Difference between revisions of "2012 AIME II Problems/Problem 10"

Revision as of 13:35, 13 August 2020

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

Solution 1

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, $n = \left(a + \frac{b}{c}\right) \left\lfloor a +\frac{b}{c} \right\rfloor \Rightarrow n = \left(a + \frac{b}{c}\right)a = a^2 + \frac{ab}{c}$

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

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

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

$a = 3 \implies\frac{b}{c} =\frac{0}{3},\frac{1}{3},\frac{2}{3}$

The pattern continues up to $a = 31$ . Note that if $a = 32$ , then $n > 1000$ . However if $a = 31$ , the largest possible $x$ is $31 + \frac{30}{31}$ , in which $n$ is still less than $1000$ . Therefore the number of positive integers for $n$ is equal to $1+2+3+...+31 = \frac{31 \cdot 32}{2} = \boxed{496}.$

Solution 2

Notice that $x\lfloor x\rfloor$ is continuous over the region $x \in [k, k+1)$ for any integer $k$ . Therefore, it takes all values in the range $[k\lfloor k\rfloor, (k+1)\lfloor k+1\rfloor) = [k^2, (k+1)k)$ over that interval. Note that if $k>32$ then $k^2 > 1000$ and if $k=31$ , the maximum value attained is $31*32 < 1000$ . It follows that the answer is $\sum_{k=1}^{31} (k+1)k-k^2 = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.$

@@ Line 11: / Line 11: @@
 Let <math>x = a + \frac{b}{c}</math> where <math>a,b,c</math> are nonnegative integers and <math>0 \le b < c</math> (essentially, <math>x</math> is a mixed number). Then,
-<cmath>n = (a + \frac{b}{c}) \lfloor a +\frac{b}{c} \rfloor \Rightarrow n = (a + \frac{b}{c})a = a^2 + \frac{ab}{c}</cmath>
+<cmath>n = \left(a + \frac{b}{c}\right) \left\lfloor a +\frac{b}{c} \right\rfloor \Rightarrow n = \left(a + \frac{b}{c}\right)a = a^2 + \frac{ab}{c}</cmath>
 Here it is sufficient for <math>\frac{ab}{c}</math> to be an integer. We can use casework to find values of <math>n</math> based on the value of <math>a</math>:

2012 AIME II (Problems • Answer Key • Resources)
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