2012 AIME II Problems/Problem 10
Contents
[hide]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
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 a,b,c are nonnegative integers and (essentially, x is a mixed number). Then,
Here it is sufficient for to be an integer. We can use casework to find values of n based on the value of a:
nothing because n is positive
The pattern continues up to . Note that if , then . However if , the largest possible x is , in which is still less than . Therefore the number of positive integers for n 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
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 |
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