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

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== Solution ==
 
== Solution ==
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<cmath>\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\\
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When <math>A=M=C=4</math> then <math>AMC+AM+AC+MC = 112</math>, and that is the greatest answer choice, so the answer is <math>\boxed{E}</math>.
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&= A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13\end{align*}</cmath>
 
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By [[AM-GM]], <math>\frac{(A+1) + (M+1) + (C+1)}{3} = 5 \ge \sqrt[3]{(A+1)(M+1)(C+1)}</math>; thus <math>(A+1)(M+1)(C+1)</math> is [[maximum|maximized]] at <math>125</math>, which occurs when <math>A=B=C=4</math>.
 
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<cmath>A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 125 - 13 = 112 \Rightarrow \mathrm{(E)}</cmath>
 
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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>\boxed{E}</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 20:58, 6 January 2015

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

When $A=M=C=4$ then $AMC+AM+AC+MC = 112$, and that is the greatest answer choice, so the answer is $\boxed{E}$.

See also

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

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