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

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

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

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

@@ 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}}

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