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

m (Solution)
m (Solution)
Line 13: Line 13:
 
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.  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=B=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 14:46, 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 \[(A+1)(M+1)(C+1)=\] \[AMC+AM+AC+MC+13\] 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 $\boxed{E}$.

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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