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

(Problem)
(Solution)
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== Solution ==
 
== Solution ==
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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It is not hard to see that
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<cmath>(A+1)(M+1)(C+1)=AMC+AM+AC+MC+A+M+C+1</cmath>
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Since <math>A+M+C=12</math>, we can rewrite this as
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<cmath>AMC+AM+AC+MC+13</cmath>
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So we wish to maximize
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<cmath>(A+1)(M+1)(C+1)-13</cmath>
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Which is largest when all the factors are equal.  Since <math>A+M+C=12</math>, we set <math>A=B=C=4</math>
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Which gives us
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<cmath>(4+1)(4+1)(4+1)-13=112</cmath>
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so the answer is <math>\boxed{E}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 14:39, 16 May 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

It is not hard to see that \[(A+1)(M+1)(C+1)=AMC+AM+AC+MC+A+M+C+1\] Since $A+M+C=12$, we can rewrite this as \[AMC+AM+AC+MC+13\] So we wish to maximize \[(A+1)(M+1)(C+1)-13\] Which is largest when all the factors are equal. Since $A+M+C=12$, we set $A=B=C=4$ Which gives us \[(4+1)(4+1)(4+1)-13=112\] 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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