2000 AMC 12 Problems/Problem 12

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Problem

Let A, M, and C be nonnegative integers such that $\displaystyle A + M + C=12$. 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

$(A + 1)(M + 1)(C + 1) = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + A + M + C + 1$
$(A + 1)(M + 1)(C + 1) = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13$

The term $(A + 1)(M + 1)(C + 1)$ is maximized when A, M, and C are closed together, which in this case would be if all of them were 4. Thus,

$125 = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13$
$A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 112 \Rightarrow \mathrm{E}$

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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