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

(Replaced content with "НИГЕР")
(Tag: Replaced)
m (Undo revision 171626 by Idontlikereggins (talk))
(Tag: Undo)
Line 1: Line 1:
НИГЕР
+
== Problem ==
 +
<!-- 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>
 +
 
 +
== Solution 1 ==
 +
It is not hard to see that
 +
<cmath>(A+1)(M+1)(C+1)=</cmath>
 +
<cmath>AMC+AM+AC+MC+A+M+C+1</cmath>
 +
Since <math>A+M+C=12</math>, we can rewrite this as
 +
<cmath>(A+1)(M+1)(C+1)=</cmath>
 +
<cmath>AMC+AM+AC+MC+13</cmath>
 +
So we wish to maximize
 +
<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=M=C=4</math>
 +
Which gives us
 +
<cmath>(4+1)(4+1)(4+1)-13=112</cmath>
 +
so the answer is <math>\boxed{\text{E}}</math>.
 +
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 <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>.
 +
 
 +
== Solution 3 ==
 +
 
 +
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>.
 +
 
 +
== Video Solution ==
 +
https://youtu.be/lxqxQhGterg
 +
 
 +
== See also ==
 +
{{AMC12 box|year=2000|num-b=11|num-a=13}}
 +
 
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Revision as of 07:46, 20 February 2022

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png