https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&feed=atom&action=history2011 AIME I Problems/Problem 7 - Revision history2024-03-29T08:41:29ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=211408&oldid=prevSosiaops: /* Solution 1 */ fixed more latex2024-01-20T01:12:16Z<p><span dir="auto"><span class="autocomment">Solution 1: </span> fixed more latex</span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></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 1==</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 1==</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><math> m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}</math>. Now, divide by <math>m^{x_0}</math> to get <math>1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}</math>. Notice that since we can choose all nonnegative <math>x_0,...,x_{2011}</math>, we can make <math>x_n-x_0</math> whatever we desire. WLOG, let <math>x_0\geq...\geq x_{2011}</math> and let <math>a_n=x_n-x_0</math>.  Notice that, also, <math>m^{a_{2011}}</math> doesn't matter if we are able to make <math> m^{a_1} +m^{a_2} + .... + m^{a_{2010}}</math> equal to <math>1-\left(\frac{1}{m}\right)^x</math> for any power of <math>x</math>. Consider <math>m=2</math>. We can achieve a sum of <math>1-\left(\frac{1}{2}\right)^x</math> by doing <math>\frac{1}{2}+\frac{1}{4}+...</math> (the "simplest" sequence). If we don't have <math>\frac{1}{2}</math>, to compensate, we need <math>2\cdot 1\frac{1}{4}</math>'s. Now, let's try to generalize. The "simplest" sequence is having <math>\frac{1}{m}</math> <math>m-1</math> times, <math>\frac{1}{m^2}</math> <math>m-1</math> times, <math>\ldots</math>. To make other sequences, we can split <math>m-1</math> <math>\frac{1}{m^i}</math>s into <math>m(m-1)</math> <math>\text{ }\frac{1}{m^{i+1}}</math>s since <math>(m-1)\cdot\frac{1}{m^{i}} = m(m-1)\cdot\frac{1}{m^{i+1}}</math>. Since we want <math>2010</math> terms, we have <math>\sum</math> <math>(m-1)\cdot m^x=2010</math>. However, since we can set <math>x</math> to be anything we want (including 0), all we care about is that <math>m-1 | 2010</math> which happens <math>\boxed{016}</math> times.</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^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}</math>. Now, divide by <math>m^{x_0}</math> to get <math>1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}</math>. Notice that since we can choose all nonnegative <math>x_0,...,x_{2011}</math>, we can make <math>x_n-x_0</math> whatever we desire. WLOG, let <math>x_0\geq...\geq x_{2011}</math> and let <math>a_n=x_n-x_0</math>.  Notice that, also, <math>m^{a_{2011}}</math> doesn't matter if we are able to make <math> m^{a_1} +m^{a_2} + .... + m^{a_{2010}}</math> equal to <math>1-\left(\frac{1}{m}\right)^x</math> for any power of <math>x</math>. Consider <math>m=2</math>. We can achieve a sum of <math>1-\left(\frac{1}{2}\right)^x</math> by doing <math>\frac{1}{2}+\frac{1}{4}+...</math> (the "simplest" sequence). If we don't have <math>\frac{1}{2}</math>, to compensate, we need <math>2\cdot 1\frac{1}{4}</math>'s. Now, let's try to generalize. The "simplest" sequence is having <math>\frac{1}{m}</math> <math>m-1</math> times, <math>\frac{1}{m^2}</math> <math>m-1</math> times, <math>\ldots</math>. To make other sequences, we can split <math>m-1</math> <math>\frac{1}{m^i}</math>s into <math>m(m-1)</math> <math>\text{ }\frac{1}{m^{i+1}}</math>s since <math>(m-1)\cdot\frac{1}{m^{i}} = m(m-1)\cdot\frac{1}{m^{i+1}}</math>. Since we want <math>2010</math> terms, we have <math>\sum</math> <math>(m-1)\cdot m^x=2010</math>. However, since we can set <math>x</math> to be anything we want (including 0), all we care about is that <math><ins class="diffchange diffchange-inline">(</ins>m-1<ins class="diffchange diffchange-inline">)\text{ }</ins>|<ins class="diffchange diffchange-inline">\text{ } </ins>2010</math> which happens <math>\boxed{016}</math> times.</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>
</table>Sosiaopshttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=211407&oldid=prevSosiaops: /* Solution 1 */ cleaned up latex and fixed terms that got swapped accidentally2024-01-20T01:11:02Z<p><span dir="auto"><span class="autocomment">Solution 1: </span> cleaned up latex and fixed terms that got swapped accidentally</span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 01:11, 20 January 2024</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;"></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 1==</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 1==</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><math> m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}</math>. Now, divide by <math>m^{x_0}</math> to get <math>1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}</math>. Notice that since we can choose all nonnegative <math>x_0,...,x_{2011}</math>, we can make <math>x_n-x_0</math> whatever we desire. WLOG, let <math>x_0\geq...\geq x_{2011}</math> and let <math>a_n=x_n-x_0</math>.  Notice that, also, <math>m^{a_{2011}}</math> doesn't matter if we are able to make <math> m^{a_1} +m^{a_2} + .... + m^{a_{2010}}</math> equal to <math>1-\left(\frac{1}{m}\right)^x</math> for any power of <math>x</math>. Consider <math>m=2</math>. We can achieve a sum of <math>1-\left(\frac{1}{2}\right)^x</math> by doing <math>\frac{1}{2}+\frac{1}{4}+...</math> (the "simplest" sequence). If we don't have <math>\frac{1}{2}</math>, to compensate, we need <math>2\cdot 1\frac{1}{4}</math>'s. Now, let's try to generalize. The "simplest" sequence is having <math>\frac{1}{m}</math> <math>m-1</math> times, <math>\frac{1}{m^2}</math> <math>m-1</math> times, <math>\ldots</math>. To make other sequences, we can split <math>m-1</math> <math>\frac{1}{m^i}</math>s into <math>m(m-1)</math> <math>\frac{1}{m^{i+1}}</math>s since <math>m\cdot\frac{1}{m^{i<del class="diffchange diffchange-inline">+1</del>}}<del class="diffchange diffchange-inline">\cdot </del>=m(m-1)\cdot\frac{1}{m^{i}}</math>. Since we want <math>2010</math> terms, we have <math>\sum</math> <math>(m-1)\cdot m^x=2010</math>. However, since we can set <math>x</math> to be anything we want (including 0), all we care about is that <math>m-1 | 2010</math> which happens <math>\boxed{016}</math> times.</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^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}</math>. Now, divide by <math>m^{x_0}</math> to get <math>1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}</math>. Notice that since we can choose all nonnegative <math>x_0,...,x_{2011}</math>, we can make <math>x_n-x_0</math> whatever we desire. WLOG, let <math>x_0\geq...\geq x_{2011}</math> and let <math>a_n=x_n-x_0</math>.  Notice that, also, <math>m^{a_{2011}}</math> doesn't matter if we are able to make <math> m^{a_1} +m^{a_2} + .... + m^{a_{2010}}</math> equal to <math>1-\left(\frac{1}{m}\right)^x</math> for any power of <math>x</math>. Consider <math>m=2</math>. We can achieve a sum of <math>1-\left(\frac{1}{2}\right)^x</math> by doing <math>\frac{1}{2}+\frac{1}{4}+...</math> (the "simplest" sequence). If we don't have <math>\frac{1}{2}</math>, to compensate, we need <math>2\cdot 1\frac{1}{4}</math>'s. Now, let's try to generalize. The "simplest" sequence is having <math>\frac{1}{m}</math> <math>m-1</math> times, <math>\frac{1}{m^2}</math> <math>m-1</math> times, <math>\ldots</math>. To make other sequences, we can split <math>m-1</math> <math>\frac{1}{m^i}</math>s into <math>m(m-1)</math> <math><ins class="diffchange diffchange-inline">\text{ }</ins>\frac{1}{m^{i+1}}</math>s since <math><ins class="diffchange diffchange-inline">(</ins>m<ins class="diffchange diffchange-inline">-1)</ins>\cdot\frac{1}{m^{i}} = m(m-1)\cdot\frac{1}{m^{i<ins class="diffchange diffchange-inline">+1</ins>}}</math>. Since we want <math>2010</math> terms, we have <math>\sum</math> <math>(m-1)\cdot m^x=2010</math>. However, since we can set <math>x</math> to be anything we want (including 0), all we care about is that <math>m-1 | 2010</math> which happens <math>\boxed{016}</math> times.</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>
</table>Sosiaopshttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=205730&oldid=prevTurtle: /* An Olympiad Problem that's (almost) the exact same (and it came before 2011, MAA) */2023-11-25T16:30:23Z<p><span dir="auto"><span class="autocomment">An Olympiad Problem that's (almost) the exact same (and it came before 2011, MAA)</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:30, 25 November 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l50" >Line 50:</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>2001 Austrian-Polish Math Individual Competition #1</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>2001 Austrian-Polish Math Individual Competition #1</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>~MSC</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>~MSC</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 style="font-weight: bold; text-decoration: none;">Skul</del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== 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>Turtlehttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=205729&oldid=prevTurtle: /* An Olympiad Problem that's (almost) the exact same (and it came before 2011, MAA) */2023-11-25T16:30:10Z<p><span dir="auto"><span class="autocomment">An Olympiad Problem that's (almost) the exact same (and it came before 2011, MAA)</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:30, 25 November 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l50" >Line 50:</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>2001 Austrian-Polish Math Individual Competition #1</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>2001 Austrian-Polish Math Individual Competition #1</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>~MSC</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>~MSC</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;">Skul</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>Turtlehttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=187953&oldid=prevHeheman at 20:10, 25 January 20232023-01-25T20:10:13Z<p></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>What is meant by <math>m</math>-chopping is taking an existing block of say <math>m^{k}</math> and turning it into <math>m</math> blocks of <math>m^{k-1}</math>. This process increases the total number of blocks by <math>m-1</math> per chop.</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>What is meant by <math>m</math>-chopping is taking an existing block of say <math>m^{k}</math> and turning it into <math>m</math> blocks of <math>m^{k-1}</math>. This process increases the total number of blocks by <math>m-1</math> per chop.</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 problem wants us to find the number of positive integers <math>m</math> where some number of chops will turn <math>1</math> block into <math>2011</math> such blocks, thus increasing the total amount by <math>2010= 2 \cdot 3 \cdot 5 \cdot 67</math>. Thus <math>m-1 | 2010</math>, and a cursory check on extreme cases will confirm that there are indeed <math>\boxed{016}</math> possible <math>m</math>s.</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 problem wants us to find the number of positive integers <math>m</math> where some number of chops will turn <math>1</math> block into <math>2011</math> such blocks, thus increasing the total amount by <math>2010= 2 \cdot 3 \cdot 5 \cdot 67</math>. Thus <math>m-1 | 2010</math>, and a cursory check on extreme cases will confirm that there are indeed <math>\boxed{016}</math> possible <math>m</math>s.</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 6 (fast, easy)==</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;">The problem is basically saying that we want to find 2011 powers of m that add together to equal another power of m. If we had a power of m, m^n, then to get to m^(n+1) or m*(m^n), we have to add m^n, m-1 times. Then when we are at m^(n+1), to get to m^(n+2), it is similar. So we have to have m^(some number) = m^n + (m-1)(m^n) + (m-1)(m^(n+1))...<=This expression has 1 + (m-1) + (m-1) + ... = 1 + k(m-1) terms, and the number of terms must be equal to 2011 (the problem said). Then 1+k(m-1) = 2011, and we get the equation k*(m-1) = 2010 = 2*3*5*67. 2010 has 16 factors, and setting m-1 = each of these factors makes <math>\boxed{016}</math> values of m.</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>==An Olympiad Problem that's (almost) the exact same (and it came before 2011, MAA)==</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>==An Olympiad Problem that's (almost) the exact same (and it came before 2011, MAA)==</div></td></tr>
</table>Hehemanhttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=161128&oldid=prevJdong2006: /* Solution 2 */2021-08-29T00:11:11Z<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 00:11, 29 August 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;"></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>Let <<del class="diffchange diffchange-inline">math</del>>P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}</<del class="diffchange diffchange-inline">math</del>>. The problem then becomes finding the number of positive integer roots <math>m</math> for which <math>P(m) = 0</math> and <math>x_0, x_1, ..., x_{2011}</math> are nonnegative integers. We plug in <math>m = 1</math> and see that <<del class="diffchange diffchange-inline">math</del>>P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010</<del class="diffchange diffchange-inline">math</del>><del class="diffchange diffchange-inline">. </del>Now, we can say that <<del class="diffchange diffchange-inline">math</del>>P(m) = (m-1)Q(m) - 2010</<del class="diffchange diffchange-inline">math</del>> for some polynomial <math>Q(m)</math> with integer coefficients. Then if <math>P(m) = 0</math>, <math>(m-1)Q(m) = 2010</math>. Thus, if <math>P(m) = 0</math>, then <math>m-1 | 2010</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>Let <<ins class="diffchange diffchange-inline">cmath</ins>>P(m) = m^{x_0} - m^{x_1} -m^{x_2} - .... - m^{x_{2011}}</<ins class="diffchange diffchange-inline">cmath</ins>>. The problem then becomes finding the number of positive integer roots <math>m</math> for which <math>P(m) = 0</math> and <math>x_0, x_1, ..., x_{2011}</math> are nonnegative integers. We plug in <math>m = 1</math> and see that <<ins class="diffchange diffchange-inline">cmath</ins>>P(1) = 1 - 1 - 1... -1 = 1-2011 = -2010</<ins class="diffchange diffchange-inline">cmath</ins>> Now, we can say that <<ins class="diffchange diffchange-inline">cmath</ins>>P(m) = (m-1)Q(m) - 2010</<ins class="diffchange diffchange-inline">cmath</ins>> for some polynomial <math>Q(m)</math> with integer coefficients. Then if <math>P(m) = 0</math>, <math>(m-1)Q(m) = 2010</math>. Thus, if <math>P(m) = 0</math>, then <math>m-1 | 2010</math> .</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>Now, we need to show that <del class="diffchange diffchange-inline">for all </del><<del class="diffchange diffchange-inline">math>m-1 | 2010</math>, <math</del>> m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}<del class="diffchange diffchange-inline">. </del></<del class="diffchange diffchange-inline">math</del>><del class="diffchange diffchange-inline">. </del>We try with the first few <math>m</math> that satisfy this.</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>Now, we need to show that <<ins class="diffchange diffchange-inline">cmath</ins>>m^{x_{0}}=\sum_{k = 1}^{2011}m^{x_{k}}<ins class="diffchange diffchange-inline">\forall m-1|2011 </ins></<ins class="diffchange diffchange-inline">cmath</ins>> We try with the first few <math>m</math> that satisfy this.</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>For <math>m = 2</math>, we see we can satisfy this if <math>x_0 = 2010</math>, <math>x_1 = 2009</math>, <math>x_2 = 2008</math>, <math>\cdots</math> , <math>x_{2008} = 2</math>, <math>x_{2009} = 1</math>, <math> x_{2010} = 0</math>, <math>x_{2011} = 0</math>, because <<del class="diffchange diffchange-inline">math</del>>2^{2009} + 2^{2008} + \cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \cdots + 2^1 + 2^1 = \cdots</<del class="diffchange diffchange-inline">math</del>> (based on the idea <math>2^n + 2^n = 2^{n+1}</math>, leading to a chain of substitutions of this kind) <<del class="diffchange diffchange-inline">math</del>>= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}</<del class="diffchange diffchange-inline">math</del>>. Thus <math>2</math> is a possible value of <math>m</math>. For other values, for example <math>m = 3</math>, we can use the same strategy, with <<del class="diffchange diffchange-inline">math</del>>x_{2011} = x_{2010} = x_{2009} = 0</<del class="diffchange diffchange-inline">math</del>>, <<del class="diffchange diffchange-inline">math</del>>x_{2008} = x_{2007} = 1</<del class="diffchange diffchange-inline">math</del>>, <<del class="diffchange diffchange-inline">math</del>>x_{2006} = x_{2005} = 2</<del class="diffchange diffchange-inline">math</del>>, <<del class="diffchange diffchange-inline">math</del>>\cdots</<del class="diffchange diffchange-inline">math</del>>, <<del class="diffchange diffchange-inline">math</del>>x_2 = x_1 = 1004</<del class="diffchange diffchange-inline">math</del>> and <<del class="diffchange diffchange-inline">math</del>>x_0 = 1005</<del class="diffchange diffchange-inline">math</del>>, because  </div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>For <math>m = 2</math>, we see we can satisfy this if <math>x_0 = 2010</math>, <math>x_1 = 2009</math>, <math>x_2 = 2008</math>, <math>\cdots</math> , <math>x_{2008} = 2</math>, <math>x_{2009} = 1</math>, <math> x_{2010} = 0</math>, <math>x_{2011} = 0</math>, because <<ins class="diffchange diffchange-inline">cmath</ins>>2^{2009} + 2^{2008} + \cdots + 2^1 + 2^0 +2^ 0 = 2^{2009} + 2^{2008} + \cdots + 2^1 + 2^1 = \cdots</<ins class="diffchange diffchange-inline">cmath</ins>> (based on the idea <math>2^n + 2^n = 2^{n+1}</math>, leading to a chain of substitutions of this kind) <<ins class="diffchange diffchange-inline">cmath</ins>>= 2^{2009} + 2^{2008} + 2^{2008} = 2^{2009} + 2^{2009} = 2^{2010}</<ins class="diffchange diffchange-inline">cmath</ins>>. Thus <math>2</math> is a possible value of <math>m</math>. For other values, for example <math>m = 3</math>, we can use the same strategy, with <<ins class="diffchange diffchange-inline">cmath</ins>>x_{2011} = x_{2010} = x_{2009} = 0</<ins class="diffchange diffchange-inline">cmath</ins>>, <<ins class="diffchange diffchange-inline">cmath</ins>>x_{2008} = x_{2007} = 1</<ins class="diffchange diffchange-inline">cmath</ins>>, <<ins class="diffchange diffchange-inline">cmath</ins>>x_{2006} = x_{2005} = 2</<ins class="diffchange diffchange-inline">cmath</ins>>, <<ins class="diffchange diffchange-inline">cmath</ins>>\cdots</<ins class="diffchange diffchange-inline">cmath</ins>>, <<ins class="diffchange diffchange-inline">cmath</ins>>x_2 = x_1 = 1004</<ins class="diffchange diffchange-inline">cmath</ins>> and <<ins class="diffchange diffchange-inline">cmath</ins>>x_0 = 1005</<ins class="diffchange diffchange-inline">cmath</ins>>, because  </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">math</del>>3^0 + 3^0 + 3^0 +3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^1+3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^2+3^2+3^2+\cdots+3^{1004} +3^{1004} <del class="diffchange diffchange-inline">= \cdots</del></<del class="diffchange diffchange-inline">math</del>></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><<ins class="diffchange diffchange-inline">cmath</ins>>3^0 + 3^0 + 3^0 +3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^1+3^1+3^1+3^2+3^2+\cdots+3^{1004} +3^{1004} = 3^2+3^2+3^2+\cdots+3^{1004} +3^{1004}</<ins class="diffchange diffchange-inline">cmath</ins>></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><<del class="diffchange diffchange-inline">math</del>>=3^{1004} +3^{1004}+3^{1004} = 3^{1005}</<del class="diffchange diffchange-inline">math</del>><del class="diffchange diffchange-inline">.</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><<ins class="diffchange diffchange-inline">cmath</ins>><ins class="diffchange diffchange-inline">=\cdots </ins>= 3^{1004} +3^{1004}+3^{1004} = 3^{1005}</<ins class="diffchange diffchange-inline">cmath</ins>></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>It's clearly seen we can use the same strategy for all <math>m-1 |2010</math>. We count all positive <math>m</math> satisfying <math>m-1 |2010</math>, and see there are <math>\boxed{016}</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>It's clearly seen we can use the same strategy for all <math>m-1 |2010</math>. We count all positive <math>m</math> satisfying <math>m-1 |2010</math>, and see there are <math>\boxed{016}</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>
</table>Jdong2006https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=132127&oldid=prevAops turtle: /* Solution 5 */2020-08-19T02:16:00Z<p><span dir="auto"><span class="autocomment">Solution 5</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 02:16, 19 August 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l40" >Line 40:</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>First of all, note that the nonnegative integer condition really does not matter, since even if we have a nonnegative power, there is always a power of <math>m</math> we can multiply to get to non-negative powers.</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>First of all, note that the nonnegative integer condition really does not matter, since even if we have a nonnegative power, there is always a power of <math>m</math> we can multiply to get to non-negative powers.</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>Now we see that our problem is just a matter of m-chopping blocks.</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>Now we see that our problem is just a matter of m-chopping blocks.</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>What is meant by <math>m</math>-chopping is taking an existing block of say <math>m^{k}</math> and turning it into <math>m</math> blocks of <math>m^{k-1}</math>. This process increases the total <del class="diffchange diffchange-inline"># </del>of blocks by <math>m-1</math> per chop.</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>What is meant by <math>m</math>-chopping is taking an existing block of say <math>m^{k}</math> and turning it into <math>m</math> blocks of <math>m^{k-1}</math>. This process increases the total <ins class="diffchange diffchange-inline">number </ins>of blocks by <math>m-1</math> per chop.</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 problem wants us to find the number of positive integers <math>m</math> where some number of chops will turn <math>1</math> block into <math>2011</math> such blocks, thus increasing the total amount by <math>2010= 2 \cdot 3 \cdot 5 \cdot 67</math>. Thus <math>m-1 | 2010</math>, and a cursory check on extreme cases will confirm that there are indeed <math>\boxed{016}</math> possible <math>m</math>s.</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 problem wants us to find the number of positive integers <math>m</math> where some number of chops will turn <math>1</math> block into <math>2011</math> such blocks, thus increasing the total amount by <math>2010= 2 \cdot 3 \cdot 5 \cdot 67</math>. Thus <math>m-1 | 2010</math>, and a cursory check on extreme cases will confirm that there are indeed <math>\boxed{016}</math> possible <math>m</math>s.</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>
</table>Aops turtlehttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=132126&oldid=prevAops turtle: /* Solution 5 */2020-08-19T02:11:30Z<p><span dir="auto"><span class="autocomment">Solution 5</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 02:11, 19 August 2020</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;"></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>First of all, note that the nonnegative integer condition really does not matter, since even if we have a nonnegative power, there is always a power of <math>m</math> we can multiply to get to non-negative powers.</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>First of all, note that the nonnegative integer condition really does not matter, since even if we have a nonnegative power, there is always a power of <math>m</math> we can multiply to get to non-negative powers.</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>Now we see that our problem is just a matter of <del class="diffchange diffchange-inline"><math>"</del>m<del class="diffchange diffchange-inline">"</math></del>-chopping blocks.</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>Now we see that our problem is just a matter of m-chopping blocks.</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>What is meant by <math>m</math>-chopping is taking an existing block of say <math>m^{k}</math> and turning it into <math>m</math> blocks of <math>m^{k-1}</math>. This process increases the total # of blocks by <math>m-1</math> per chop.</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>What is meant by <math>m</math>-chopping is taking an existing block of say <math>m^{k}</math> and turning it into <math>m</math> blocks of <math>m^{k-1}</math>. This process increases the total # of blocks by <math>m-1</math> per chop.</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 problem wants us to find the number of positive integers <math>m</math> where some number of chops will turn <math>1</math> block into <math>2011</math> such blocks, thus increasing the total amount by <math>2010= 2 \cdot 3 \cdot 5 \cdot 67</math>. Thus <math>m-1 | 2010</math>, and a cursory check on extreme cases will confirm that there are indeed <math>\boxed{016}</math> possible <math>m</math>s.</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 problem wants us to find the number of positive integers <math>m</math> where some number of chops will turn <math>1</math> block into <math>2011</math> such blocks, thus increasing the total amount by <math>2010= 2 \cdot 3 \cdot 5 \cdot 67</math>. Thus <math>m-1 | 2010</math>, and a cursory check on extreme cases will confirm that there are indeed <math>\boxed{016}</math> possible <math>m</math>s.</div></td></tr>
</table>Aops turtlehttps://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=127462&oldid=prevHi13: /* Solution 1 */2020-07-04T19:14:57Z<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;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 19:14, 4 July 2020</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;"></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 1==</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 1==</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><math> m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}</math>. Now, divide by <math>m^{x_0}</math> to get <math>1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}</math>. Notice that since we can choose all nonnegative <math>x_0,...,x_{<del class="diffchange diffchange-inline">2010</del>}</math>, we can make <math>x_n-x_0</math> whatever we desire. WLOG, let <math>x_0\geq...\geq x_{2011}</math> and let <math>a_n=x_n-x_0</math>.  Notice that, also, <math>m^{a_{2011}}</math> doesn't matter if we are able to make <math> m^{a_1} +m^{a_2} + .... + m^{a_{<del class="diffchange diffchange-inline">2011</del>}}</math> equal to <math>1-\left(\frac{1}{m}\right)^x</math> for any power of <math>x</math>. Consider <math>m=2</math>. We can achieve a sum of <math>1-\left(\frac{1}{2}\right)^x</math> by doing <math>\frac{1}{2}+\frac{1}{4}+...</math> (the "simplest" sequence). If we don't have <math>\frac{1}{2}</math>, to compensate, we need <math>2\cdot 1\frac{1}{4}</math>'s. Now, let's try to generalize. The "simplest" sequence is having <math>\frac{1}{m}</math> <math>m-1</math> times, <math>\frac{1}{m^2}</math> <math>m-1</math> times, <math>\ldots</math>. To make other sequences, we can split <math>m-1</math> <math>\frac{1}{m^i}</math>s into <math>m(m-1)</math> <math>\frac{1}{m^{i+1}}</math>s since <math>m\cdot\frac{1}{m^{i+1}}\cdot =m(m-1)\cdot\frac{1}{m^{i}}</math>. Since we want <math>2010</math> terms, we have <math>\sum</math> <math>(m-1)\cdot m^x=2010</math>. However, since we can set <math>x</math> to be anything we want (including 0), all we care about is that <math>m-1 | 2010</math> which happens <math>\boxed{016}</math> times.</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^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}</math>. Now, divide by <math>m^{x_0}</math> to get <math>1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}</math>. Notice that since we can choose all nonnegative <math>x_0,...,x_{<ins class="diffchange diffchange-inline">2011</ins>}</math>, we can make <math>x_n-x_0</math> whatever we desire. WLOG, let <math>x_0\geq...\geq x_{2011}</math> and let <math>a_n=x_n-x_0</math>.  Notice that, also, <math>m^{a_{2011}}</math> doesn't matter if we are able to make <math> m^{a_1} +m^{a_2} + .... + m^{a_{<ins class="diffchange diffchange-inline">2010</ins>}}</math> equal to <math>1-\left(\frac{1}{m}\right)^x</math> for any power of <math>x</math>. Consider <math>m=2</math>. We can achieve a sum of <math>1-\left(\frac{1}{2}\right)^x</math> by doing <math>\frac{1}{2}+\frac{1}{4}+...</math> (the "simplest" sequence). If we don't have <math>\frac{1}{2}</math>, to compensate, we need <math>2\cdot 1\frac{1}{4}</math>'s. Now, let's try to generalize. The "simplest" sequence is having <math>\frac{1}{m}</math> <math>m-1</math> times, <math>\frac{1}{m^2}</math> <math>m-1</math> times, <math>\ldots</math>. To make other sequences, we can split <math>m-1</math> <math>\frac{1}{m^i}</math>s into <math>m(m-1)</math> <math>\frac{1}{m^{i+1}}</math>s since <math>m\cdot\frac{1}{m^{i+1}}\cdot =m(m-1)\cdot\frac{1}{m^{i}}</math>. Since we want <math>2010</math> terms, we have <math>\sum</math> <math>(m-1)\cdot m^x=2010</math>. However, since we can set <math>x</math> to be anything we want (including 0), all we care about is that <math>m-1 | 2010</math> which happens <math>\boxed{016}</math> times.</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>
</table>Hi13https://artofproblemsolving.com/wiki/index.php?title=2011_AIME_I_Problems/Problem_7&diff=127461&oldid=prevHi13: /* Solution 1 */2020-07-04T19:13:44Z<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;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 19:13, 4 July 2020</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;"></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 1==</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 1==</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><math> m^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}</math>. Now, divide by <math>m^{x_0}</math> to get <math>1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}</math>. Notice that since we can choose all nonnegative <math>x_0,...,x_{<del class="diffchange diffchange-inline">2011</del>}</math>, we can make <math>x_n-x_0</math> whatever we desire. WLOG, let <math>x_0\geq...\geq x_{2011}</math> and let <math>a_n=x_n-x_0</math>.  Notice that, also, <math>m^{a_{2011}}</math> doesn't matter if we are able to make <math> m^{a_1} +m^{a_2} + .... + m^{a_{2011}}</math> equal to <math>1-\left(\frac{1}{m}\right)^x</math> for any power of <math>x</math>. Consider <math>m=2</math>. We can achieve a sum of <math>1-\left(\frac{1}{2}\right)^x</math> by doing <math>\frac{1}{2}+\frac{1}{4}+...</math> (the "simplest" sequence). If we don't have <math>\frac{1}{2}</math>, to compensate, we need <math>2\cdot 1\frac{1}{4}</math>'s. Now, let's try to generalize. The "simplest" sequence is having <math>\frac{1}{m}</math> <math>m-1</math> times, <math>\frac{1}{m^2}</math> <math>m-1</math> times, <math>\ldots</math>. To make other sequences, we can split <math>m-1</math> <math>\frac{1}{m^i}</math>s into <math>m(m-1)</math> <math>\frac{1}{m^{i+1}}</math>s since <math>m\cdot\frac{1}{m^{i+1}}\cdot =m(m-1)\cdot\frac{1}{m^{i}}</math>. Since we want <math>2010</math> terms, we have <math>\sum</math> <math>(m-1)\cdot m^x=2010</math>. However, since we can set <math>x</math> to be anything we want (including 0), all we care about is that <math>m-1 | 2010</math> which happens <math>\boxed{016}</math> times.</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^{x_0}= m^{x_1} +m^{x_2} + .... + m^{x_{2011}}</math>. Now, divide by <math>m^{x_0}</math> to get <math>1= m^{x_1-x_0} +m^{x_2-x_0} + .... + m^{x_{2011}-x_0}</math>. Notice that since we can choose all nonnegative <math>x_0,...,x_{<ins class="diffchange diffchange-inline">2010</ins>}</math>, we can make <math>x_n-x_0</math> whatever we desire. WLOG, let <math>x_0\geq...\geq x_{2011}</math> and let <math>a_n=x_n-x_0</math>.  Notice that, also, <math>m^{a_{2011}}</math> doesn't matter if we are able to make <math> m^{a_1} +m^{a_2} + .... + m^{a_{2011}}</math> equal to <math>1-\left(\frac{1}{m}\right)^x</math> for any power of <math>x</math>. Consider <math>m=2</math>. We can achieve a sum of <math>1-\left(\frac{1}{2}\right)^x</math> by doing <math>\frac{1}{2}+\frac{1}{4}+...</math> (the "simplest" sequence). If we don't have <math>\frac{1}{2}</math>, to compensate, we need <math>2\cdot 1\frac{1}{4}</math>'s. Now, let's try to generalize. The "simplest" sequence is having <math>\frac{1}{m}</math> <math>m-1</math> times, <math>\frac{1}{m^2}</math> <math>m-1</math> times, <math>\ldots</math>. To make other sequences, we can split <math>m-1</math> <math>\frac{1}{m^i}</math>s into <math>m(m-1)</math> <math>\frac{1}{m^{i+1}}</math>s since <math>m\cdot\frac{1}{m^{i+1}}\cdot =m(m-1)\cdot\frac{1}{m^{i}}</math>. Since we want <math>2010</math> terms, we have <math>\sum</math> <math>(m-1)\cdot m^x=2010</math>. However, since we can set <math>x</math> to be anything we want (including 0), all we care about is that <math>m-1 | 2010</math> which happens <math>\boxed{016}</math> times.</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>
</table>Hi13