2011 AIME II Problems/Problem 15: Difference between revisions
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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} | 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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Revision as of 17:17, 31 March 2011
Problem:
Let 
. A real number 
is chosen at random from the interval 
. The probability that 
is equal to 
, where 
, 
, 
, 
, and 
are positive integers, and none of , , or is divisible by the square of a prime. Find 
.
Solution:
Good luck people.
