2011 AIME II Problems/Problem 15
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[hide]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 1
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 .
P.S. You don't need to calculate all the values of P(x) calculated by the above solution. Some very simple modular arithmetic eliminates a large portion of the numbers. The time saved is not that much if you are already at your mathcounts prime.
Solution 2
Make the substitution , so We're looking for solutions to with the new bounds . Since the left side is an integer, it must be that is a perfect square. For simplicity, write and Since , it must be that , which gives solutions , respectively. But this gives us three cases to check:
Case 1: .
In this case, we have Case 2: .
In this case, we have Case 3:
In this case, we have To finish, the total length of the interval from which we choose is . The total length of the success intervals is which means the probability is AIMEifying the answer gives .
See also
2011 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by [[2011 AIME II Problems/Problem {{{num-a}}}|Problem {{{num-a}}}]] | |
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