AIME 2011b Problem 15 PaperAnswer Let P(x)=x2−3x−9P(x) = x^2 - 3x - 9P(x)=x2−3x−9. A real number xxx is chosen at random from the interval 5≤x≤155 \le x \le 155≤x≤15. The probability that ⌊P(x)⌋=P(⌊x⌋)\left\lfloor\sqrt{P(x)}\right\rfloor = \sqrt{P(\left\lfloor x \right\rfloor)}⌊P(x)⌋=P(⌊x⌋) is equal to a+b+c−de\frac{\sqrt{a} + \sqrt{b} + \sqrt{c} - d}{e}ea+b+c−d , where aaa, bbb, ccc, ddd, and eee are positive integers. Find a+b+c+d+ea + b + c + d + ea+b+c+d+e. Dataset text AIME 1983 to 2026. Quoted with source. Show the answer 850 Solutions No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label. Add a solution PreviousProblem 14 Locator aime/2011b/p15 · Tag 42SYPT335B · Edit this page on GitHub · Keys: j next, k previous