2000 AMC 10 Problems/Problem 20
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, M must be 3, and C must be 4. . So the answer is
Solution 3
According to our knowledge in AM-GM, the closer numbers are, the value of their product is bigger. Assume that , we can get the set which the answer is ~bluesoul
Video Solution
https://www.youtube.com/watch?v=Vdou0LpTlzY&t=22s
https://www.youtube.com/watch?v=ECzZPuuQMbE ~David
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.