https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&feed=atom&action=history2006 AIME I Problems/Problem 15 - Revision history2024-03-29T00:17:48ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=200127&oldid=prevMartin2001: /* Solution 4 */2023-10-24T23:31:39Z<p><span dir="auto"><span class="autocomment">Solution 4</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:31, 24 October 2023</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>=== Solution 4 ===</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 4 ===</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>Playing around with a couple numbers, we see that we can generate the sequence <math>0, 3, -6, 3, -6, \cdots</math>, and we can also generate the sequence <math>3, <del class="diffchange diffchange-inline">-</del>6, 9, 12, \cdots</math> after each <math>-6</math> value. Thus, we will apply this to try and find some bounds. We can test if the first <math>1000</math> pairs of numbers each sum up to <math>-3</math>, and the rest form an arithmetic sequence, if the first <math>990</math> pairs sum up to <math>-3</math>, and so on. When we get to <math>980</math>, we find that <math>980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303</math>. If we shift the number of pairs up by <math>1</math>, we get <math>981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}</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>Playing around with a couple numbers, we see that we can generate the sequence <math>0, 3, -6, 3, -6, \cdots</math>, and we can also generate the sequence <math>3, 6, 9, 12, \cdots</math> after each <math>-6</math> value. Thus, we will apply this to try and find some bounds. We can test if the first <math>1000</math> pairs of numbers each sum up to <math>-3</math>, and the rest form an arithmetic sequence, if the first <math>990</math> pairs sum up to <math>-3</math>, and so on. When we get to <math>980</math>, we find that <math>980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303</math>. If we shift the number of pairs up by <math>1</math>, we get <math>981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}</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>~ Spacesam</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>~ Spacesam</div></td></tr>
</table>Martin2001https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=200125&oldid=prevMartin2001: /* Solution 4 */2023-10-24T23:22:06Z<p><span dir="auto"><span class="autocomment">Solution 4</span></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 23:22, 24 October 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l55" >Line 55:</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>=== Solution 4 ===</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 4 ===</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>Playing around with a couple numbers, we see that we can generate the sequence <math>0, 3, -6, 3, -6, \cdots</math>, and we can also generate the sequence <math>3, 6, 9, 12, \cdots</math> after each <math>-6</math> value. Thus, we will apply this to try and find some bounds. We can test if the first <math>1000</math> pairs of numbers each sum up to <math>-3</math>, and the rest form an arithmetic sequence, if the first <math>990</math> pairs sum up to <math>-3</math>, and so on. When we get to <math>980</math>, we find that <math>980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303</math>. If we shift the number of pairs up by <math>1</math>, we get <math>981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}</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>Playing around with a couple numbers, we see that we can generate the sequence <math>0, 3, -6, 3, -6, \cdots</math>, and we can also generate the sequence <math>3, <ins class="diffchange diffchange-inline">-</ins>6, 9, 12, \cdots</math> after each <math>-6</math> value. Thus, we will apply this to try and find some bounds. We can test if the first <math>1000</math> pairs of numbers each sum up to <math>-3</math>, and the rest form an arithmetic sequence, if the first <math>990</math> pairs sum up to <math>-3</math>, and so on. When we get to <math>980</math>, we find that <math>980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303</math>. If we shift the number of pairs up by <math>1</math>, we get <math>981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}</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>~ Spacesam</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>~ Spacesam</div></td></tr>
</table>Martin2001https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=185117&oldid=prevRyanjwang: /* Solution 4 */2022-12-29T02:23:58Z<p><span dir="auto"><span class="autocomment">Solution 4</span></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 02:23, 29 December 2022</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>=== Solution 4 ===</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 4 ===</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>Playing around with a couple numbers, we see that we can generate the sequence <math>0, 3, -6, 3, -6, \cdots</math>, and we can also generate the sequence <math>3, 6, 9, 12, \cdots</math> after each <math>-6</math> value. Thus, we will apply this to try and find some bounds. We can test if the first <math>1000</math> pairs of numbers each sum up to <math>-3</math>, and the rest form an arithmetic sequence, if the first <math>990</math> pairs sum up to <math>-3</math>, and so on. When we get to <math>980</math>, we find that <math>980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303</math>. If we shift the number of pairs up by <math>1</math>, we get <math>981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}</math>. ~ Spacesam</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>Playing around with a couple numbers, we see that we can generate the sequence <math>0, 3, -6, 3, -6, \cdots</math>, and we can also generate the sequence <math>3, 6, 9, 12, \cdots</math> after each <math>-6</math> value. Thus, we will apply this to try and find some bounds. We can test if the first <math>1000</math> pairs of numbers each sum up to <math>-3</math>, and the rest form an arithmetic sequence, if the first <math>990</math> pairs sum up to <math>-3</math>, and so on. When we get to <math>980</math>, we find that <math>980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303</math>. If we shift the number of pairs up by <math>1</math>, we get <math>981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}</math>.</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> </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>~ Spacesam</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>== See also ==</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>== See also ==</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>{{AIME box|year=2006|n=I|num-b=14|after=Last Question}}</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>{{AIME box|year=2006|n=I|num-b=14|after=Last Question}}</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>{{MAA Notice}}</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>{{MAA Notice}}</div></td></tr>
</table>Ryanjwanghttps://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=185116&oldid=prevRyanjwang: /* Solution 4 */2022-12-29T02:23:18Z<p><span dir="auto"><span class="autocomment">Solution 4</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
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<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 02:23, 29 December 2022</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>=== Solution 4 ===</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 4 ===</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>Playing around with a couple numbers, we see that we can generate the sequence <math>0, 3, -6, 3, -6, \cdots</math>, and we can also generate the sequence <math>3, 6, 9, 12, \cdots</math> after each <math>-6</math> value. Thus, we will apply this to try and find some bounds. We can test if the first <math>1000</math> pairs of numbers each sum up to <math>-3</math>, and the rest form an arithmetic sequence, if the first <math>990</math> pairs sum up to <math>-3</math>, and so on. When we get to <math>980</math>, we find that <math>980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303</math>. If we shift the number of pairs up by <math>1</math>, we get <math>981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}</math>. <del class="diffchange diffchange-inline">- </del>Spacesam</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>Playing around with a couple numbers, we see that we can generate the sequence <math>0, 3, -6, 3, -6, \cdots</math>, and we can also generate the sequence <math>3, 6, 9, 12, \cdots</math> after each <math>-6</math> value. Thus, we will apply this to try and find some bounds. We can test if the first <math>1000</math> pairs of numbers each sum up to <math>-3</math>, and the rest form an arithmetic sequence, if the first <math>990</math> pairs sum up to <math>-3</math>, and so on. When we get to <math>980</math>, we find that <math>980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303</math>. If we shift the number of pairs up by <math>1</math>, we get <math>981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}</math>. <ins class="diffchange diffchange-inline">~ </ins>Spacesam</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>== See also ==</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>== See also ==</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>{{AIME box|year=2006|n=I|num-b=14|after=Last Question}}</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>{{AIME box|year=2006|n=I|num-b=14|after=Last Question}}</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>{{MAA Notice}}</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>{{MAA Notice}}</div></td></tr>
</table>Ryanjwanghttps://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=155936&oldid=prevPinkpig: /* Solution 2 */2021-06-14T18:48:28Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></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 18:48, 14 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l28" >Line 28:</td>
<td colspan="2" class="diff-lineno">Line 28:</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 2 ===</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 2 ===</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>First, we state that <del class="diffchange diffchange-inline">if </del><math>x_{i - 1}\ge0</math>, <math>|x_i| = |x_{i - 1}| + 3</math> and iff <math>x_{i - 1} < 0</math>, <math>|x_i| = |x_{i - 1}| - 3</math>. Now suppose <math>x_i = x_j</math> for some <math>0\le i < j\le2006</math>. Now, this means that <math>|x_i| = |x_j|</math>, and so the number of positive numbers in the set <math>\{x_i,x_{i + 1},\ldots,x_{j - 1}\}</math> equals the number of negative numbers. Now pair the numbers in this list up in the following way: Whenever a positive and a negative number are adjacent in this progression, pair them up and remove them from this list. We claim that every pair will sum to -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>First, we state that <ins class="diffchange diffchange-inline">iff </ins><math>x_{i - 1}\ge0</math>, <math>|x_i| = |x_{i - 1}| + 3</math> and iff <math>x_{i - 1} < 0</math>, <math>|x_i| = |x_{i - 1}| - 3</math>. Now suppose <math>x_i = x_j</math> for some <math>0\le i < j\le2006</math>. Now, this means that <math>|x_i| = |x_j|</math>, and so the number of positive numbers in the set <math>\{x_i,x_{i + 1},\ldots,x_{j - 1}\}</math> equals the number of negative numbers. Now pair the numbers in this list up in the following way: Whenever a positive and a negative number are adjacent in this progression, pair them up and remove them from this list. We claim that every pair will sum to -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>If the positive number comes first, then the negative number will have a magnitude three greater, so this is true. If the negative number comes first, then the positive number will have magnitude three smaller, and this will also be true. Now let us examine what happens when we remove those two from the sequence. WLOG, let the numbers be <math>x_k</math> and <math>x_{k + 1}</math>. Since one is positive and the other is negative, <math>|x_{k + 2}| = |x_{k + 1}|\pm3 = |x_k|\pm3\mp3 = |x_k| = |x_{k - 1} + 3|</math>. So the new sequence works under the same criteria as the old one. In this way, we can pair all of the numbers up in this subsequence so the sums of the pairs are -3. Thus, the average of these numbers will be -3/2 for all subsequences that start and end with the same number (not including one of those).</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>If the positive number comes first, then the negative number will have a magnitude three greater, so this is true. If the negative number comes first, then the positive number will have magnitude three smaller, and this will also be true. Now let us examine what happens when we remove those two from the sequence. WLOG, let the numbers be <math>x_k</math> and <math>x_{k + 1}</math>. Since one is positive and the other is negative, <math>|x_{k + 2}| = |x_{k + 1}|\pm3 = |x_k|\pm3\mp3 = |x_k| = |x_{k - 1} + 3|</math>. So the new sequence works under the same criteria as the old one. In this way, we can pair all of the numbers up in this subsequence so the sums of the pairs are -3. Thus, the average of these numbers will be -3/2 for all subsequences that start and end with the same number (not including one of those).</div></td></tr>
</table>Pinkpighttps://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=155935&oldid=prevPinkpig: /* Solution 2 */2021-06-14T18:48:04Z<p><span dir="auto"><span class="autocomment">Solution 2</span></span></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 18:48, 14 June 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l28" >Line 28:</td>
<td colspan="2" class="diff-lineno">Line 28:</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 2 ===</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 2 ===</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>First, we state that <del class="diffchange diffchange-inline">iff </del><math>x_{i - 1}\ge0</math>, <math>|x_i| = |x_{i - 1}| + 3</math> and iff <math>x_{i - 1} < 0</math>, <math>|x_i| = |x_{i - 1}| - 3</math>. Now suppose <math>x_i = x_j</math> for some <math>0\le i < j\le2006</math>. Now, this means that <math>|x_i| = |x_j|</math>, and so the number of positive numbers in the set <math>\{x_i,x_{i + 1},\ldots,x_{j - 1}\}</math> equals the number of negative numbers. Now pair the numbers in this list up in the following way: Whenever a positive and a negative number are adjacent in this progression, pair them up and remove them from this list. We claim that every pair will sum to -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>First, we state that <ins class="diffchange diffchange-inline">if </ins><math>x_{i - 1}\ge0</math>, <math>|x_i| = |x_{i - 1}| + 3</math> and iff <math>x_{i - 1} < 0</math>, <math>|x_i| = |x_{i - 1}| - 3</math>. Now suppose <math>x_i = x_j</math> for some <math>0\le i < j\le2006</math>. Now, this means that <math>|x_i| = |x_j|</math>, and so the number of positive numbers in the set <math>\{x_i,x_{i + 1},\ldots,x_{j - 1}\}</math> equals the number of negative numbers. Now pair the numbers in this list up in the following way: Whenever a positive and a negative number are adjacent in this progression, pair them up and remove them from this list. We claim that every pair will sum to -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>If the positive number comes first, then the negative number will have a magnitude three greater, so this is true. If the negative number comes first, then the positive number will have magnitude three smaller, and this will also be true. Now let us examine what happens when we remove those two from the sequence. WLOG, let the numbers be <math>x_k</math> and <math>x_{k + 1}</math>. Since one is positive and the other is negative, <math>|x_{k + 2}| = |x_{k + 1}|\pm3 = |x_k|\pm3\mp3 = |x_k| = |x_{k - 1} + 3|</math>. So the new sequence works under the same criteria as the old one. In this way, we can pair all of the numbers up in this subsequence so the sums of the pairs are -3. Thus, the average of these numbers will be -3/2 for all subsequences that start and end with the same number (not including one of those).</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>If the positive number comes first, then the negative number will have a magnitude three greater, so this is true. If the negative number comes first, then the positive number will have magnitude three smaller, and this will also be true. Now let us examine what happens when we remove those two from the sequence. WLOG, let the numbers be <math>x_k</math> and <math>x_{k + 1}</math>. Since one is positive and the other is negative, <math>|x_{k + 2}| = |x_{k + 1}|\pm3 = |x_k|\pm3\mp3 = |x_k| = |x_{k - 1} + 3|</math>. So the new sequence works under the same criteria as the old one. In this way, we can pair all of the numbers up in this subsequence so the sums of the pairs are -3. Thus, the average of these numbers will be -3/2 for all subsequences that start and end with the same number (not including one of those).</div></td></tr>
</table>Pinkpighttps://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=152943&oldid=prevShihan: /* Solution 3 */2021-04-29T17:41:33Z<p><span dir="auto"><span class="autocomment">Solution 3</span></span></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 17:41, 29 April 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l47" >Line 47:</td>
<td colspan="2" class="diff-lineno">Line 47:</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><math>{x_{2007}}^2 = 6\left(x_1 + x_2 + \cdots + x_{2006}\right) + 9\cdot{2007}</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><math>{x_{2007}}^2 = 6\left(x_1 + x_2 + \cdots + x_{2006}\right) + 9\cdot{2007}</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>So <math>|x_1 + x_2 + \cdots + x_{2006}| = \<del class="diffchange diffchange-inline">frac {</del>\left|{x_{2007}}^2 - 9\cdot{2007}\right|<del class="diffchange diffchange-inline">}{6}</del></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>So <math>|x_1 + x_2 + \cdots + x_{2006}| = \<ins class="diffchange diffchange-inline">tfrac 16 </ins>\left|{x_{2007}}^2 - 9\cdot{2007}\right|</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>We know <math>3\ |\ x_{2007}</math> and we want to minimize <math>\left|{x_{2007}}^2 - 9\cdot{2007}\right|</math>, so <math>x_{2007}</math> must be <math>3\cdot{45}</math> for it to be minimal (<math>45^2 = 2025</math> which is closest to <math>2007</math>). We can achieve this with <math>x_k=3k</math> till <math>x_{45}=135</math> and then alternating <math>x_{46}=132</math>, <math>x_{47}=135</math> and so on ... Then <math>x_{2k}=132</math> and <math>x_{2k+1}=135</math> for all <math>k>22</math>. Since <math>2007</math> is odd, we have <math>x_{2007}=135</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>We know <math>3\ |\ x_{2007}</math> and we want to minimize <math>\left|{x_{2007}}^2 - 9\cdot{2007}\right|</math>, so <math>x_{2007}</math> must be <math>3\cdot{45}</math> for it to be minimal (<math>45^2 = 2025</math> which is closest to <math>2007</math>). We can achieve this with <math>x_k=3k</math> till <math>x_{45}=135</math> and then alternating <math>x_{46}=132</math>, <math>x_{47}=135</math> and so on ... Then <math>x_{2k}=132</math> and <math>x_{2k+1}=135</math> for all <math>k>22</math>. Since <math>2007</math> is odd, we have <math>x_{2007}=135</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>This means that <math>|x_1 + x_2 + \cdots + x_{2006}| = \left|<del class="diffchange diffchange-inline">\frac {</del>9(2025 - 2007)<del class="diffchange diffchange-inline">}{6}</del>\right| = \boxed{027}</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>This means that <math>|x_1 + x_2 + \cdots + x_{2006}| = <ins class="diffchange diffchange-inline">\tfrac 16 </ins>\left|9(2025 - 2007)\right| = \boxed{027}</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>=== Solution 4 ===</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 4 ===</div></td></tr>
</table>Shihanhttps://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=152942&oldid=prevShihan: /* Solution 3 */2021-04-29T17:39:11Z<p><span dir="auto"><span class="autocomment">Solution 3</span></span></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 17:39, 29 April 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l49" >Line 49:</td>
<td colspan="2" class="diff-lineno">Line 49:</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>|x_1 + x_2 + \cdots + x_{2006}| = \frac {\left|{x_{2007}}^2 - 9\cdot{2007}\right|}{6}</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>|x_1 + x_2 + \cdots + x_{2006}| = \frac {\left|{x_{2007}}^2 - 9\cdot{2007}\right|}{6}</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>We know <math>3\ |\ x_{2007}</math> and we want to minimize <math>\left|{x_{2007}}^2 - 9\cdot{2007}\right|</math>, so <math>x_{2007}</math> must be <math>3\cdot{45}</math> for it to be minimal (<math>45^2 = 2025</math> which is closest to <math>2007</math>). We can achieve this with <math><del class="diffchange diffchange-inline">x_0</del>=<del class="diffchange diffchange-inline">0, x_1</del>=<del class="diffchange diffchange-inline">-3</del>, <del class="diffchange diffchange-inline">x_2</del>=<del class="diffchange diffchange-inline">0</del></math> and so on <del class="diffchange diffchange-inline">till </del><math>x_{<del class="diffchange diffchange-inline">1962</del>}=<del class="diffchange diffchange-inline">0</del></math><del class="diffchange diffchange-inline">; then we can choose </del><math>x_{<del class="diffchange diffchange-inline">1962</del>+<del class="diffchange diffchange-inline">k</del>}=<del class="diffchange diffchange-inline">3k</del></math><del class="diffchange diffchange-inline">; since </del><math>2007 <del class="diffchange diffchange-inline">= 1962 + 45</del></math> <del class="diffchange diffchange-inline">to get </del><math>x_{2007}<del class="diffchange diffchange-inline">=3\cdot 45 </del>= 135</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>We know <math>3\ |\ x_{2007}</math> and we want to minimize <math>\left|{x_{2007}}^2 - 9\cdot{2007}\right|</math>, so <math>x_{2007}</math> must be <math>3\cdot{45}</math> for it to be minimal (<math>45^2 = 2025</math> which is closest to <math>2007</math>). We can achieve this with <math><ins class="diffchange diffchange-inline">x_k</ins>=<ins class="diffchange diffchange-inline">3k</math> till <math>x_{45}=135</math> and then alternating <math>x_{46}</ins>=<ins class="diffchange diffchange-inline">132</math></ins>, <ins class="diffchange diffchange-inline"><math>x_{47}</ins>=<ins class="diffchange diffchange-inline">135</ins></math> and so on <ins class="diffchange diffchange-inline">... Then </ins><math>x_{<ins class="diffchange diffchange-inline">2k</ins>}=<ins class="diffchange diffchange-inline">132</ins></math> <ins class="diffchange diffchange-inline">and </ins><math>x_{<ins class="diffchange diffchange-inline">2k</ins>+<ins class="diffchange diffchange-inline">1</ins>}=<ins class="diffchange diffchange-inline">135</math> for all <math>k>22</ins></math><ins class="diffchange diffchange-inline">. Since </ins><math>2007</math> <ins class="diffchange diffchange-inline">is odd, we have </ins><math>x_{2007}=135</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>This means that <math>|x_1 + x_2 + \cdots + x_{2006}| = \left|\frac {9(2025 - 2007)}{6}\right| = \boxed{027}</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>This means that <math>|x_1 + x_2 + \cdots + x_{2006}| = \left|\frac {9(2025 - 2007)}{6}\right| = \boxed{027}</math></div></td></tr>
</table>Shihanhttps://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=152941&oldid=prevShihan: /* Solution 3 */2021-04-29T17:35:54Z<p><span dir="auto"><span class="autocomment">Solution 3</span></span></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 17:35, 29 April 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l49" >Line 49:</td>
<td colspan="2" class="diff-lineno">Line 49:</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>|x_1 + x_2 + \cdots + x_{2006}| = \frac {\left|{x_{2007}}^2 - 9\cdot{2007}\right|}{6}</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>|x_1 + x_2 + \cdots + x_{2006}| = \frac {\left|{x_{2007}}^2 - 9\cdot{2007}\right|}{6}</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>We know <math>3\ |\ x_{2007}</math> and we want to minimize <math>\left|{x_{2007}}^2 - 9\cdot{2007}\right|</math>, so <math>x_{2007}</math> must be <math>3\cdot{45}</math> for it to be minimal (<math>45^2 = 2025</math> which is closest to <math>2007</math>)<del class="diffchange diffchange-inline">. To acheive this we need <math>x_{2007}=135</math></del>. We can <del class="diffchange diffchange-inline">get </del><math>x_0=0, x_1=-3, x_2=0</math> and so on till <math>x_{1962}=0</math>; then we can choose <math>x_{1962+k}=3k</math>; since <math>2007 = 1962 + 45</math> to get <math>x_{2007}=3\cdot 45 = 135</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>We know <math>3\ |\ x_{2007}</math> and we want to minimize <math>\left|{x_{2007}}^2 - 9\cdot{2007}\right|</math>, so <math>x_{2007}</math> must be <math>3\cdot{45}</math> for it to be minimal (<math>45^2 = 2025</math> which is closest to <math>2007</math>). We can <ins class="diffchange diffchange-inline">achieve this with </ins><math>x_0=0, x_1=-3, x_2=0</math> and so on till <math>x_{1962}=0</math>; then we can choose <math>x_{1962+k}=3k</math>; since <math>2007 = 1962 + 45</math> to get <math>x_{2007}=3\cdot 45 = 135</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>This means that <math>|x_1 + x_2 + \cdots + x_{2006}| = \left|\frac {9(2025 - 2007)}{6}\right| = \boxed{027}</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>This means that <math>|x_1 + x_2 + \cdots + x_{2006}| = \left|\frac {9(2025 - 2007)}{6}\right| = \boxed{027}</math></div></td></tr>
</table>Shihanhttps://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_15&diff=152938&oldid=prevShihan: /* Solution 3 */2021-04-29T17:26:54Z<p><span dir="auto"><span class="autocomment">Solution 3</span></span></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 17:26, 29 April 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l49" >Line 49:</td>
<td colspan="2" class="diff-lineno">Line 49:</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>|x_1 + x_2 + \cdots + x_{2006}| = \frac {\left|{x_{2007}}^2 - 9\cdot{2007}\right|}{6}</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>|x_1 + x_2 + \cdots + x_{2006}| = \frac {\left|{x_{2007}}^2 - 9\cdot{2007}\right|}{6}</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>We know <math>3\ |\ x_{2007}</math> and we want to minimize <math>\left|{x_{2007}}^2 - 9\cdot{2007}\right|</math>, so <math>x_{2007}</math> must be <math>3\cdot{45}</math> for it to be minimal (<math>45^2 = 2025</math> which is closest to <math>2007</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>We know <math>3\ |\ x_{2007}</math> and we want to minimize <math>\left|{x_{2007}}^2 - 9\cdot{2007}\right|</math>, so <math>x_{2007}</math> must be <math>3\cdot{45}</math> for it to be minimal (<math>45^2 = 2025</math> which is closest to <math>2007</math>)<ins class="diffchange diffchange-inline">. To acheive this we need <math>x_{2007}=135</math>. We can get <math>x_0=0, x_1=-3, x_2=0</math> and so on till <math>x_{1962}=0</math>; then we can choose <math>x_{1962+k}=3k</math>; since <math>2007 = 1962 + 45</math> to get <math>x_{2007}=3\cdot 45 = 135</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 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>This means that <math>|x_1 + x_2 + \cdots + x_{2006}| = \left|\frac {9(2025 - 2007)}{6}\right| = \boxed{027}</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>This means that <math>|x_1 + x_2 + \cdots + x_{2006}| = \left|\frac {9(2025 - 2007)}{6}\right| = \boxed{027}</math></div></td></tr>
</table>Shihan