2000 AMC 12 Problem 12

Attempt Problem 12 of the 2000 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 12 solutions, or check the answer key.

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12.

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

6262

7272

9292

102102

112112

Answer: E
Concepts:factoringAM-GM Inequalityoptimization
Difficulty rating: 1650
Solution:

Observe that AMC+AM+MC+CA=(A+1)(M+1)(C+1)(A+M+C)1. \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,A + M + C = 12, this equals (A+1)(M+1)(C+1)13.(A + 1)(M + 1)(C + 1) - 13. The three factors sum to 15,15, so their product is maximized when each equals 5,5, giving 53=125.5^3 = 125.

The maximum value is 12513=112.125 - 13 = 112.

Thus, the correct answer is E.

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