Difference between revisions of "2016 AIME I Problems"
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==Problem 11== | ==Problem 11== | ||
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| + | Let <math>P(x)</math> be a nonzero polynomial such that <math>(x-1)P(x+1)=(x+2)P(x)</math> for every real <math>x</math>, and <math>\left(P(2)\right)^2 = P(3)</math>. Then <math>P(\tfrac72)=\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
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[[2016 AIME I Problems/Problem 11 | Solution]] | [[2016 AIME I Problems/Problem 11 | Solution]] | ||
Revision as of 12:34, 4 March 2016
| 2016 AIME I (Answer Key) | AoPS Contest Collections • PDF | ||
|
Instructions
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| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
to 
, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.Contents
Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Let 
be a nonzero polynomial such that 
for every real 
, and 
. Then 
, where 
and 
are relatively prime positive integers. Find 
.
Problem 12
Problem 13
Problem 14
Problem 15
| 2016 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by 2015 AIME II |
Followed by 2016 AIME II | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
