2012 AIME II Problems/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 .
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
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
Bounding gives . Thus there are a total of possible values for , for each value of . Checking, we see , so there are such values for .
After a bit of experimenting, we let . We claim that I (the integer part of ) = . (Prove it yourself using contradiction !) so now we get that . This implies that solutions exist iff , or for all natural numbers of the form where . Hence, 1 solution exists for ! 2 for and so on. Therefore our final answer is
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