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

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== See also ==
 
== See also ==
*[[2000 AMC 12]]
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{{AMC12 box|year=2000|num-b=11|num-a=13}}
  
{{AMC12 box|year=2000|num-b=11|num-a=13}}
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[[Category:Introductory Algebra Problems]]

Revision as of 18:18, 4 March 2007

Problem

Let A, M, and C be nonnegative integers such that $\displaystyle A + M + C=12$. 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

$(A + 1)(M + 1)(C + 1) = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + A + M + C + 1$
$(A + 1)(M + 1)(C + 1) = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13$

The term $(A + 1)(M + 1)(C + 1)$ is maximized when A, M, and C are closed together, which in this case would be if all of them were 4. Thus,

$125 = A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C + 13$
$A \cdot M \cdot C + A \cdot M + M \cdot C + A\cdot C = 112 \Rightarrow \mathrm{E}$

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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