Difference between revisions of "2000 AMC 10 Problems/Problem 20"
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==Solution 2 == | ==Solution 2 == | ||
Notice that if we want to maximize <math>AMC + AM + MC + AC</math>, we want A, M, and C to be as close as possible. For example, if <math>A = 7, B = 2,</math> and <math>C=1,</math> then the expression would have a much smaller value than if we were to substitute <math>A = 4, B = 5</math>, and <math>C = 1</math>. So to make A, B, and C as close together as possible, we divide <math>\frac{10}{3}</math> to get <math>3</math>. Therefore, A must be 3, B must be 3, and C must be 4. <math>AMC + AM + MC + AC = 36 + 12 + 12 + 9 = 69</math>. So the answer is <math>\boxed{\textbf{(C)}\ 69.}</math> | Notice that if we want to maximize <math>AMC + AM + MC + AC</math>, we want A, M, and C to be as close as possible. For example, if <math>A = 7, B = 2,</math> and <math>C=1,</math> then the expression would have a much smaller value than if we were to substitute <math>A = 4, B = 5</math>, and <math>C = 1</math>. So to make A, B, and C as close together as possible, we divide <math>\frac{10}{3}</math> to get <math>3</math>. Therefore, A must be 3, B must be 3, and C must be 4. <math>AMC + AM + MC + AC = 36 + 12 + 12 + 9 = 69</math>. So the answer is <math>\boxed{\textbf{(C)}\ 69.}</math> | ||
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==Video Solution== | ==Video Solution== |
Revision as of 18:38, 6 July 2020
Problem
Let , , and be nonnegative integers such that . What is the maximum value of ?
Solution 1
The trick is to realize that the sum is similar to the product . If we multiply , we get We know that , therefore and Now consider the maximal value of this expression. Suppose that some two of , , and differ by at least . Then this triple is not optimal. (To see this, WLOG let We can then increase the value of by changing and .)
Therefore the maximum is achieved when is a rotation of . The value of in this case is and thus the maximum of is
Solution 2
Notice that if we want to maximize , we want A, M, and C to be as close as possible. For example, if and then the expression would have a much smaller value than if we were to substitute , and . So to make A, B, and C as close together as possible, we divide to get . Therefore, A must be 3, B must be 3, and C must be 4. . So the answer is
Video Solution
https://www.youtube.com/watch?v=Vdou0LpTlzY&t=22s
See Also
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.