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

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== Solution ==
 
== Solution ==
<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\\
+
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>.
&= A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13\end{align*}</cmath>
 
 
 
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>.
 
 
 
<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>\boxed{E}</math>.
 
  
 
== See also ==
 
== See also ==

Revision as of 21:58, 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 (ProblemsAnswer KeyResources)
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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