2012 AIME II Problems/Problem 10
Contents
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
Solution 3
Bounding gives . Thus there are a total of possible values for , for each value of . Checking, we see , so there are such values for .
Solution 4
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
Video Solution
~MathProblemSolvingSkills.com
See Also
2009 AIME I Problems/Problem 6
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.