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

m (Solution)
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<cmath>(4+1)(4+1)(4+1)-13=112</cmath>
 
<cmath>(4+1)(4+1)(4+1)-13=112</cmath>
 
so the answer is <math>\boxed{\text{E}}</math>.
 
so the answer is <math>\boxed{\text{E}}</math>.
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 +
Another solution:
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 +
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 shape 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>.
  
 
== See also ==
 
== See also ==

Revision as of 22:21, 15 February 2017

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 \[(A+1)(M+1)(C+1)=\] \[AMC+AM+AC+MC+13\] 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 $A=B=C=4$ Which gives us \[(4+1)(4+1)(4+1)-13=112\] so the answer is $\boxed{\text{E}}$.

Another solution:

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 shape 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 (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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