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2000 AMC 12 Problems/Problem 12: Difference between revisions

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So we wish to maximize
So we wish to maximize
<cmath>(A+1)(M+1)(C+1)-13</cmath>
<cmath>(A+1)(M+1)(C+1)-13</cmath>
Which is largest when all the factors are equal (consequence of AM-GM).  Since <math>A+M+C=12</math>, we set <math>A=B=C=4</math>
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>
Which gives us  
Which gives us  
<cmath>(4+1)(4+1)(4+1)-13=112</cmath>
<cmath>(4+1)(4+1)(4+1)-13=112</cmath>

Revision as of 13:01, 24 July 2018

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 $\boxed{\text{E}}$.

Solution 2

If you know that to maximize your result you 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 $\boxed{\text{E}}$.

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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