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
We know that x cannot be irrational because the product of a rational number and an irrational number is irrational (but n is an integer). Therefore x 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 based on the value of a:
a = 0 --> nothing because n is positive
a = 1 --> b/c = 0/1
a = 2 --> b/c = 0/2, 1/2
a = 3 --> b/c = 0/3, 1/3, 2/3
The pattern continues up to a = 31. Note that if a = 32, then n > 1000. However if a = 31, the largest possible x is 31 + 30/31, in which n is still less than 1000. Therefore the number of positive integers for n is equal to 1+2+3+...+31 = 31*32/2 = 496.
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 |