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

Revision as of 22:05, 31 May 2025

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 $\sum_{k=1}^{31} (k+1)k-k^2 = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.$

Solution 3

Bounding gives $x^2\le n<x^2+x$ . Thus there are a total of $x$ possible values for $n$ , for each value of $x^2$ . Checking, we see $31^2+31=992<1000$ , so there are $1+2+3+...+31= \boxed{496}$ such values for $n$ .

Solution 4

After a bit of experimenting, we let $n=l^2+s, s < 2n+1$ . We claim that I (the integer part of $x$ ) = $l$ . (Prove it yourself using contradiction !) so now we get that $x=l+\frac{s}{l}$ . This implies that solutions exist iff $s<l$ , or for all natural numbers of the form $l^2+s$ where $s<l$ . Hence, 1 solution exists for $l=1$ ! 2 for $l=2$ and so on. Therefore our final answer is $31+30+\dots+1= \boxed{496}

=== Solution 5 ===

Notice we can write$ (Error compiling LaTeX. Unknown error_msg)floor(x) $as$ x - frac(x) $. Since$ n $is a positive integer less than$ 1000 $,$ 1 \leq x(x - frac(x)) \leq 999 $. We now have$ 1 \leq x^{2} - x*frac(x) \leq 999 $. Since$ x*frac(x) $ranges from$ 0 $to$ x $, we can rewrite as$ 1 \leq x^{2} \leq 999 + x $. Notice all$ x $that satisfy this inequality range from$ 1 $to$ 32 $. Now every range from$ y $to$ y+1 $for$ x $result in exactly$ y $solutions that give a positive integer solution for$ n $. The first range is$ 1 $to$ 2 $and the last range is$ 31 $to$ 32 $. Thus, by using our above lemma, we can find the total number of solutions by$ 1+2+…+31 = \frac{31*32}{2} = \boxed{496}$.

~ilikemath247365

Video Solution

2012 AIME II #10

~MathProblemSolvingSkills.com

@@ Line 37: / Line 37: @@
 Hence, 1 solution exists for <math>l=1</math>! 2 for <math>l=2</math> and so on. Therefore our final answer is <math>31+30+\dots+1= \boxed{496}
-== Solution 5 ==
+=== Solution 5 ===
 Notice we can write </math>floor(x)<math> as </math>x - frac(x)<math>. Since </math>n<math> is a positive integer less than </math>1000<math>, </math>1 \leq x(x - frac(x)) \leq 999<math>. We now have </math>1 \leq x^{2} - x*frac(x) \leq 999<math>. Since </math>x*frac(x)<math> ranges from </math>0<math> to </math>x<math>, we can rewrite as </math>1 \leq x^{2} \leq 999 + x<math>. Notice all </math>x<math> that satisfy this inequality range from </math>1<math> to </math>32<math>. Now every range from </math>y<math> to </math>y+1<math> for </math>x<math> result in exactly </math>y<math> solutions that give a positive integer solution for </math>n<math>. The first range is </math>1<math> to </math>2<math> and the last range is </math>31<math> to </math>32<math>. Thus, by using our above lemma, we can find the total number of solutions by </math>1+2+…+31 = \frac{31*32}{2} = \boxed{496}$.

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

Page

Toolbox

Search