Difference between revisions of "2000 AMC 12 Problems/Problem 12"
(→Solution) |
m (→Solution) |
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Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
It is not hard to see that | It is not hard to see that | ||
− | <cmath>(A+1)(M+1)(C+1)=AMC+AM+AC+MC+A+M+C+1</cmath> | + | <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 | Since <math>A+M+C=12</math>, we can rewrite this as | ||
<cmath>AMC+AM+AC+MC+13</cmath> | <cmath>AMC+AM+AC+MC+13</cmath> |
Revision as of 14:40, 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 Since , we can rewrite this as So we wish to maximize Which is largest when all the factors are equal. Since , we set Which gives us 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.