https://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&feed=atom&action=history2013 USAMO Problems/Problem 5 - Revision history2024-03-29T14:11:27ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=197918&oldid=prevAlphacapture: /* Solution */ Added Solution 22023-09-07T04:31:08Z<p><span dir="auto"><span class="autocomment">Solution: </span> Added Solution 2</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 04:31, 7 September 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l2" >Line 2:</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>Given positive integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.</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>Given positive integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.</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>== Solution ==</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>== Solution <ins class="diffchange diffchange-inline">1 </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>This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.</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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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</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;">== Solution 2 ==</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;">This is a rephrasing of the above solution.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins 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;">It is enough to solve the problem when <math>m,n</math> are replaced by <math>km,kn</math> for any positive integer <math>k</math>. In particular, by taking <math>k=2^a5^b</math> for appropriate values of <math>a,b</math>, we may assume <math>n=10^sn_1</math> where <math>n_1</math> is relatively prime to 10.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Furthermore, adding or removing trailing zeros from <math>m</math> and <math>n</math> doesn't affect the claim, so we may further assume <math>\gcd(n,10)=1</math> and that <math>m</math> has a xillion trailing zeros (enough to make <math>m</math> way bigger than <math>n</math>, and also so that <math>m</math> has at least one trailing zero).</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">For clarity of exposition, we will also multiply <math>m,n</math> by a small number to make the units digit of <math>n</math> be <math>1</math> (though this is not necessary for the solution to work).</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">The point is that, for any positive integer <math>X</math>, most nonzero digits appear the same number of times in <math>X</math> and <math>X+999\dots999</math> if there are enough <math>9</math>s; In particular, if the units digits of <math>X</math> is <math>1</math>, then all nonzero digits appear the same number of times as long as there are at least as many <math>9</math>s as digits in <math>X</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">So we will pick <math>c</math> to satisfy:</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;">* <math>mc=nc+999\dots999</math> where the number of <math>9</math>s is more than the number of digits of <math>nc</math></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">* The units digit of <math>nc</math> is <math>1</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Because we made the units of <math>n</math> to be <math>1</math>, the second condition is equivalent to making the units digit of <math>c</math> to be <math>1</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">The first condition is equivalent to <math>(m-n)c=999\dots999</math>. Because <math>m</math> has at least one trailing 0, the units digit of <math>m-n</math> is 9, so <math>\gcd(m-n,10)=1</math> and there is some <math>c</math> so that <math>(m-n)c=999\dots999</math>, and the units digit of <math>c</math> must be <math>1</math> which agrees with the other condition.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Finally, as <math>m\gg n, (m-n)c\approx mc\gg nc</math> so it is possible to make the number of <math>9</math>s more than the number of digits in <math>nc</math>.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 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>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
</table>Alphacapturehttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=192938&oldid=prevIntegralarefun: complete reformatting2023-05-10T22:27:41Z<p>complete reformatting</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 22:27, 10 May 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td class='diff-marker'>−</td><td style="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>Given <del class="diffchange diffchange-inline">postive </del>integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.</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">== Problem ==</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">{{MAA Notice}}</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>Given <ins class="diffchange diffchange-inline">positive </ins>integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">==Solution==</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 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 ==</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>This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.</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;"></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>Without Loss of Generality, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</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>Without Loss of Generality, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</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><del style="font-weight: bold; text-decoration: none;"></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>Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.</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>Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.</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;"></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>Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another.</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>Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td 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">This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</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">{{MAA Notice}}</ins></div></td></tr>
</table>Integralarefunhttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=130525&oldid=prevNegia at 05:05, 4 August 20202020-08-04T05:05:21Z<p></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 05:05, 4 August 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l7" >Line 7:</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="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">WLOG</del>, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</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><ins class="diffchange diffchange-inline">Without Loss of Generality</ins>, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</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;"></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>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=130517&oldid=prevNegia: /* Solution */2020-08-04T04:50:03Z<p><span dir="auto"><span class="autocomment">Solution</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 04:50, 4 August 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l8" >Line 8:</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>WLOG, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</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>WLOG, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</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 class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 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>Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.</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>Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.</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 class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 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>Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another.</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>Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another.</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 class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</div></td></tr>
</table>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=130515&oldid=prevNegia: /* Solution */2020-08-04T04:49:43Z<p><span dir="auto"><span class="autocomment">Solution</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 04:49, 4 August 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l7" >Line 7:</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;"></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>WLOG, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</math><del class="diffchange diffchange-inline">. Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number</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>WLOG, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</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 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;">Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.</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="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>Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another. This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</div></td></tr>
</table>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=130514&oldid=prevNegia: /* Solution */2020-08-04T04:48:43Z<p><span dir="auto"><span class="autocomment">Solution</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 04:48, 4 August 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l6" >Line 6:</td>
<td colspan="2" class="diff-lineno">Line 6:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.</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>WLOG, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</math>. Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\<del class="diffchange diffchange-inline">phi</del>(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.</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> </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>WLOG, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</math>. Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\<ins class="diffchange diffchange-inline">varphi</ins>(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.</div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 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>Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another. This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another. This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</div></td></tr>
</table>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=61517&oldid=prevLightest at 16:56, 18 April 20142014-04-18T16:56:45Z<p></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:56, 18 April 2014</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l1" >Line 1:</td>
<td colspan="2" class="diff-lineno">Line 1:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Given postive integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.</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>Given postive integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
<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==</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">WLOG, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</math>. Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\phi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.</ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;"></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another. This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</ins></div></td></tr>
</table>Lightesthttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=53372&oldid=prevTenniskidperson3 at 21:59, 3 July 20132013-07-03T21:59:34Z<p></p>
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</table>Tenniskidperson3https://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=52735&oldid=prevYuler1818181818 at 23:43, 11 May 20132013-05-11T23:43:46Z<p></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:43, 11 May 2013</td>
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<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Given postive integers and , prove that there is a positive integer such that the numbers and have the same number of occurrences of each non-zero digit when written in base ten.</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>Given postive integers <ins class="diffchange diffchange-inline"><math>m</math> </ins>and <ins class="diffchange diffchange-inline"><math>n</math></ins>, prove that there is a positive integer <ins class="diffchange diffchange-inline"><math>c</math> </ins>such that the numbers <ins class="diffchange diffchange-inline"><math>cm</math> </ins>and <ins class="diffchange diffchange-inline"><math>cn</math> </ins>have the same number of occurrences of each non-zero digit when written in base ten.</div></td></tr>
</table>Yuler1818181818https://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=52699&oldid=prevYuler1818181818: Created page with "Given postive integers and , prove that there is a positive integer such that the numbers and have the same number of occurrences of each non-zero digit when written in base ten."2013-05-11T22:49:21Z<p>Created page with "Given postive integers and , prove that there is a positive integer such that the numbers and have the same number of occurrences of each non-zero digit when written in base ten."</p>
<p><b>New page</b></p><div>Given postive integers and , prove that there is a positive integer such that the numbers and have the same number of occurrences of each non-zero digit when written in base ten.</div>Yuler1818181818