Difference between revisions of "2000 AMC 12 Problems/Problem 12"

Revision as of 00:02, 3 February 2009

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 $\mathrm{E}$ .

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

@@ Line 11: / Line 11: @@
 <cmath>A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 125 - 13 = 112 \Rightarrow \mathrm{(E)}</cmath>
+Alternatively, a quick AMC heuristic method would be to consider the equality case since that often gives a maximum or minimum. So <math>A=M=C=4</math> gives <math>AMC+AM+AC+MC = 112</math>, and that is the greatest answer choice, so the answer is <math>\mathrm{E}</math>.
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Difference between revisions of "2000 AMC 12 Problems/Problem 12"

Revision as of 00:02, 3 February 2009

Problem

Solution

See also