2000 AMC 10 第 20 题

Question

设

A

、

M

和

C

是非负整数，且

A + M + C = 10

。求下面表达式的最大值：

\begin{aligned} &A \cdot M \cdot C + A \cdot M \\ &\quad {}+ M \cdot C + C \cdot A \end{aligned}\text{？}

Let

A,

M,

and

C

be nonnegative integers such that

A + M + C = 10.

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}

49

59

69

79

89

给每个变量加

1

，并展开

(A+1)(M+1)(C+1)

。

Add

1

to each variable and expand

(A+1)(M+1)(C+1)

原式等于

(A+1)(M+1)(C+1) - 11

，所以要最大化三个和为

13

的正整数的乘积。

The expression equals

(A+1)(M+1)(C+1) - 11,

so maximize a product of three positive integers summing to

13

注意

\begin{gathered} A \cdot M \cdot C + AM + MC + CA \\ = (A+1)(M+1)(C+1) \\ {}- (A + M + C) - 1 \\ = (A+1)(M+1)(C+1) - 11 \end{gathered}\text{。}

现在要最大化三个正整数的乘积，它们的和为

13

。最平均的分配是

4, 4, 5

，乘积为

4 \cdot 4 \cdot 5 = 80

。

因此原式的最大值为

80 - 11 = 69

。

所以正确答案是 C。

Notice that

\begin{gathered} A \cdot M \cdot C + AM + MC + CA \\ = (A+1)(M+1)(C+1) \\ {}- (A + M + C) - 1 \\ = (A+1)(M+1)(C+1) - 11. \end{gathered}

We maximize a product of three positive integers summing to

13.

The most balanced split is

4, 4, 5,

giving

4 \cdot 4 \cdot 5 = 80.

The maximum is

80 - 11 = 69.

Thus, the correct answer is C.

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