2000 AMC 12 Problems/Problem 12: Difference between revisions
No edit summary |
|||
| Line 25: | Line 25: | ||
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>. | 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 == | == See also == | ||
Revision as of 21:05, 18 July 2021
Problem
Let 
and 
be nonnegative integers such that 
. What is the maximum value of 
?
Solution 1
It is not hard to see that
![\[(A+1)(M+1)(C+1)=\]](http://latex.artofproblemsolving.com/e/1/a/e1a20a664fc92825a76c41415d24fe902a9e4fb1.png)
![\[AMC+AM+AC+MC+A+M+C+1\]](http://latex.artofproblemsolving.com/7/b/6/7b615f683ed979dc734b1fc7c95d4b8765e5d094.png)
Since 
, we can rewrite this as
![\[AMC+AM+AC+MC+13\]](http://latex.artofproblemsolving.com/7/9/f/79f3f8b8cb3dee37405c83c414548b1ebd2a63c5.png)
So we wish to maximize
![\[(A+1)(M+1)(C+1)-13\]](http://latex.artofproblemsolving.com/a/3/f/a3f9f9ef8289341a9a4f3e43f0aa4199a2d28b58.png)
Which is largest when all the factors are equal (consequence of AM-GM). Since , we set 
Which gives us
![\[(4+1)(4+1)(4+1)-13=112\]](http://latex.artofproblemsolving.com/8/c/c/8cc110b16d54e7fd301a03ee24ee13a2f40e0e05.png)
so the answer is 
.
Solution 2 (Nonrigorous)
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 polygon make it a square) then you can try to make 
and as close as possible. In this case, they would all be equal to 
, so 
, giving you the answer of .
Solution 3
Assume 
, 
, and are equal to . Since the resulting value of 
will be 
and this is the largest answer choice, our answer is 
.
Video Solution
See also
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
