https://artofproblemsolving.com/wiki/index.php?title=2003_Indonesia_MO_Problems/Problem_3&feed=atom&action=history2003 Indonesia MO Problems/Problem 3 - Revision history2024-03-29T14:10:56ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2003_Indonesia_MO_Problems/Problem_3&diff=119625&oldid=prevRockmanex3: Added answer confirmation2020-03-17T15:50:25Z<p>Added answer confirmation</p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><br></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><br></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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> <ins class="diffchange diffchange-inline"> 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>.</ins></div></td></tr>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><br></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div><br></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>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></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>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> <ins class="diffchange diffchange-inline"> 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>.</ins></div></td></tr>
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</table>Rockmanex3https://artofproblemsolving.com/wiki/index.php?title=2003_Indonesia_MO_Problems/Problem_3&diff=97147&oldid=prevRockmanex3: Solution to Problem 3 -- floors and ceilings2018-08-11T03:55:53Z<p>Solution to Problem 3 -- floors and ceilings</p>
<p><b>New page</b></p><div>==Problem==<br />
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Find all real solutions of the equation <math>\lfloor x^2 \rfloor + \lceil x^2 \rceil = 2003</math>.<br />
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[Note: For any real number <math>\alpha</math>, <math>\lfloor \alpha \rfloor</math> is the largest integer less than or equal to <math>\alpha</math>, and <math>\lceil \alpha \rceil</math> denote the smallest integer more than or equal to <math>\alpha</math>.]<br />
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==Solution==<br />
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There are two cases to consider -- one where <math>x</math> is positive and one where <math>x</math> is negative.<br />
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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><br />
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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><br />
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In interval notation, the solutions of the equation are <math>\boxed{(-\sqrt{1002}, -\sqrt{1001}) \cup (\sqrt{1001}, \sqrt{1002})}.</math><br />
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==See Also==<br />
{{Indonesia MO box|year=2003|num-b=2|num-a=4}}<br />
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[[Category:Intermediate Algebra Problems]]</div>Rockmanex3