Difference between revisions of "2000 AMC 12 Problems/Problem 12"
m (→Solution) |
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| 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 13:46, 16 May 2015
Problem
Let 
and 
be nonnegative integers such that 
. What is the maximum value of 
?
Solution
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 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
