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.