公式末尾零最优化

难度评级：2300

解答：

因子 $2$ 比因子 $5$ 充足，所以 $n$ 等于 $a!\,b!\,c!$ 中因子 $5$ 的个数，即 $n = \sum_{k \ge 1}\left(\left\lfloor \tfrac{a}{5^k}\right\rfloor + \left\lfloor \tfrac{b}{5^k}\right\rfloor + \left\lfloor \tfrac{c}{5^k}\right\rfloor\right).$

对每个 $k$ ， $\lfloor a/5^k \rfloor + \lfloor b/5^k \rfloor$ $+ \lfloor c/5^k \rfloor$ $\ge \lfloor 2006/5^k \rfloor - 2$ 。对 $k = 1, 2, 3, 4$ 求和（因为 $2006 \lt 5^5$ ），得到 $\begin{aligned} &n \ge (401 + 80 + 16 + 3) \\ &\quad {}- 4 \cdot 2 = 492. \end{aligned}$

等号可以达到，例如取 $a = b = 624$ 、 $c = 758$ 。所以最小值为 $492$ 。

因此，正确答案是 B。

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 $k,$ $\lfloor a/5^k \rfloor + \lfloor b/5^k \rfloor$ $+ \lfloor c/5^k \rfloor$ $\ge \lfloor 2006/5^k \rfloor - 2.$ 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.

2006 AMC 12B 第 22 题

答案：B

难度评级：2300

解答：

其他年份的第 22 题