Difference between revisions of "2000 AMC 12 Problems/Problem 12"
(→Problem) |
(→Solution) |
||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
− | + | It is not hard to see that | |
+ | <cmath>(A+1)(M+1)(C+1)=AMC+AM+AC+MC+A+M+C+1</cmath> | ||
+ | Since <math>A+M+C=12</math>, we can rewrite this as | ||
+ | <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. Since <math>A+M+C=12</math>, we set <math>A=B=C=4</math> | ||
+ | Which gives us | ||
+ | <cmath>(4+1)(4+1)(4+1)-13=112</cmath> | ||
+ | so the answer is <math>\boxed{E}</math>. | ||
== See also == | == See also == |
Revision as of 14:39, 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.