2012 AIME II Problems/Problem 10

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 to find values of n based on the value of a:

$a = 0 \implies$ nothing because n is positive

$a = 1 \implies \frac{b}{c} = \frac{0}{1}$

$a = 2 \implies \frac{b}{c} = \frac{0}{2},\frac{1}{2}$

$a = 3 \implies\frac{b}{c} =\frac{0}{3},\frac{1}{3},\frac{1}{3}$

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 $1+2+3+...+31 = \frac{31*32}{2} = \boxed{496.}$

2012 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2012 AIME II Problems/Problem 10

Problem 10

Solution

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