Difference between revisions of "2000 AMC 10 Problems/Problem 20"
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==Solution== | ==Solution== | ||
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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 <math>AMC + AM + AC + MC + A + M + C + 1</math>. | ||
+ | |||
+ | We know that <math>(A+M+C)=10</math>, therefore <math>(A+1)(M+1)(C+1) = (AMC + AM + AC + MC) + 11</math>. | ||
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+ | Therefore the maximum value of <math>AMC+AM+MC+CA</math> is equal to the maximum value of <math>(A+1)(M+1)(C+1)-11</math>. Now we will find this maximum. | ||
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+ | Suppose that some two of <math>A</math>, <math>M</math>, and <math>C</math> differ by at least 2. | ||
+ | Then this triple <math>(A,M,C)</math> is surely not optimal. | ||
+ | Proof: 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\gets A-1</math> and <math>C\gets C+1</math>. | ||
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+ | Therefore the maximum is achieved in the cases where <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 + AC + MC</math> is $80-11 = \boxed{69}. | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=19|num-a=21}} | {{AMC10 box|year=2000|num-b=19|num-a=21}} |
Revision as of 11:18, 11 January 2009
Problem
Let , , and be nonnegative integers such that . What is the maximum value of ?
Solution
The trick is to realize that the sum is similar to the product .
If we multiply , we get .
We know that , therefore .
Therefore the maximum value of is equal to the maximum value of . Now we will find this maximum.
Suppose that some two of , , and differ by at least 2. Then this triple is surely not optimal. Proof: WLOG let . We can then increase the value of by changing and .
Therefore the maximum is achieved in the cases where is a rotation of . The value of in this case is . And thus the maximum of is $80-11 = \boxed{69}.
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 |