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)
Printable version | AoPS Contest CollectionsPDF

Instructions

  1. This is a 15-question, 3-hour examination. All answers are integers ranging from $000$ to $999$, inclusive. Your score will be the number of correct answers; i.e., there is neither partial credit nor a penalty for wrong answers.
  2. No aids other than scratch paper, graph paper, ruler, compass, and protractor are permitted. In particular, calculators and computers are not permitted.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Problem 1

Solution

Problem 2

Solution

Problem 3

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Let $P(x)$ be a nonzero polynomial such that $(x-1)P(x+1)=(x+2)P(x)$ for every real $x$, and $\left(P(2)\right)^2 = P(3)$. Then $P(\tfrac72)=\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

2016 AIME I (ProblemsAnswer KeyResources)
Preceded by
2015 AIME II
Followed by
2016 AIME II
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All AIME Problems and Solutions

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