2011 AIME II Problems/Problem 15

Problem

Let . A real number is chosen at random from the interval . The probability that is equal to , where , , , , and are positive integers. Find .

Solution

Table of values of P([x]):

P(5) = 1 P(6) = 9 P(7) = 19 P(8) = 31 P(9) = 45 P(10) = 61 P(11) = 79 P(12) = 99 P(13) = 121 P(14) = 145 P(15) = 171

In order for [√P(x)] = √P([x]) to hold, √P([x]) must be an integer and hence P([x]) must be a perfect square. This limits x to 5 < x < 6 or 6 < x < 7 or 13 < x < 14 since, from the table above, those are the only values of x for which P([x]) is an integer. However, in order for √P(x) to be rounded down to √P([x]), P(x) must not be greater than the next perfect square after P([x]) (for 5 < x < 6, etc.). Note that in all the cases the next value of P(x) always passes the next perfect square after P([x]), so in no cases will all values of x in the said intervals work. Now, we consider the three difference cases.

5 < x < 6: P(x) must not be greater than the first perfect square after 1, which is 4. Since P(x) is increasing for x > 5, we just need to find where P(x) = 4 and the values that will work will be 5 < x < root.

x^2 - 3x - 9 = 4 x = (3 + √61)/2

So in this case, the only values that will work are 5 < x < (3 + √61)/2.

6 < x < 7: P(x) must not be greater than the first perfect square after 9, which is 16.

x^2 - 3x - 9 = 16 x = (3 + √109)/2

So in this case, the only values that will work are 6 < x < (3 + √109)/2.

13 < x < 14: P(x) must not be greater than the first perfect square after 121, which is 144.

x^2 - 3x - 9 = 144 x = (3 + √721)/2

So in this case, the only values that will work are 13 < x < (3 + √721)/2.

Now, we find the length of the working intervals and divide it by the length of the total interval, 15 - 5 = 10:

(((3 + √61)/2 - 5) + ((3 + √109)/2 - 6) + ((3 + √721)/2 - 13))/10 = (√61 + √109 + √721 - 39)/20

So the answer is 61 + 109 + 721 + 39 + 20 = 950.