Difference between revisions of "2000 AMC 10 Problems/Problem 20"
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− | The trick is to realize that the sum <math>AMC+AM+MC+CA</math> is similar to the product <math>(A+1)(M+1)(C+1)</math>. If we multiply <math>(A+1)(M+1)(C+1)</math>, we get <cmath>(A+1)(M+1)(C+1) = AMC + AM + AC + MC + A + M + C + 1.</cmath> We know that <math>A+M+C=10</math>, therefore <math>(A+1)(M+1)(C+1) = (AMC + AM + MC + CA) + 11</math> and <cmath>AMC + AM + MC + CA = (A+1)(M+1)(C+1) - 11.</cmath> Now consider the maximal value of this expression. Suppose that some two of <math>A</math>, <math>M</math>, and <math>C</math> differ by at least <math>2</math>. Then this triple <math>(A,M,C)</math> is not optimal. (To see this, | + | The trick is to realize that the sum <math>AMC+AM+MC+CA</math> is similar to the product <math>(A+1)(M+1)(C+1)</math>. If we multiply <math>(A+1)(M+1)(C+1)</math>, we get <cmath>(A+1)(M+1)(C+1) = AMC + AM + AC + MC + A + M + C + 1.</cmath> We know that <math>A+M+C=10</math>, therefore <math>(A+1)(M+1)(C+1) = (AMC + AM + MC + CA) + 11</math> and <cmath>AMC + AM + MC + CA = (A+1)(M+1)(C+1) - 11.</cmath> Now consider the maximal value of this expression. Suppose that some two of <math>A</math>, <math>M</math>, and <math>C</math> differ by at least <math>2</math>. Then this triple <math>(A,M,C)</math> is not optimal. (To see this, WLOG let <math>A\geq C+2.</math> We can then increase the value of <math>(A+1)(M+1)(C+1)</math> by changing <math>A \to A-1</math> and <math>C \to C+1</math>.) |
Therefore the maximum is achieved when <math>(A,M,C)</math> is a rotation of <math>(3,3,4)</math>. The value of <math>(A+1)(M+1)(C+1)</math> in this case is <math>4\cdot 4\cdot 5=80,</math> and thus the maximum of <math>AMC + AM + MC + CA</math> is <math>80-11 = \boxed{\textbf{(C)}\ 69}.</math> | Therefore the maximum is achieved when <math>(A,M,C)</math> is a rotation of <math>(3,3,4)</math>. The value of <math>(A+1)(M+1)(C+1)</math> in this case is <math>4\cdot 4\cdot 5=80,</math> and thus the maximum of <math>AMC + AM + MC + CA</math> is <math>80-11 = \boxed{\textbf{(C)}\ 69}.</math> |
Revision as of 08:45, 18 November 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 |
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