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2011 AIME II Problems/Problem 15: Difference between revisions

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Solution:
Solution:


Good luck people.
Good luck.

Revision as of 17:25, 31 March 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 $\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{d}}{e}$ , where $a$, $b$, $c$, $d$, and $e$ are positive integers, and none of $a$, $b$, or $c$ is divisible by the square of a prime. Find $a + b + c + d + e$.

Solution:

Good luck.

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