2000 AMC 12 Problems/Problem 12

Revision as of 01:02, 3 February 2009 by Duelist (talk | contribs) (Solution)

Problem

Let $A, M,$ and $C$ be nonnegative integers such that $A + M + C=12$. What is the maximum value of $A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C$?

$\mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 }$

Solution

\begin{align*}(A + 1)(M + 1)(C + 1) &= A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + A + M + C + 1\\ &= A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13\end{align*}

By AM-GM, $\frac{(A+1) + (M+1) + (C+1)}{3} = 5 \ge \sqrt[3]{(A+1)(M+1)(C+1)}$; thus $(A+1)(M+1)(C+1)$ is maximized at $125$, which occurs when $A=B=C=4$.

\[A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 125 - 13 = 112 \Rightarrow \mathrm{(E)}\]

Alternatively, a quick AMC heuristic method would be to consider the equality case since that often gives a maximum or minimum. So $A=M=C=4$ gives $AMC+AM+AC+MC = 112$, and that is the greatest answer choice, so the answer is $\mathrm{E}$.

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 12 Problems and Solutions
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