2012 AIME II Problems/Problem 10

Revision as of 19:09, 31 March 2012 by Aimesolver (talk | contribs) (Solution)

Problem 10

Find the number of positive integers $n$ less than $1000$ for which there exists a positive real number $x$ such that $n=x\lfloor x \rfloor$.

Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.


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 $x = a + \frac{b}{c}$ where a,b,c are nonnegative integers and $0 \le b < c$ (essentially, x is a mixed number). Then,


$n = (a + \frac{b}{c}) \lfloor a +\frac{b}{c} \rfloor \Rightarrow n = (a + \frac{b}{c})a = a^2 + \frac{ab}{c}$ Here it is sufficient for $\frac{ab}{c}$ 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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All AIME Problems and Solutions
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