Difference between revisions of "2000 AMC 12 Problems/Problem 12"
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<cmath>(4+1)(4+1)(4+1)-13=112</cmath> | <cmath>(4+1)(4+1)(4+1)-13=112</cmath> | ||
so the answer is <math>\boxed{\text{E}}</math>. | 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) == | == Solution 2 (Nonrigorous) == | ||
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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>. | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/lxqxQhGterg | ||
== See also == | == See also == |
Latest revision as of 04:40, 29 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 Since , we can rewrite this as So we wish to maximize Which is largest when all the factors are equal (consequence of AM-GM). Since , we set Which gives us so the answer is . 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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.