https://artofproblemsolving.com/wiki/index.php?title=1981_IMO_Problems/Problem_4&feed=atom&action=history1981 IMO Problems/Problem 4 - Revision history2024-03-29T08:40:08ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1981_IMO_Problems/Problem_4&diff=45769&oldid=prevBappa1971: /* Alternate Solution */2012-03-28T13:43:55Z<p><span dir="auto"><span class="autocomment">Alternate Solution</span></span></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;"><div>Which implies, <math>p< n \le 2p</math>.</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>Which implies, <math>p< n \le 2p</math>.</div></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;"></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="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>As, <math>n-1 \ge p,3,2</math>, there must be a multiple of <math>p</math>, a multiple of <math>3</math> and  a multiple of <math>2</math> in the set, <math>\{m-1,m-2,\cdots,m-n+1}</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>As, <math>n-1 \ge p,3,2</math>, there must be a multiple of <math>p</math>, a multiple of <math>3</math> and  a multiple of <math>2</math> in the set, <math>\{m-1,m-2,\cdots,m-n+1<ins class="diffchange diffchange-inline">\</ins>}</math>.</div></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;"></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>So, <math>2p\ |\ \mbox{lcm}(m-1,m-2,\cdots,m-n+1)</math> and <math>3p\ |\ \mbox{lcm}(m-1,m-2,\cdots,m-n+1)</math>.</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>So, <math>2p\ |\ \mbox{lcm}(m-1,m-2,\cdots,m-n+1)</math> and <math>3p\ |\ \mbox{lcm}(m-1,m-2,\cdots,m-n+1)</math>.</div></td></tr>
</table>Bappa1971https://artofproblemsolving.com/wiki/index.php?title=1981_IMO_Problems/Problem_4&diff=44913&oldid=prevBappa1971 at 09:18, 20 February 20122012-02-20T09:18:40Z<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 09:18, 20 February 2012</td>
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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>Q.E.D.</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>Q.E.D.</div></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;"></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="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>{<del class="diffchange diffchange-inline">{alternate solutions}</del>}</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><ins class="diffchange diffchange-inline">== Alternate Solution ==</ins></div></td></tr>
<tr><td colspan="2"> </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><ins class="diffchange diffchange-inline">Let, for some <math>n</math> and <math>m</math> with <math>m>n</math>, <math>m\ |\ \mbox</ins>{<ins class="diffchange diffchange-inline">lcm</ins>}<ins class="diffchange diffchange-inline">(m-1,m-2,\cdots,m-n+1)\ |\ (m-1)(m-2)\cdots(m-n+1)</math>.</ins></div></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;"></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 colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;">We can trivially check that, there is no such <math>m</math> for <math>n=3</math>, only <math>m=3</math> works for <math>n=4</math> and <math>m=3,8</math> works for <math>n=5</math>.</ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;">Now, consider, <math>n>5</math>. By [http://en.wikipedia.org/wiki/Bertrand's_postulate Bertrand's postulate] there is a prime <math>p</math> such that <math>2 \le \left\lfloor\frac{n}{2}\right\rfloor < p < 2\left\lfloor\frac{n}{2}\right\rfloor</math>.</ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;">Which implies, <math>p< n \le 2p</math>.</ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;">As, <math>n-1 \ge p,3,2</math>, there must be a multiple of <math>p</math>, a multiple of <math>3</math> and  a multiple of <math>2</math> in the set, <math>\{m-1,m-2,\cdots,m-n+1}</math>.</ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;">So, <math>2p\ |\ \mbox{lcm}(m-1,m-2,\cdots,m-n+1)</math> and <math>3p\ |\ \mbox{lcm}(m-1,m-2,\cdots,m-n+1)</math>.</ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;">So, <math>m=2p</math> and <math>m=3p</math> both work for <math>n>5</math>.</ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;">So,</ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;">There exists solution for all <math>n>3</math>,</ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;">Only one Solution for <math>n=4</math>.</ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </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><ins style="font-weight: bold; text-decoration: none;">Q.E.D.</ins></div></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>{{IMO box|num-b=3|num-a=5|year=1981}}</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>{{IMO box|num-b=3|num-a=5|year=1981}}</div></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;"></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>[[Category:Olympiad Number Theory Problems]]</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>[[Category:Olympiad Number Theory Problems]]</div></td></tr>
</table>Bappa1971https://artofproblemsolving.com/wiki/index.php?title=1981_IMO_Problems/Problem_4&diff=18776&oldid=prevTemperal: template2007-10-26T00:46:38Z<p>template</p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 00:46, 26 October 2007</td>
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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>== Problem ==</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>== Problem ==</div></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;"></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="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>(a) For which values of <math> <del class="diffchange diffchange-inline">\displaystyle </del>n>2</math> is there a [[set]] of <math><del class="diffchange diffchange-inline">\displaystyle </del>n</math> consecutive [[positive integer]]s such that the largest number in the set is a [[divisor]] of the [[least common multiple]] of the remaining <math><del class="diffchange diffchange-inline">\displaystyle </del>n-1</math> numbers?</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>(a) For which values of <math>n>2</math> is there a [[set]] of <math>n</math> consecutive [[positive integer]]s such that the largest number in the set is a [[divisor]] of the [[least common multiple]] of the remaining <math>n-1</math> numbers?</div></td></tr>
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<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>(b) For which values of <math><del class="diffchange diffchange-inline">\displaystyle </del>n>2</math> is there exactly one set having the stated property?</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>(b) For which values of <math>n>2</math> is there exactly one set having the stated property?</div></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;"></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>== Solution ==</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>== Solution ==</div></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;"></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="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>Let <math> k = \prod p_i^{e_i} </math> be the greatest element of the set, written in its [[prime factorization]].  Then <math><del class="diffchange diffchange-inline">\displaystyle </del>k</math> divides the least common multiple of the other elements of the set if and only if the set has [[cardinality]] at least <math><del class="diffchange diffchange-inline">\displaystyle </del>\max \{ p_i^{e_i} \}</math>, since for any of the <math>p_i^{e_i}</math>, we must go down at least to <math>k-p_i^{e_i}</math> to obtain another [[multiple]] of <math>p_i^{e_i}</math>.  In particular, there is no set of cardinality 3 satisfying our conditions, because each number greater than or equal to 3 must be divisible by a number that is greater than two and is a power of a prime.</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>Let <math> k = \prod p_i^{e_i} </math> be the greatest element of the set, written in its [[prime factorization]].  Then <math>k</math> divides the least common multiple of the other elements of the set if and only if the set has [[cardinality]] at least <math>\max \{ p_i^{e_i} \}</math>, since for any of the <math>p_i^{e_i}</math>, we must go down at least to <math>k-p_i^{e_i}</math> to obtain another [[multiple]] of <math>p_i^{e_i}</math>.  In particular, there is no set of cardinality 3 satisfying our conditions, because each number greater than or equal to 3 must be divisible by a number that is greater than two and is a power of a prime.</div></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;"></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="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 <math> <del class="diffchange diffchange-inline">\displaystyle </del>n > 3 </math>, we may let <math> <del class="diffchange diffchange-inline">\displaystyle </del>k = \mbox{lcm} (n-1, n-2) = (n-1)(n-2) </math>, since all the <math><del class="diffchange diffchange-inline">\displaystyle </del>p_i^{e_i}</math> must clearly be less than <math><del class="diffchange diffchange-inline">\displaystyle </del>n</math> and this product must also be greater than <math><del class="diffchange diffchange-inline">\displaystyle </del>n</math> if <math><del class="diffchange diffchange-inline">\displaystyle </del>n</math> is at least 4.  For <math> <del class="diffchange diffchange-inline">\displaystyle </del>n > 4 </math>, we may also let <math> <del class="diffchange diffchange-inline">\displaystyle </del>k = \mbox{lcm} (n-2, n-3) = (n-2)(n-3)</math>, for the same reasons.  However, for <math> <del class="diffchange diffchange-inline">\displaystyle </del>n = 4 </math>, this does not work, and indeed no set works other than <math> <del class="diffchange diffchange-inline">\displaystyle </del>\{ 3,4,5,6 \} </math>.  To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.</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 <math>n > 3 </math>, we may let <math>k = \mbox{lcm} (n-1, n-2) = (n-1)(n-2) </math>, since all the <math>p_i^{e_i}</math> must clearly be less than <math>n</math> and this product must also be greater than <math>n</math> if <math>n</math> is at least 4.  For <math>n > 4 </math>, we may also let <math>k = \mbox{lcm} (n-2, n-3) = (n-2)(n-3)</math>, for the same reasons.  However, for <math>n = 4 </math>, this does not work, and indeed no set works other than <math>\{ 3,4,5,6 \} </math>.  To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.</div></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;"></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>Q.E.D.</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>Q.E.D.</div></td></tr>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l15" >Line 15:</td>
<td colspan="2" class="diff-lineno">Line 15:</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>{{alternate solutions}}</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>{{alternate solutions}}</div></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;"></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="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>== <del class="diffchange diffchange-inline">Resources =</del>=</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><ins class="diffchange diffchange-inline">{{IMO box|num-b</ins>=<ins class="diffchange diffchange-inline">3|num-a</ins>=<ins class="diffchange diffchange-inline">5|year</ins>=1981<ins class="diffchange diffchange-inline">}}</ins></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> </div></td><td colspan="2"> </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><del class="diffchange diffchange-inline">* [[</del>1981 <del class="diffchange diffchange-inline">IMO Problems]]</del></div></td><td colspan="2"> </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><del class="diffchange diffchange-inline">* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366641#p366641 Discussion on AoPS/MathLinks]</del></div></td><td colspan="2"> </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> </div></td><td colspan="2"> </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;"></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>[[Category:Olympiad Number Theory Problems]]</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>[[Category:Olympiad Number Theory Problems]]</div></td></tr>
</table>Temperalhttps://artofproblemsolving.com/wiki/index.php?title=1981_IMO_Problems/Problem_4&diff=10491&oldid=prevJBL at 21:09, 29 October 20062006-10-29T21:09:37Z<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 21:09, 29 October 2006</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</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>== Problem ==</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>== Problem ==</div></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;"></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="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>(a) For which values of <math> \displaystyle n>2</math> is there a set of <math>\displaystyle n</math> consecutive positive <del class="diffchange diffchange-inline">integers </del>such that the largest number in the set is a divisor of the least common multiple of the remaining <math>\displaystyle n-1</math> numbers?</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>(a) For which values of <math> \displaystyle n>2</math> is there a <ins class="diffchange diffchange-inline">[[</ins>set<ins class="diffchange diffchange-inline">]] </ins>of <math>\displaystyle n</math> consecutive <ins class="diffchange diffchange-inline">[[</ins>positive <ins class="diffchange diffchange-inline">integer]]s </ins>such that the largest number in the set is a <ins class="diffchange diffchange-inline">[[</ins>divisor<ins class="diffchange diffchange-inline">]] </ins>of the <ins class="diffchange diffchange-inline">[[</ins>least common multiple<ins class="diffchange diffchange-inline">]] </ins>of the remaining <math>\displaystyle n-1</math> numbers?</div></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;"></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>(b) For which values of <math>\displaystyle n>2</math> is there exactly one set having the stated property?</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>(b) For which values of <math>\displaystyle n>2</math> is there exactly one set having the stated property?</div></td></tr>
<tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l7" >Line 7:</td>
<td colspan="2" class="diff-lineno">Line 7:</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>== Solution ==</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>== Solution ==</div></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;"></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="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>Let <math> k = \prod p_i^{e_i} </math> be the greatest element of the set.  Then <math>\displaystyle k</math> divides the least common multiple of the other elements of the set <del class="diffchange diffchange-inline">[[iff]]. </del>the set has [[cardinality]] <del class="diffchange diffchange-inline">of </del>at least <math>\displaystyle \max \{ p_i^{e_i} \}</math>, since for any of the <math>p_i^{e_i}</math>, we must go down at least to <math>k-p_i^{e_i}</math> to obtain another multiple of <math>p_i^{e_i}</math>.  In particular, there is no set of cardinality 3 satisfying our conditions, because each number greater than or equal to 3 must be divisible by a number that is greater than two and is a power of a prime.</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>Let <math> k = \prod p_i^{e_i} </math> be the greatest element of the set<ins class="diffchange diffchange-inline">, written in its [[prime factorization]]</ins>.  Then <math>\displaystyle k</math> divides the least common multiple of the other elements of the set <ins class="diffchange diffchange-inline">if and only if </ins>the set has [[cardinality]] at least <math>\displaystyle \max \{ p_i^{e_i} \}</math>, since for any of the <math>p_i^{e_i}</math>, we must go down at least to <math>k-p_i^{e_i}</math> to obtain another <ins class="diffchange diffchange-inline">[[</ins>multiple<ins class="diffchange diffchange-inline">]] </ins>of <math>p_i^{e_i}</math>.  In particular, there is no set of cardinality 3 satisfying our conditions, because each number greater than or equal to 3 must be divisible by a number that is greater than two and is a power of a prime.</div></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;"></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>For <math> \displaystyle n > 3 </math>, we may let <math> \displaystyle k = \mbox{lcm} (n-1, n-2) = (n-1)(n-2) </math>, since all the <math>\displaystyle p_i^{e_i}</math> must clearly be less than <math>\displaystyle n</math> and this product must also be greater than <math>\displaystyle n</math> if <math>\displaystyle n</math> is at least 4.  For <math> \displaystyle n > 4 </math>, we may also let <math> \displaystyle k = \mbox{lcm} (n-2, n-3) = (n-2)(n-3)</math>, for the same reasons.  However, for <math> \displaystyle n = 4 </math>, this does not work, and indeed no set works other than <math> \displaystyle \{ 3,4,5,6 \} </math>.  To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.</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>For <math> \displaystyle n > 3 </math>, we may let <math> \displaystyle k = \mbox{lcm} (n-1, n-2) = (n-1)(n-2) </math>, since all the <math>\displaystyle p_i^{e_i}</math> must clearly be less than <math>\displaystyle n</math> and this product must also be greater than <math>\displaystyle n</math> if <math>\displaystyle n</math> is at least 4.  For <math> \displaystyle n > 4 </math>, we may also let <math> \displaystyle k = \mbox{lcm} (n-2, n-3) = (n-2)(n-3)</math>, for the same reasons.  However, for <math> \displaystyle n = 4 </math>, this does not work, and indeed no set works other than <math> \displaystyle \{ 3,4,5,6 \} </math>.  To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.</div></td></tr>
</table>JBLhttps://artofproblemsolving.com/wiki/index.php?title=1981_IMO_Problems/Problem_4&diff=10479&oldid=prevBoy Soprano II: /* Solution */ typo or Freudian slip2006-10-29T20:37:47Z<p><span dir="auto"><span class="autocomment">Solution: </span> typo or Freudian slip</span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class="diff-marker" />
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<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:37, 29 October 2006</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l7" >Line 7:</td>
<td colspan="2" class="diff-lineno">Line 7:</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>== Solution ==</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>== Solution ==</div></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;"></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="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>Let <math> k = \prod p_i^{e_i} </math> be the greatest element of the set.  Then <math>\displaystyle k</math> divides the least common multiple of the other elements of the set [[iff]]. the set has [[cardinality]] of at least <math>\displaystyle \max \{ p_i^{e_i} \}</math>, since for any of the <math>p_i^{e_i}</math>, we must go down at least to <math>k-p_i^{e_i}</math> to obtain another multiple of <math>p_i^{e_i}</math>.  In particular, there is no set of cardinality 3 satisfying our conditions, because each number greater than or equal <del class="diffchange diffchange-inline">two </del>3 must be divisible by a number that is greater than two and is a power of a prime.</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>Let <math> k = \prod p_i^{e_i} </math> be the greatest element of the set.  Then <math>\displaystyle k</math> divides the least common multiple of the other elements of the set [[iff]]. the set has [[cardinality]] of at least <math>\displaystyle \max \{ p_i^{e_i} \}</math>, since for any of the <math>p_i^{e_i}</math>, we must go down at least to <math>k-p_i^{e_i}</math> to obtain another multiple of <math>p_i^{e_i}</math>.  In particular, there is no set of cardinality 3 satisfying our conditions, because each number greater than or equal <ins class="diffchange diffchange-inline">to </ins>3 must be divisible by a number that is greater than two and is a power of a prime.</div></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;"></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>For <math> \displaystyle n > 3 </math>, we may let <math> \displaystyle k = \mbox{lcm} (n-1, n-2) = (n-1)(n-2) </math>, since all the <math>\displaystyle p_i^{e_i}</math> must clearly be less than <math>\displaystyle n</math> and this product must also be greater than <math>\displaystyle n</math> if <math>\displaystyle n</math> is at least 4.  For <math> \displaystyle n > 4 </math>, we may also let <math> \displaystyle k = \mbox{lcm} (n-2, n-3) = (n-2)(n-3)</math>, for the same reasons.  However, for <math> \displaystyle n = 4 </math>, this does not work, and indeed no set works other than <math> \displaystyle \{ 3,4,5,6 \} </math>.  To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.</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>For <math> \displaystyle n > 3 </math>, we may let <math> \displaystyle k = \mbox{lcm} (n-1, n-2) = (n-1)(n-2) </math>, since all the <math>\displaystyle p_i^{e_i}</math> must clearly be less than <math>\displaystyle n</math> and this product must also be greater than <math>\displaystyle n</math> if <math>\displaystyle n</math> is at least 4.  For <math> \displaystyle n > 4 </math>, we may also let <math> \displaystyle k = \mbox{lcm} (n-2, n-3) = (n-2)(n-3)</math>, for the same reasons.  However, for <math> \displaystyle n = 4 </math>, this does not work, and indeed no set works other than <math> \displaystyle \{ 3,4,5,6 \} </math>.  To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.</div></td></tr>
</table>Boy Soprano IIhttps://artofproblemsolving.com/wiki/index.php?title=1981_IMO_Problems/Problem_4&diff=10478&oldid=prevBoy Soprano II: /* Solution */ typo2006-10-29T20:35:38Z<p><span dir="auto"><span class="autocomment">Solution: </span> typo</span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:35, 29 October 2006</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l7" >Line 7:</td>
<td colspan="2" class="diff-lineno">Line 7:</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>== Solution ==</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>== Solution ==</div></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;"></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="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>Let <math> k = \prod p_i^{e_i} </math> be the greatest element of the set.  Then <math>\displaystyle k</math> divides the least common multiple of the other elements of the set [[iff]]. the set has [[cardinality]] of at least <math>\displaystyle \max \{ p_i^{e_i} \}</math>, since for any of the <math>p_i^{e_i}</math>, we must go down at least to <math>k-p_i^{e_i}</math> to obtain another multiple of <math>p_i^{e_i}</math>.  In particular, there is no set of cardinality 3 satisfying our <del class="diffchange diffchange-inline">conidtions</del>, because each number greater than or equal two 3 must be divisible by a number that is greater than two and is a power of a prime.</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>Let <math> k = \prod p_i^{e_i} </math> be the greatest element of the set.  Then <math>\displaystyle k</math> divides the least common multiple of the other elements of the set [[iff]]. the set has [[cardinality]] of at least <math>\displaystyle \max \{ p_i^{e_i} \}</math>, since for any of the <math>p_i^{e_i}</math>, we must go down at least to <math>k-p_i^{e_i}</math> to obtain another multiple of <math>p_i^{e_i}</math>.  In particular, there is no set of cardinality 3 satisfying our <ins class="diffchange diffchange-inline">conditions</ins>, because each number greater than or equal two 3 must be divisible by a number that is greater than two and is a power of a prime.</div></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;"></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>For <math> \displaystyle n > 3 </math>, we may let <math> \displaystyle k = \mbox{lcm} (n-1, n-2) = (n-1)(n-2) </math>, since all the <math>\displaystyle p_i^{e_i}</math> must clearly be less than <math>\displaystyle n</math> and this product must also be greater than <math>\displaystyle n</math> if <math>\displaystyle n</math> is at least 4.  For <math> \displaystyle n > 4 </math>, we may also let <math> \displaystyle k = \mbox{lcm} (n-2, n-3) = (n-2)(n-3)</math>, for the same reasons.  However, for <math> \displaystyle n = 4 </math>, this does not work, and indeed no set works other than <math> \displaystyle \{ 3,4,5,6 \} </math>.  To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.</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>For <math> \displaystyle n > 3 </math>, we may let <math> \displaystyle k = \mbox{lcm} (n-1, n-2) = (n-1)(n-2) </math>, since all the <math>\displaystyle p_i^{e_i}</math> must clearly be less than <math>\displaystyle n</math> and this product must also be greater than <math>\displaystyle n</math> if <math>\displaystyle n</math> is at least 4.  For <math> \displaystyle n > 4 </math>, we may also let <math> \displaystyle k = \mbox{lcm} (n-2, n-3) = (n-2)(n-3)</math>, for the same reasons.  However, for <math> \displaystyle n = 4 </math>, this does not work, and indeed no set works other than <math> \displaystyle \{ 3,4,5,6 \} </math>.  To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.</div></td></tr>
</table>Boy Soprano IIhttps://artofproblemsolving.com/wiki/index.php?title=1981_IMO_Problems/Problem_4&diff=10477&oldid=prevBoy Soprano II at 20:33, 29 October 20062006-10-29T20:33:15Z<p></p>
<p><b>New page</b></p><div>== Problem ==<br />
<br />
(a) For which values of <math> \displaystyle n>2</math> is there a set of <math>\displaystyle n</math> consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining <math>\displaystyle n-1</math> numbers?<br />
<br />
(b) For which values of <math>\displaystyle n>2</math> is there exactly one set having the stated property?<br />
<br />
== Solution ==<br />
<br />
Let <math> k = \prod p_i^{e_i} </math> be the greatest element of the set. Then <math>\displaystyle k</math> divides the least common multiple of the other elements of the set [[iff]]. the set has [[cardinality]] of at least <math>\displaystyle \max \{ p_i^{e_i} \}</math>, since for any of the <math>p_i^{e_i}</math>, we must go down at least to <math>k-p_i^{e_i}</math> to obtain another multiple of <math>p_i^{e_i}</math>. In particular, there is no set of cardinality 3 satisfying our conidtions, because each number greater than or equal two 3 must be divisible by a number that is greater than two and is a power of a prime.<br />
<br />
For <math> \displaystyle n > 3 </math>, we may let <math> \displaystyle k = \mbox{lcm} (n-1, n-2) = (n-1)(n-2) </math>, since all the <math>\displaystyle p_i^{e_i}</math> must clearly be less than <math>\displaystyle n</math> and this product must also be greater than <math>\displaystyle n</math> if <math>\displaystyle n</math> is at least 4. For <math> \displaystyle n > 4 </math>, we may also let <math> \displaystyle k = \mbox{lcm} (n-2, n-3) = (n-2)(n-3)</math>, for the same reasons. However, for <math> \displaystyle n = 4 </math>, this does not work, and indeed no set works other than <math> \displaystyle \{ 3,4,5,6 \} </math>. To prove this, we simply note that for any integer not equal to 6 and greater than 4 must have some power-of-a-prime factor greater than 3.<br />
<br />
Q.E.D.<br />
<br />
{{alternate solutions}}<br />
<br />
== Resources ==<br />
<br />
* [[1981 IMO Problems]]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366641#p366641 Discussion on AoPS/MathLinks]<br />
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[[Category:Olympiad Number Theory Problems]]</div>Boy Soprano II