2000 AMC 12 Problem 12

Below is the professionally curated solution for Problem 12 of the 2000 AMC 12, 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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Concepts:factoringAM-GM Inequalityoptimization

Difficulty rating: 1650

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?A \cdot M \cdot C + A \cdot M + M \cdot C + C \cdot A?

6262

7272

9292

102102

112112

Solution:

Observe that AMC+AM+MC+CA=(A+1)(M+1)(C+1)(A+M+C)1. AMC + AM + MC + CA = (A + 1)(M + 1)(C + 1) - (A + M + C) - 1.

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.

Problem 12 in Other Years