2012 AIME II Problems/Problem 10
Problem 10
Find the number of positive integers less than for which there exists a positive real number such that .
Note: is the greatest integer less than or equal to .
Solution
Solution 1
We know that cannot be irrational because the product of a rational number and an irrational number is irrational (but is an integer). Therefore is rational.
Let where are nonnegative integers and (essentially, is a mixed number). Then,
Here it is sufficient for to be an integer. We can use casework to find values of based on the value of :
nothing because n is positive
The pattern continues up to . Note that if , then . However if , the largest possible is , in which is still less than . Therefore the number of positive integers for is equal to
Solution 2
Notice that is continuous over the region for any integer . Therefore, it takes all values in the range over that interval. Note that if then and if , the maximum value attained is . It follows that the answer is
See Also
2009 AIME I Problems/Problem 6
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