Difference between revisions of "2011 AIME II Problems/Problem 15"

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Problem:
 
Problem:
  
Let <math>P(x) = x^2 - 3x - 9</math>. A real number <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> is equal to <math>\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}}{e}</math> , where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are positive integers, and none of <math>a</math>, <math>b</math>, or <math>c</math> is divisible by the square of a prime. Find <math>a + b + c + d + e</math>.
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Let <math>P(x) = x^2 - 3x - 9</math>. A real number <math>x</math> is chosen at random from the interval <math>5 \le x \le 15</math>. The probability that <math>\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}</math> is equal to <math>\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}</math> , where <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math>, and <math>e</math> are positive integers. Find <math>a + b + c + d + e</math>.
  
 
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Solution:
 
Solution:
  
Good luck.
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Good luck. Or not.

Revision as of 16:00, 3 April 2011

Problem:

Let $P(x) = x^2 - 3x - 9$. A real number $x$ is chosen at random from the interval $5 \le x \le 15$. The probability that $\lfloor\sqrt{P(x)}\rfloor = \sqrt{P(\lfloor x \rfloor)}$ is equal to $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}$ , where $a$, $b$, $c$, $d$, and $e$ are positive integers. Find $a + b + c + d + e$.


Solution:

Good luck. Or not.

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