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

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=== Solution 2===
 
=== Solution 2===
 
Notice that <math>x\lfloor x\rfloor</math> is continuous over the region <math>x \in [k, k+1)</math> for any integer <math>k</math>. Therefore, it takes all values in the range <math>[k\lfloor k\rfloor, (k+1)\lfloor k+1\rfloor) = [k^2, (k+1)k)</math> over that interval. Note that if <math>k>32</math> then <math>k^2 > 1000</math> and if <math>k=31</math>, the maximum value attained is <math>31*32 < 1000</math>. It follows that the answer is <math> \sum_{k=1}^{31} (k+1)k-k^2  = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.</math>
 
Notice that <math>x\lfloor x\rfloor</math> is continuous over the region <math>x \in [k, k+1)</math> for any integer <math>k</math>. Therefore, it takes all values in the range <math>[k\lfloor k\rfloor, (k+1)\lfloor k+1\rfloor) = [k^2, (k+1)k)</math> over that interval. Note that if <math>k>32</math> then <math>k^2 > 1000</math> and if <math>k=31</math>, the maximum value attained is <math>31*32 < 1000</math>. It follows that the answer is <math> \sum_{k=1}^{31} (k+1)k-k^2  = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.</math>
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=== Solution 3===
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Bounding gives <math>x^2\le n<x^2+x</math>. Thus there are a total of <math>x</math> possible values for <math>n</math>, for each value of <math>x^2</math>. Checking, we see <math>31^2+31=992<1000</math>, so there are <cmath>1+2+3+...+31= \boxed{496}</cmath> such values for <math>n</math>.
  
 
=== Solution 4===
 
=== Solution 4===
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Bounding gives <math>x^2\le n<x^2+x</math>. Thus there are a total of <math>x</math> possible values for <math>n</math>, for each value of <math>x^2</math>. Checking, we see <math>31^2+31=992<1000</math>, so there are <cmath>1+2+3+...+32= \boxed{496}</cmath> such values for <math>n</math>.
 
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== Solution 3 ==
 
  
 
After a bit of experimenting, we let <math>n=l^2+s, s < 2n+1</math>. We claim that I (the integer part of <math>x</math>) = <math>l</math> . (Prove it yourself using contradiction !) so now we get that <math>x=l+\frac{s}{l}</math>. This implies that solutions exist iff <math>s<l</math>, or for all natural numbers of the form <math>l^2+s</math> where <math>s<l</math>.  
 
After a bit of experimenting, we let <math>n=l^2+s, s < 2n+1</math>. We claim that I (the integer part of <math>x</math>) = <math>l</math> . (Prove it yourself using contradiction !) so now we get that <math>x=l+\frac{s}{l}</math>. This implies that solutions exist iff <math>s<l</math>, or for all natural numbers of the form <math>l^2+s</math> where <math>s<l</math>.  
 
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}</math>
 
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}</math>
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=== Video Solution ===
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[https://youtu.be/GqiNJuOsl1w?si=yd68PahrwevAa6fn 2012 AIME II #10]
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~MathProblemSolvingSkills.com
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== See Also ==
 
== See Also ==
 
[[2009 AIME I Problems/Problem 6|2009 AIME I Problems/Problem 6]]
 
[[2009 AIME I Problems/Problem 6|2009 AIME I Problems/Problem 6]]

Latest revision as of 15:37, 24 July 2024

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 $31+30+\dots+1= \boxed{496}$


Video Solution

2012 AIME II #10

~MathProblemSolvingSkills.com


See Also

2009 AIME I Problems/Problem 6

2012 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AIME Problems and Solutions

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