AMC10 2000 A

AMC10 2000 A · Q20

AMC10 2000 A · Q20. It mainly tests Algebra misc, Casework.

Let $A$, $M$, and $C$ be nonnegative integers such that $A + M + C = 10$. What is the maximum value of $A \cdot M \cdot C + A \cdot M + M \cdot C + C \cdot A$?

设$A$、$M$、$C$是非负整数且$A+M+C=10$。求$A\cdot M\cdot C+A\cdot M+M\cdot C+C\cdot A$的最大值。

(A) 49 49

(B) 59 59

(D) 79 79

(E) 89 89

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Note that $AMC + AM + MC + CA = (A+1)(M+1)(C+1) - (A+M+C) - 1 = pqr - 11,$ where $p$, $q$, and $r$ are positive integers whose sum is $13$. A case-by-case analysis shows that $pqr$ is largest when two of the numbers $p$, $q$, $r$ are $4$ and the third is $5$. Thus the answer is $4\cdot 4\cdot 5 - 11 = 69$.

答案（C）：注意到 $AMC + AM + MC + CA = (A+1)(M+1)(C+1) - (A+M+C) - 1 = pqr - 11,$ 其中 $p$、$q$、$r$ 是和为 $13$ 的正整数。逐一分类分析可知，当 $p,q,r$ 中有两个为 $4$、另一个为 $5$ 时，$pqr$ 取得最大值。因此答案为 $4\cdot 4\cdot 5 - 11 = 69$。

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