Question

设 $A$ 、 $M$ 和 $C$ 是非负整数，且 $A + M + C = 12$ 。求下式的最大值： $\begin{aligned} &A \cdot M \cdot C + A \cdot M \\ &\quad {}+ M \cdot C + C \cdot A? \end{aligned}$

Let $A,$ $M,$ and $C$ be nonnegative integers such that $A + M + C = 12.$ What is the maximum value of $\begin{aligned} &A \cdot M \cdot C + A \cdot M \\ &\quad {}+ M \cdot C + C \cdot A? \end{aligned}$

$62$

$72$

$92$

$102$

$112$

答案：E

知识点：因式分解算术-几何平均不等式最优化

难度评级：1650

解答：

注意 $\begin{aligned} &AMC + AM + MC + CA \\ &\quad = (A + 1)(M + 1)(C + 1) \\ &\quad {}- (A + M + C) - 1. \end{aligned}$

因为 $A + M + C = 12$ ，原式等于 $(A + 1)(M + 1)(C + 1) - 13$ 。这三个因数的和为 $15$ ，乘积在它们都等于 $5$ 时最大，等于 $5^3 = 125$ 。

最大值为 $125 - 13 = 112$ 。

因此，正确答案是 E。

Observe that $\begin{aligned} &AMC + AM + MC + CA \\ &\quad = (A + 1)(M + 1)(C + 1) \\ &\quad {}- (A + M + C) - 1. \end{aligned}$

Since $A + M + C = 12,$ this equals $(A + 1)(M + 1)(C + 1) - 13.$ The three factors sum to $15,$ so their product is maximized when each equals $5,$ giving $5^3 = 125.$

The maximum value is $125 - 13 = 112.$

Thus, the correct answer is E.

2000 AMC 12 第 12 题

答案：E

难度评级：1650

解答：

其他年份的第 12 题