Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 Indonesia MO Problems/Problem 3"

(Solution to Problem 3 -- floors and ceilings)
 
m (Added answer confirmation)
 
Line 10: Line 10:
  
 
<br>
 
<br>
−
For the positive case, if <math>x = \sqrt{a},</math> then the equation results in <math>2a = 2003.</math>  Since the equation does not have an integral solution, <math>x \ne \sqrt{a}.</math>  If we let <math>\sqrt{a} < x < \sqrt{a+1}.</math>  That means <math>a + a + 1 = 2003,</math> and solving the equation yields <math>a = 1001.</math>
+
For the positive case, if <math>x = \sqrt{a},</math> then the equation results in <math>2a = 2003.</math>  Since the equation does not have an integral solution, <math>x \ne \sqrt{a}.</math>  If we let <math>\sqrt{a} < x < \sqrt{a+1}.</math>  That means <math>a + a + 1 = 2003,</math> and solving the equation yields <math>a = 1001.</math> For confirmation, <math>1001 \le \lfloor x^2 \rfloor < 1002</math> and <math>1001 < \lceil x^2 \rceil \le 1002</math>, so <math>2002 < \lfloor x^2 \rfloor + \lceil x^2 \rceil < 2004</math>.  Since <math>\lfloor x^2 \rfloor + \lceil x^2 \rceil</math> can only have integral values, <math>\lfloor x^2 \rfloor + \lceil x^2 \rceil = 2003</math>.
  
 
<br>
 
<br>
−
For the negative case, if <math>x = -\sqrt{a},</math> then the equation results in <math>2a = 2003.</math>  This also does not have an integral solution, so <math>x \ne \sqrt{a}.</math>  If we let <math>-\sqrt{a+1} < x < -\sqrt{a}.</math>  That means <math>a+1 + a = 2003,</math> and this equation also yields <math>a = 1001.</math>
+
For the negative case, if <math>x = -\sqrt{a},</math> then the equation results in <math>2a = 2003.</math>  This also does not have an integral solution, so <math>x \ne \sqrt{a}.</math>  If we let <math>-\sqrt{a+1} < x < -\sqrt{a}.</math>  That means <math>a+1 + a = 2003,</math> and this equation also yields <math>a = 1001.</math> For confirmation, <math>1001 < \lfloor x^2 \rfloor \le 1002</math> and <math>1001 \le \lceil x^2 \rceil < 1002</math>, so <math>2002 < \lfloor x^2 \rfloor + \lceil x^2 \rceil < 2004</math>.  Since <math>\lfloor x^2 \rfloor + \lceil x^2 \rceil</math> can only have integral values, <math>\lfloor x^2 \rfloor + \lceil x^2 \rceil = 2003</math>.
  
 
<br>
 
<br>

Latest revision as of 11:50, 17 March 2020

Problem

Find all real solutions of the equation $\lfloor x^2 \rfloor + \lceil x^2 \rceil = 2003$.

[Note: For any real number $\alpha$, $\lfloor \alpha \rfloor$ is the largest integer less than or equal to $\alpha$, and $\lceil \alpha \rceil$ denote the smallest integer more than or equal to $\alpha$.]

Solution

There are two cases to consider -- one where $x$ is positive and one where $x$ is negative.


For the positive case, if $x = \sqrt{a},$ then the equation results in $2a = 2003.$ Since the equation does not have an integral solution, $x \ne \sqrt{a}.$ If we let $\sqrt{a} < x < \sqrt{a+1}.$ That means $a + a + 1 = 2003,$ and solving the equation yields $a = 1001.$ For confirmation, $1001 \le \lfloor x^2 \rfloor < 1002$ and $1001 < \lceil x^2 \rceil \le 1002$, so $2002 < \lfloor x^2 \rfloor + \lceil x^2 \rceil < 2004$. Since $\lfloor x^2 \rfloor + \lceil x^2 \rceil$ can only have integral values, $\lfloor x^2 \rfloor + \lceil x^2 \rceil = 2003$.


For the negative case, if $x = -\sqrt{a},$ then the equation results in $2a = 2003.$ This also does not have an integral solution, so $x \ne \sqrt{a}.$ If we let $-\sqrt{a+1} < x < -\sqrt{a}.$ That means $a+1 + a = 2003,$ and this equation also yields $a = 1001.$ For confirmation, $1001 < \lfloor x^2 \rfloor \le 1002$ and $1001 \le \lceil x^2 \rceil < 1002$, so $2002 < \lfloor x^2 \rfloor + \lceil x^2 \rceil < 2004$. Since $\lfloor x^2 \rfloor + \lceil x^2 \rceil$ can only have integral values, $\lfloor x^2 \rfloor + \lceil x^2 \rceil = 2003$.


In interval notation, the solutions of the equation are $\boxed{(-\sqrt{1002}, -\sqrt{1001}) \cup (\sqrt{1001}, \sqrt{1002})}.$

See Also

2003 Indonesia MO (Problems)
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed by
Problem 4
All Indonesia MO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_Indonesia_MO_Problems/Problem_3&oldid=119625"