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

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== Problem ==
 
== Problem ==
Let <math>A, M,</math> and <math>C</math> be [[nonnegative integer]]s such that <math>A + M + C=12</math>. What is the maximum value of <math>A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C</math>?
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<!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>A, M,</math> and <math>C</math> be [[nonnegative integer]]s such that <math>A + M + C=12</math>. What is the maximum value of <math>A \cdot M \cdot C + A \cdot M + M \cdot C + A \cdot C</math>?<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude>
  
 
<math> \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 }  </math>
 
<math> \mathrm{(A) \ 62 } \qquad \mathrm{(B) \ 72 } \qquad \mathrm{(C) \ 92 } \qquad \mathrm{(D) \ 102 } \qquad \mathrm{(E) \ 112 }  </math>
  
== Solution ==
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== Solution 1 ==
<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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It is not hard to see that
&= A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13\end{align*}</cmath>
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<cmath>(A+1)(M+1)(C+1)=</cmath>
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<cmath>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>(A+1)(M+1)(C+1)=</cmath>
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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 (consequence of AM-GM).  Since <math>A+M+C=12</math>, we set <math>A=M=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{\text{E}}</math>.
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I wish you understand this problem and can use it in other problems.
  
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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== Solution 2 (Nonrigorous) ==
  
<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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If you know that to maximize your result you <math>\textit{usually}</math> have to make the numbers as close together as possible, (for example to maximize area for a polygon make it a square) then you can try to make <math>A,M</math> and <math>C</math> as close as possible. In this case, they would all be equal to <math>4</math>, so <math>AMC+AM+AC+MC=64+16+16+16=112</math>, giving you the answer of <math>\boxed{\text{E}}</math>.
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== Solution 3 ==
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Assume <math>A</math>, <math>M</math>, and <math>C</math> are equal to <math>4</math>. Since the resulting value of <math>AMC+AM+AC+MC</math> will be <math>112</math> and this is the largest answer choice, our answer is <math>\boxed{\textbf{(E) }112}</math>.
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== Video Solution ==
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https://youtu.be/lxqxQhGterg
  
 
== See also ==
 
== See also ==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 03:40, 29 July 2021

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 1

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 \[(A+1)(M+1)(C+1)=\] \[AMC+AM+AC+MC+13\] So we wish to maximize \[(A+1)(M+1)(C+1)-13\] Which is largest when all the factors are equal (consequence of AM-GM). Since $A+M+C=12$, we set $A=M=C=4$ Which gives us \[(4+1)(4+1)(4+1)-13=112\] so the answer is $\boxed{\text{E}}$. I wish you understand this problem and can use it in other problems.

Solution 2 (Nonrigorous)

If you know that to maximize your result you $\textit{usually}$ have to make the numbers as close together as possible, (for example to maximize area for a polygon make it a square) then you can try to make $A,M$ and $C$ as close as possible. In this case, they would all be equal to $4$, so $AMC+AM+AC+MC=64+16+16+16=112$, giving you the answer of $\boxed{\text{E}}$.

Solution 3

Assume $A$, $M$, and $C$ are equal to $4$. Since the resulting value of $AMC+AM+AC+MC$ will be $112$ and this is the largest answer choice, our answer is $\boxed{\textbf{(E) }112}$.

Video Solution

https://youtu.be/lxqxQhGterg

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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