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 A⋅M⋅C+A⋅M+M⋅C+C⋅A? \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
Small hint:
AMC+AM+MC+CAAMC + AM + MC + CA =(A+1)(M+1)(C+1)= (A + 1)(M + 1)(C + 1) −(A+M+C)−1- (A + M + C) - 1
Big hint:
The three factors A+1,M+1,C+1A+1, M+1, C+1 sum to 15,15, so their product is largest when they are equal
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 125−13=112.125 - 13 = 112.
Thus, the correct answer is E.
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