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 :
In order for to hold, must be an integer and hence must be a perfect square. This limits to or or since, from the table above, those are the only values of for which is an perfect square. However, in order for to be rounded down to , must be less than the next perfect square after (for the said intervals). Now, we consider the three cases:
Case :
must be less than the first perfect square after , which is , i.e.:
(because implies )
Since is increasing for , we just need to find the value where , which will give us the working range .
So in this case, the only values that will work are .
Case :
must be less than the first perfect square after , which is .
So in this case, the only values that will work are .
Case :
must be less than the first perfect square after , which is .
So in this case, the only values that will work are .
Now, we find the length of the working intervals and divide it by the length of the total interval, :
Thus, the answer is .
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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