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1987 AIME Problems/Problem 12 - Revision history
2024-03-29T11:32:58Z
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Brackie1331: /* Solution 1 */
2023-12-28T23:48:15Z
<p><span dir="auto"><span class="autocomment">Solution 1</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:48, 28 December 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>In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.</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>In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.</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><math>m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3</math>.  Since <math>r < \frac{1}{1000}</math> and <math>m - n^3 = r(3n^2 + 3nr + r^2)</math> is an integer, we must have that <math>3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000</math>.  This means that the smallest possible <math>n</math> should be less than 1000<del class="diffchange diffchange-inline">. (It is more evident once you plug in r=1/1000. Do you see why?)</del>. In particular, <math>3nr + r^2</math> should be less than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>.  <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>.  Since we want to minimize <math>n</math>, we take <math>n = 19</math>.  Then for any positive value of <math>r</math>, <math>3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000</math>, so it is possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>.  However, we still have to make sure a sufficiently small <math>r</math> exists.   </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><math>m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3</math>.  Since <math>r < \frac{1}{1000}</math> and <math>m - n^3 = r(3n^2 + 3nr + r^2)</math> is an integer, we must have that <math>3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000</math>.  This means that the smallest possible <math>n</math> should be less than 1000. In particular, <math>3nr + r^2</math> should be less than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>.  <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>.  Since we want to minimize <math>n</math>, we take <math>n = 19</math>.  Then for any positive value of <math>r</math>, <math>3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000</math>, so it is possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>.  However, we still have to make sure a sufficiently small <math>r</math> exists.   </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>In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to ensure a small enough <math>r</math>.  The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>.  Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set.  The answer is  <math>\boxed{019}</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>In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to ensure a small enough <math>r</math>.  The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>.  Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set.  The answer is  <math>\boxed{019}</math>.</div></td></tr>
</table>
Brackie1331
https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_12&diff=208583&oldid=prev
Brackie1331: /* Solution 1 */
2023-12-28T23:47:41Z
<p><span dir="auto"><span class="autocomment">Solution 1</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;">Revision as of 23:47, 28 December 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l4" >Line 4:</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>In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.</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>In order to keep <math>m</math> as small as possible, we need to make <math>n</math> as small as possible.</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><math>m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3</math>.  Since <math>r < \frac{1}{1000}</math> and <math>m - n^3 = r(3n^2 + 3nr + r^2)</math> is an integer, we must have that <math>3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000</math>.  This means that the smallest possible <math>n</math> should be <del class="diffchange diffchange-inline">quite a bit smaller </del>than 1000. In particular, <math>3nr + r^2</math> should be less than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>.  <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>.  Since we want to minimize <math>n</math>, we take <math>n = 19</math>.  Then for any positive value of <math>r</math>, <math>3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000</math>, so it is possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>.  However, we still have to make sure a sufficiently small <math>r</math> exists.   </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><math>m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3</math>.  Since <math>r < \frac{1}{1000}</math> and <math>m - n^3 = r(3n^2 + 3nr + r^2)</math> is an integer, we must have that <math>3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000</math>.  This means that the smallest possible <math>n</math> should be <ins class="diffchange diffchange-inline">less </ins>than 1000<ins class="diffchange diffchange-inline">. (It is more evident once you plug in r=1/1000. Do you see why?)</ins>. In particular, <math>3nr + r^2</math> should be less than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>.  <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>.  Since we want to minimize <math>n</math>, we take <math>n = 19</math>.  Then for any positive value of <math>r</math>, <math>3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000</math>, so it is possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>.  However, we still have to make sure a sufficiently small <math>r</math> exists.   </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>In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to ensure a small enough <math>r</math>.  The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>.  Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set.  The answer is  <math>\boxed{019}</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>In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to ensure a small enough <math>r</math>.  The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>.  Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set.  The answer is  <math>\boxed{019}</math>.</div></td></tr>
</table>
Brackie1331
https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_12&diff=178803&oldid=prev
Resources: /* Solution 2 */
2022-10-06T21:41:57Z
<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 21:41, 6 October 2022</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l18" >Line 18:</td>
<td colspan="2" class="diff-lineno">Line 18:</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>Why is it <math>(n + \frac{1}{1000})^3 - n^3 > 1</math> and not greater than or equal to? - awesomediabrine</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>Why is it <math>(n + \frac{1}{1000})^3 - n^3 > 1</math> and not greater than or equal to? - awesomediabrine</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;">Because if its equal to, then there is no integer in between the two values. - resources</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>== Solution 3 (Similar to 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 3 (Similar to Solution 2) ==</div></td></tr>
</table>
Resources
https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_12&diff=168979&oldid=prev
Hyprox1413: /* Solution 4 (Calculus) */ n to $n$
2021-12-31T19:41:13Z
<p><span dir="auto"><span class="autocomment">Solution 4 (Calculus): </span> n to $n$</span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 19:41, 31 December 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;"><div>and</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>and</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><cmath>\sqrt[3]{n^3+1} - n > \left.\frac{d\sqrt[3]{x}}{dx}\right|_{x=n^3+1} = \frac{1}{3\sqrt[3]{(n^3+1)^2}}</cmath></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><cmath>\sqrt[3]{n^3+1} - n > \left.\frac{d\sqrt[3]{x}}{dx}\right|_{x=n^3+1} = \frac{1}{3\sqrt[3]{(n^3+1)^2}}</cmath></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>From this, it is clear that <math>n = \boxed{019}</math> is the smallest n for which LHS will be less than <math>\frac{1}{1000}</math>. ~ Hyprox1413</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>From this, it is clear that <math>n = \boxed{019}</math> is the smallest <ins class="diffchange diffchange-inline"><math></ins>n<ins class="diffchange diffchange-inline"></math> </ins>for which LHS will be less than <math>\frac{1}{1000}</math>. ~ Hyprox1413</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>
</table>
Hyprox1413
https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_12&diff=168978&oldid=prev
Hyprox1413: added solution with calculus
2021-12-31T19:38:34Z
<p>added solution with calculus</p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 19:38, 31 December 2021</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>Since <math>r</math> is less than <math>1/1000</math>, we have <math>\sqrt[3]{m} < n + \frac{1}{1000}</math>. Notice that since we want <math>m</math> minimized, <math>n</math> should also be minimized. Also, <math>n^3</math> should be as close as possible, but not exceeding <math>m</math>. This means <math>m</math> should be set to <math>n^3+1</math>. Substituting and simplifying, we get <cmath>\sqrt[3]{n^3+1} < n + \frac{1}{1000}</cmath> <cmath>n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}</cmath></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>Since <math>r</math> is less than <math>1/1000</math>, we have <math>\sqrt[3]{m} < n + \frac{1}{1000}</math>. Notice that since we want <math>m</math> minimized, <math>n</math> should also be minimized. Also, <math>n^3</math> should be as close as possible, but not exceeding <math>m</math>. This means <math>m</math> should be set to <math>n^3+1</math>. Substituting and simplifying, we get <cmath>\sqrt[3]{n^3+1} < n + \frac{1}{1000}</cmath> <cmath>n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}</cmath></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>The last two terms in the right side can be ignored in the calculation because they are too small. This results in <math>1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}</math>. The minimum positive integer <math>n</math> that satisfies this is <math>\boxed{019}</math>. ~ Hb10</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>The last two terms in the right side can be ignored in the calculation because they are too small. This results in <math>1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}</math>. The minimum positive integer <math>n</math> that satisfies this is <math>\boxed{019}</math>. ~ Hb10</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;">== Solution 4 (Calculus) ==</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;">Note that the cube root is increasing for positive reals while its derivative is decreasing, so linear approximation gives</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;"><cmath>\sqrt[3]{n^3+1} - n < \left.\frac{d\sqrt[3]{x}}{dx}\right|_{x=n^3} = \frac{1}{3n^2}</cmath></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;">and</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;"><cmath>\sqrt[3]{n^3+1} - n > \left.\frac{d\sqrt[3]{x}}{dx}\right|_{x=n^3+1} = \frac{1}{3\sqrt[3]{(n^3+1)^2}}</cmath></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;">From this, it is clear that <math>n = \boxed{019}</math> is the smallest n for which LHS will be less than <math>\frac{1}{1000}</math>. ~ Hyprox1413</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>== 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>
</table>
Hyprox1413
https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_12&diff=157147&oldid=prev
Awesomediabrine: /* Solution 2 */
2021-07-01T18:06:17Z
<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 18:06, 1 July 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l17" >Line 17:</td>
<td colspan="2" class="diff-lineno">Line 17:</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>Trying values of <math>n</math>, we see that the smallest value of <math>n</math> that works is <math>\boxed{019}</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>Trying values of <math>n</math>, we see that the smallest value of <math>n</math> that works is <math>\boxed{019}</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>Why is it <math>(n + \frac{1}{1000})^3 - n^3 > 1</math> and not <del class="diffchange diffchange-inline">greator </del>than or equal to? - awesomediabrine</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>Why is it <math>(n + \frac{1}{1000})^3 - n^3 > 1</math> and not <ins class="diffchange diffchange-inline">greater </ins>than or equal to? - awesomediabrine</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 3 (Similar to 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 3 (Similar to Solution 2) ==</div></td></tr>
</table>
Awesomediabrine
https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_12&diff=157146&oldid=prev
Awesomediabrine: /* Solution 2 */
2021-07-01T18:05:48Z
<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 18:05, 1 July 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l16" >Line 16:</td>
<td colspan="2" class="diff-lineno">Line 16:</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>Trying values of <math>n</math>, we see that the smallest value of <math>n</math> that works is <math>\boxed{019}</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>Trying values of <math>n</math>, we see that the smallest value of <math>n</math> that works is <math>\boxed{019}</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><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;">Why is it <math>(n + \frac{1}{1000})^3 - n^3 > 1</math> and not greator than or equal to? - awesomediabrine</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>== Solution 3 (Similar to 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 3 (Similar to Solution 2) ==</div></td></tr>
</table>
Awesomediabrine
https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_12&diff=135804&oldid=prev
Hb10: /* Solution 3 (Similar to Solution 2) */
2020-10-25T21:07:55Z
<p><span dir="auto"><span class="autocomment">Solution 3 (Similar to 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 21:07, 25 October 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l19" >Line 19:</td>
<td colspan="2" class="diff-lineno">Line 19:</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 3 (Similar to 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 3 (Similar to Solution 2) ==</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>Since <math>r</math> is less than <math>1/1000</math>, we have <math>\sqrt[3]{m} < n + \frac{1}{1000}</math>. Notice that since we want <math>m</math> minimized, <math>n</math> should also be minimized. Also, <math>n^3</math> should be as close as possible, but not exceeding <math>m</math>. This means <math>m</math> should be set to <math>n^3+1</math>. Substituting and simplifying, we get <cmath>\sqrt[3]{n^3+1} < n + \frac{1}{1000}</cmath> <cmath>n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}</cmath></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>Since <math>r</math> is less than <math>1/1000</math>, we have <math>\sqrt[3]{m} < n + \frac{1}{1000}</math>. Notice that since we want <math>m</math> minimized, <math>n</math> should also be minimized. Also, <math>n^3</math> should be as close as possible, but not exceeding <math>m</math>. This means <math>m</math> should be set to <math>n^3+1</math>. Substituting and simplifying, we get <cmath>\sqrt[3]{n^3+1} < n + \frac{1}{1000}</cmath> <cmath>n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}</cmath></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><del class="diffchange diffchange-inline">Note that the </del>last two terms in the right side can be ignored in the calculation because they are too small. This results in <math>1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}</math>. The minimum positive integer <math>n</math> that satisfies this is <math>\boxed{019}</math>. ~ Hb10</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">The </ins>last two terms in the right side can be ignored in the calculation because they are too small. This results in <math>1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}</math>. The minimum positive integer <math>n</math> that satisfies this is <math>\boxed{019}</math>. ~ Hb10</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>
</table>
Hb10
https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_12&diff=135803&oldid=prev
Hb10 at 21:07, 25 October 2020
2020-10-25T21:07:21Z
<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<col class="diff-marker" />
<col class="diff-content" />
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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 21:07, 25 October 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l18" >Line 18:</td>
<td colspan="2" class="diff-lineno">Line 18:</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 3 (Similar to 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 3 (Similar to 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>Since <math>r</math> is less than <math>1/1000</math>, we have <math>\sqrt[3]{m} < n + \frac{1}{1000}</math>. Notice that since we want <math>m</math> minimized, <math>n</math> should also be minimized. Also, <math>n^3</math> should be as close as possible <del class="diffchange diffchange-inline">to </del><math>m</math>. This means <math>m</math> should be set to <math>n^3+1</math>. Substituting and simplifying, we get <cmath>\sqrt[3]{n^3+1} < n + \frac{1}{1000}</cmath> <cmath>n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}</cmath></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>Since <math>r</math> is less than <math>1/1000</math>, we have <math>\sqrt[3]{m} < n + \frac{1}{1000}</math>. Notice that since we want <math>m</math> minimized, <math>n</math> should also be minimized. Also, <math>n^3</math> should be as close as possible<ins class="diffchange diffchange-inline">, but not exceeding </ins><math>m</math>. This means <math>m</math> should be set to <math>n^3+1</math>. Substituting and simplifying, we get <cmath>\sqrt[3]{n^3+1} < n + \frac{1}{1000}</cmath> <cmath>n^3+1 < n^3+\frac{3}{1000}n^2+\frac{3}{1000^2}n+\frac{1}{1000^3}</cmath></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>Note that the last two terms in the right side can be ignored in the calculation because they are too small. This results in <math>1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}</math>. The minimum positive integer <math>n</math> that satisfies this is <math>\boxed{019}</math>. ~ Hb10</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>Note that the last two terms in the right side can be ignored in the calculation because they are too small. This results in <math>1 < \frac{3}{1000}n^2 \Rightarrow n^2 > \frac{1000}{3}</math>. The minimum positive integer <math>n</math> that satisfies this is <math>\boxed{019}</math>. ~ Hb10</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>
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Hb10
https://artofproblemsolving.com/wiki/index.php?title=1987_AIME_Problems/Problem_12&diff=129012&oldid=prev
N828335: /* Solution 1 */
2020-07-24T02:20:04Z
<p><span dir="auto"><span class="autocomment">Solution 1</span></span></p>
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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 02:20, 24 July 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l6" >Line 6:</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><math>m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3</math>.  Since <math>r < \frac{1}{1000}</math> and <math>m - n^3 = r(3n^2 + 3nr + r^2)</math> is an integer, we must have that <math>3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000</math>.  This means that the smallest possible <math>n</math> should be quite a bit smaller than 1000. In particular, <math>3nr + r^2</math> should be less than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>.  <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>.  Since we want to minimize <math>n</math>, we take <math>n = 19</math>.  Then for any positive value of <math>r</math>, <math>3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000</math>, so it is possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>.  However, we still have to make sure a sufficiently small <math>r</math> exists.   </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>m = (n + r)^3 = n^3 + 3n^2r + 3nr^2 + r^3</math>.  Since <math>r < \frac{1}{1000}</math> and <math>m - n^3 = r(3n^2 + 3nr + r^2)</math> is an integer, we must have that <math>3n^2 + 3nr + r^2 \geq \frac{1}{r} > 1000</math>.  This means that the smallest possible <math>n</math> should be quite a bit smaller than 1000. In particular, <math>3nr + r^2</math> should be less than 1, so <math>3n^2 > 999</math> and <math>n > \sqrt{333}</math>.  <math>18^2 = 324 < 333 < 361 = 19^2</math>, so we must have <math>n \geq 19</math>.  Since we want to minimize <math>n</math>, we take <math>n = 19</math>.  Then for any positive value of <math>r</math>, <math>3n^2 + 3nr + r^2 > 3\cdot 19^2 > 1000</math>, so it is possible for <math>r</math> to be less than <math>\frac{1}{1000}</math>.  However, we still have to make sure a sufficiently small <math>r</math> exists.   </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>In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to ensure a small enough <math>r</math>.  The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>.  Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set.  The answer is  <math>019</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>In light of the equation <math>m - n^3 = r(3n^2 + 3nr + r^2)</math>, we need to choose <math>m - n^3</math> as small as possible to ensure a small enough <math>r</math>.  The smallest possible value for <math>m - n^3</math> is 1, when <math>m = 19^3 + 1</math>.  Then for this value of <math>m</math>, <math>r = \frac{1}{3n^2 + 3nr + r^2} < \frac{1}{1000}</math>, and we're set.  The answer is  <math><ins class="diffchange diffchange-inline">\boxed{</ins>019<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>== 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>
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N828335