https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&feed=atom&action=history1983 AIME Problems/Problem 13 - Revision history2024-03-28T17:24:48ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=132360&oldid=prevVqbc: Formatting fixes2020-08-22T18:29:08Z<p>Formatting fixes</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>=== Solution 5 ===</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 5 ===</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"> </del>Let <math>\mathbb{N}_n := \{1, 2, 3, \dots n\}</math>. Let the alternating sum of a certain subset of <math>S</math> of <math>\mathbb{N}_n</math> be <math>\xi(S),</math> and let <cmath>\mathcal{A}(\mathbb{N}_n) := \sum_{S \subseteq \mathbb{N}_n} \xi(S).</cmath> We see that <cmath>\mathcal{A}(\mathbb{N}_n) = \sum_{S \subseteq \mathbb{N}_n} \xi(S) = \sum_{n \in S, S \subseteq \mathbb{N}_n} \xi(S) + \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = \mathcal{A}(\mathbb{N}_{n-1}) + \sum_{n \in S, S \subseteq \mathbb{N}_n} \left( n - \xi(S - \{n\}) \right),</cmath> as if <math>n \in S,</math> <math>n</math> is the largest element in <math>S.</math> Now, we know that <cmath>\sum_{n \in S, S \subseteq \mathbb{N}_n} n - \xi(S - \{n\}) = \sum_{S \subseteq \mathbb{N}_{n-1}} n - \xi(S) = n \cdot 2^{n-1} - \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = n \cdot 2^{n-1} - \mathcal{A}(\mathbb{N}_{n-1}),</cmath> so <cmath>\mathcal{A}(\mathbb{N}_{n}) = n \cdot 2^{n-1}.</cmath>  Thus, our answer (which is the <math>n = 7</math> case) is <math>\mathcal{A}(\mathbb{N}_{7}) = 7 \cdot 2^6 = \boxed{448}.</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 <math>\mathbb{N}_n := \{1, 2, 3, \dots n\}</math>. Let the alternating sum of a certain subset of <math>S</math> of <math>\mathbb{N}_n</math> be <math>\xi(S),</math> and let <cmath>\mathcal{A}(\mathbb{N}_n) := \sum_{S \subseteq \mathbb{N}_n} \xi(S).</cmath> We see that <cmath>\mathcal{A}(\mathbb{N}_n) = \sum_{S \subseteq \mathbb{N}_n} \xi(S) = \sum_{n \in S, S \subseteq \mathbb{N}_n} \xi(S) + \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = \mathcal{A}(\mathbb{N}_{n-1}) + \sum_{n \in S, S \subseteq \mathbb{N}_n} \left( n - \xi(S - \{n\}) \right),</cmath> as if <math>n \in S,</math> <math>n</math> is the largest element in <math>S.</math> Now, we know that <cmath>\sum_{n \in S, S \subseteq \mathbb{N}_n} n - \xi(S - \{n\}) = \sum_{S \subseteq \mathbb{N}_{n-1}} n - \xi(S) = n \cdot 2^{n-1} - \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = n \cdot 2^{n-1} - \mathcal{A}(\mathbb{N}_{n-1}),</cmath> so <cmath>\mathcal{A}(\mathbb{N}_{n}) = n \cdot 2^{n-1}.</cmath>  Thus, our answer (which is the <math>n = 7</math> case) is <math>\mathcal{A}(\mathbb{N}_{7}) = 7 \cdot 2^6 = \boxed{448}.</math></div></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;"><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>Vqbchttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=129938&oldid=prevLilavigne: /* Problem */2020-07-31T12:28:07Z<p><span dir="auto"><span class="autocomment">Problem</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Problem ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Problem ==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="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><!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>For <math>\{1, 2, 3, \ldots, n\}</math> and each of its non-empty subsets a unique '''alternating sum''' <del class="diffchange diffchange-inline">sum </del>is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for <math>\{1, 2, 3, 6,9\}</math> is <math>9-6+3-2+1=5</math> and for <math>\{5\}</math> it is simply <math>5</math>. Find the sum of all such alternating sums for <math>n=7</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude></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><!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>For <math>\{1, 2, 3, \ldots, n\}</math> and each of its non-empty subsets a unique '''alternating sum''' is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for <math>\{1, 2, 3, 6,9\}</math> is <math>9-6+3-2+1=5</math> and for <math>\{5\}</math> it is simply <math>5</math>. Find the sum of all such alternating sums for <math>n=7</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td></tr>
</table>Lilavignehttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=129443&oldid=prevApratrip: /* Solution 5 */2020-07-27T03:53:11Z<p><span dir="auto"><span class="autocomment">Solution 5</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></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 5 ===</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 5 ===</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;"><div>  Let <math>\mathbb{N}_n := \{1, 2, 3, \dots n\}</math>. Let the alternating sum of a certain subset of <math>S</math> of <math>\mathbb{N}_n</math> be <math>\xi(S),</math> and let <cmath>\mathcal{A}(\mathbb{N}_n) := \sum_{S \subseteq \mathbb{N}_n} \xi(S).</cmath> We see that <cmath>\mathcal{A}(\mathbb{N}_n) = \sum_{S \subseteq \mathbb{N}_n} \xi(S) = \sum_{n \in S, S \subseteq \mathbb{N}_n} \xi(S) + \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = \mathcal{A}(\mathbb{N}_{n-1}) + \sum_{n \in S, S \subseteq \mathbb{N}_n} \left( n - \xi(S - \{n\}) \right),</cmath> as if <math>n \in S,</math> <math>n</math> is the largest element in <math>S.</math> Now, we know that <cmath>\sum_{n \in S, S \subseteq \mathbb{N}_n} n - \xi(S - \{n\}) = \sum_{S \subseteq \mathbb{N}_{n-1}} n - \xi(S) = n \cdot 2^{n-1} - \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = n \cdot 2^{n-1} - \mathcal{A}(\mathbb{N}_{n-1}),</cmath> so <cmath>\mathcal{A}(\mathbb{N}_{n}) = n \cdot 2^{n-1}.</cmath>  Thus, our answer (which is the <math>n = 7</math> case) is <math>\mathcal{A}(\mathbb{N}_{7}) = 7 \cdot 2^6 = \boxed{448}.</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>  Let <math>\mathbb{N}_n := \{1, 2, 3, \dots n\}</math>. Let the alternating sum of a certain subset of <math>S</math> of <math>\mathbb{N}_n</math> be <math>\xi(S),</math> and let <cmath>\mathcal{A}(\mathbb{N}_n) := \sum_{S \subseteq \mathbb{N}_n} \xi(S).</cmath> We see that <cmath>\mathcal{A}(\mathbb{N}_n) = \sum_{S \subseteq \mathbb{N}_n} \xi(S) = \sum_{n \in S, S \subseteq \mathbb{N}_n} \xi(S) + \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = \mathcal{A}(\mathbb{N}_{n-1}) + \sum_{n \in S, S \subseteq \mathbb{N}_n} \left( n - \xi(S - \{n\}) \right),</cmath> as if <math>n \in S,</math> <math>n</math> is the largest element in <math>S.</math> Now, we know that <cmath>\sum_{n \in S, S \subseteq \mathbb{N}_n} n - \xi(S - \{n\}) = \sum_{S \subseteq \mathbb{N}_{n-1}} n - \xi(S) = n \cdot 2^{n-1} - \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = n \cdot 2^{n-1} - \mathcal{A}(\mathbb{N}_{n-1}),</cmath> so <cmath>\mathcal{A}(\mathbb{N}_{n}) = n \cdot 2^{n-1}.</cmath>  Thus, our answer (which is the <math>n = 7</math> case) is <math>\mathcal{A}(\mathbb{N}_{7}) = 7 \cdot 2^6 = \boxed{448}.</math></div></td></tr>
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</table>Apratriphttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=129442&oldid=prevApratrip: /* Solution */2020-07-27T03:52:33Z<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></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>The same thing happens to any positive integer less than 7. This is because the determination of a positive or negative contribution is dependent on the number of larger numbers in front of it(For example, the sign of 3 is dependent on the presence of 4, 5, 6, and 7 in the subset). If the number of larger numbers is even, it gives in a positive copy while odd produces its negative. We know that the frequencies of these two cases occurring are the same because <math>0=(1-1)^{n}=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...\binom{n}{n}</math> via the Binomial Theorem. Therefore, all positive integers less than 7 will not have any effect and our sum will be <math>\boxed{448}</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>The same thing happens to any positive integer less than 7. This is because the determination of a positive or negative contribution is dependent on the number of larger numbers in front of it(For example, the sign of 3 is dependent on the presence of 4, 5, 6, and 7 in the subset). If the number of larger numbers is even, it gives in a positive copy while odd produces its negative. We know that the frequencies of these two cases occurring are the same because <math>0=(1-1)^{n}=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...\binom{n}{n}</math> via the Binomial Theorem. Therefore, all positive integers less than 7 will not have any effect and our sum will be <math>\boxed{448}</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 5 ===</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;"> Let <math>\mathbb{N}_n := \{1, 2, 3, \dots n\}</math>. Let the alternating sum of a certain subset of <math>S</math> of <math>\mathbb{N}_n</math> be <math>\xi(S),</math> and let <cmath>\mathcal{A}(\mathbb{N}_n) := \sum_{S \subseteq \mathbb{N}_n} \xi(S).</cmath> We see that <cmath>\mathcal{A}(\mathbb{N}_n) = \sum_{S \subseteq \mathbb{N}_n} \xi(S) = \sum_{n \in S, S \subseteq \mathbb{N}_n} \xi(S) + \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = \mathcal{A}(\mathbb{N}_{n-1}) + \sum_{n \in S, S \subseteq \mathbb{N}_n} \left( n - \xi(S - \{n\}) \right),</cmath> as if <math>n \in S,</math> <math>n</math> is the largest element in <math>S.</math> Now, we know that <cmath>\sum_{n \in S, S \subseteq \mathbb{N}_n} n - \xi(S - \{n\}) = \sum_{S \subseteq \mathbb{N}_{n-1}} n - \xi(S) = n \cdot 2^{n-1} - \sum_{S \subseteq \mathbb{N}_{n-1}} \xi(S) = n \cdot 2^{n-1} - \mathcal{A}(\mathbb{N}_{n-1}),</cmath> so <cmath>\mathcal{A}(\mathbb{N}_{n}) = n \cdot 2^{n-1}.</cmath>  Thus, our answer (which is the <math>n = 7</math> case) is <math>\mathcal{A}(\mathbb{N}_{7}) = 7 \cdot 2^6 = \boxed{448}.</math></ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== 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>Apratriphttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=123419&oldid=prevYukrant1: /* Solution 2 (almost the same as Solution 1) */2020-06-01T13:51:40Z<p><span dir="auto"><span class="autocomment">Solution 2 (almost the same as 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 13:51, 1 June 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l13" >Line 13:</td>
<td colspan="2" class="diff-lineno">Line 13:</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 (almost the same as 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 2 (almost the same as Solution 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;"></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>Consider a given subset <math>T</math> of <math>S</math> that contains <math>7</math>; then there is a subset <math>T'</math> which contains all the elements of <math>T</math> except for <math>7</math>, and only those elements . Since each element of <math>T'</math> has one fewer element preceding it than it does in <math>T</math>, their signs are opposite. Thus the sum of the alternating sums of <math>T</math> and <math>T'</math> is equal to 7. There are <math>2^6</math> subsets containing 7, so our answer is <math>7 <del class="diffchange diffchange-inline">* </del>2^6 = \boxed{448}</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>Consider a given subset <math>T</math> of <math>S</math> that contains <math>7</math>; then there is a subset <math>T'</math> which contains all the elements of <math>T</math> except for <math>7</math>, and only those elements . Since each element of <math>T'</math> has one fewer element preceding it than it does in <math>T</math>, their signs are opposite. Thus the sum of the alternating sums of <math>T</math> and <math>T'</math> is equal to 7. There are <math>2^6</math> subsets containing 7, so our answer is <math>7 <ins class="diffchange diffchange-inline">\cdot </ins>2^6 = \boxed{448}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>=== Solution 3 ===</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>=== Solution 3 ===</div></td></tr>
</table>Yukrant1https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=121091&oldid=prevSpeedstorm: /* Solution 1 */2020-04-17T20:47:33Z<p><span dir="auto"><span class="autocomment">Solution 1</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 20:47, 17 April 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l9" >Line 9:</td>
<td colspan="2" class="diff-lineno">Line 9:</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>Then the alternating sum of <math>S</math>, plus the alternating sum of <math>S \cup \{7\}</math>, is <math>7</math>. This is because, since <math>7</math> is the largest element, when we take an alternating sum, each number in <math>S</math> ends up with the opposite sign of each corresponding element of <math>S\cup \{7\}</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>Then the alternating sum of <math>S</math>, plus the alternating sum of <math>S \cup \{7\}</math>, is <math>7</math>. This is because, since <math>7</math> is the largest element, when we take an alternating sum, each number in <math>S</math> ends up with the opposite sign of each corresponding element of <math>S\cup \{7\}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Because there are <math>2^{6} <del class="diffchange diffchange-inline">- 1 </del>= <del class="diffchange diffchange-inline">63</del></math> of these pairs of sets <del class="diffchange diffchange-inline">(subtracting <math>1</math> to exclude the empty set)</del>, the sum of all possible subsets of our given set is <math><del class="diffchange diffchange-inline">63 </del>\cdot 7</math><del class="diffchange diffchange-inline">. However</del>, <del class="diffchange diffchange-inline">we forgot to include the subset that only contains <math>7</math>, so the </del>answer <del class="diffchange diffchange-inline">is </del><math><del class="diffchange diffchange-inline">64 \cdot 7=</del>\boxed{448}</math><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>Because there are <math>2^{6}=<ins class="diffchange diffchange-inline">64</ins></math> of these pairs of sets, the sum of all possible subsets of our given set is <math><ins class="diffchange diffchange-inline">64 </ins>\cdot 7</math>, <ins class="diffchange diffchange-inline">giving an </ins>answer <ins class="diffchange diffchange-inline">of </ins><math>\boxed{448}</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> </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">Note: The empty set is the same as the subset that only includes <math>7</math> so you could have just left it as <math>2^{6}</math> pairs of sets</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>=== Solution 2 (almost the same as 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 2 (almost the same as Solution 1) ===</div></td></tr>
</table>Speedstormhttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=113523&oldid=prevVvluo: /* Solution 3 */2019-12-27T04:32:43Z<p><span dir="auto"><span class="autocomment">Solution 3</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 04:32, 27 December 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l20" >Line 20:</td>
<td colspan="2" class="diff-lineno">Line 20:</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>Denote the desired total of all alternating sums of an <math>n</math>-element set as <math>S_n</math>. We are looking for <math>S_7</math>. Notice that all alternating sums of an <math>n</math>-element set are also alternating sums of an <math>n+1</math>-element set. However, when we go from an <math>n</math> to <math>n+1</math> element set, for each subset with the new element, we are adding the new element and subtracting one of the alternating sums of the <math>n</math>-element set. There are <math>2^n</math> subsets of an <math>n+1</math>-element set that includes the new element, giving us the relationship <math>S_{n+1} = S_n + 2^n(n+1) - S_n = 2^n(n+1)</math>. When <math>n = 6</math>, we therefore get <math>S_ 7 = 2^6(7) = \boxed{448}</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>Denote the desired total of all alternating sums of an <math>n</math>-element set as <math>S_n</math>. We are looking for <math>S_7</math>. Notice that all alternating sums of an <math>n</math>-element set are also alternating sums of an <math>n+1</math>-element set. However, when we go from an <math>n</math> to <math>n+1</math> element set, for each subset with the new element, we are adding the new element and subtracting one of the alternating sums of the <math>n</math>-element set. There are <math>2^n</math> subsets of an <math>n+1</math>-element set that includes the new element, giving us the relationship <math>S_{n+1} = S_n + 2^n(n+1) - S_n = 2^n(n+1)</math>. When <math>n = 6</math>, we therefore get <math>S_ 7 = 2^6(7) = \boxed{448}</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 4 ===</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;">We analyze all the numbers from 1 to 7 separately to see where the number contributes its positive or negative to the sum of the alternating sums. Whenever 7 appears, which it does 64 times, it contributes a positive because it is always first. This gives a net gain of <math>7 \cdot 64=448</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;">If we look at when 6 appears, which it also does 64 times, whether it comes as positive or negative depends on the presence of 7. Half of the subsets with 6 have 7 resulting in subtracting 6 each time, while the other half does not have 7 adding 6 each time, so these contributions of sixes cancel each other out giving a net gain of 0.</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 same thing happens to any positive integer less than 7. This is because the determination of a positive or negative contribution is dependent on the number of larger numbers in front of it(For example, the sign of 3 is dependent on the presence of 4, 5, 6, and 7 in the subset). If the number of larger numbers is even, it gives in a positive copy while odd produces its negative. We know that the frequencies of these two cases occurring are the same because <math>0=(1-1)^{n}=\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-...\binom{n}{n}</math> via the Binomial Theorem. Therefore, all positive integers less than 7 will not have any effect and our sum will be <math>\boxed{448}</math>.</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== 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>Vvluohttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=106113&oldid=prevTost: /* Solution 1 */2019-06-04T17:46:03Z<p><span dir="auto"><span class="autocomment">Solution 1</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 17:46, 4 June 2019</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l10" >Line 10:</td>
<td colspan="2" class="diff-lineno">Line 10:</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>Because there are <math>2^{6} - 1 = 63</math> of these pairs of sets (subtracting <math>1</math> to exclude the empty set), the sum of all possible subsets of our given set is <math>63 \cdot 7</math>. However, we forgot to include the subset that only contains <math>7</math>, so the answer is <math>64 \cdot 7=\boxed{448}</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>Because there are <math>2^{6} - 1 = 63</math> of these pairs of sets (subtracting <math>1</math> to exclude the empty set), the sum of all possible subsets of our given set is <math>63 \cdot 7</math>. However, we forgot to include the subset that only contains <math>7</math>, so the answer is <math>64 \cdot 7=\boxed{448}</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;">Note: The empty set is the same as the subset that only includes <math>7</math> so you could have just left it as <math>2^{6}</math> pairs of sets.</ins></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>=== Solution 2 (almost the same as 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 2 (almost the same as Solution 1) ===</div></td></tr>
</table>Tosthttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=103014&oldid=prevSevenoptimus: Cleaned up the solutions2019-02-16T00:02:35Z<p>Cleaned up the solutions</p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 00:02, 16 February 2019</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><!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>For <math>\{1, 2, 3, \ldots, n\}</math> and each of its non-empty subsets a unique '''alternating sum''' sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for <math>\{1, 2, 3, 6,9\}</math> is <math>9-6+3-2+1=5</math> and for <math>\{5\}</math> it is simply <math>5</math>. Find the sum of all such alternating sums for <math>n=7</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude></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><!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>For <math>\{1, 2, 3, \ldots, n\}</math> and each of its non-empty subsets a unique '''alternating sum''' sum is defined as follows. Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for <math>\{1, 2, 3, 6,9\}</math> is <math>9-6+3-2+1=5</math> and for <math>\{5\}</math> it is simply <math>5</math>. Find the sum of all such alternating sums for <math>n=7</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude></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 1==</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">== 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> </div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">=</ins>== Solution 1 <ins class="diffchange diffchange-inline">=</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>Let <math>S</math> be a non-[[empty set | empty]] [[subset]] of <math>\{1,2,3,4,5,6\}</math>.  </div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Let <math>S</math> be a non-[[empty set | empty]] [[subset]] of <math>\{1,2,3,4,5,6\}</math>.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Then the alternating sum of <math>S</math> plus the alternating sum of <math>S</math> <del class="diffchange diffchange-inline">with 7 included </del>is 7. <del class="diffchange diffchange-inline">In mathematical terms</del>, <math><del class="diffchange diffchange-inline">S+ (S\cup 7)=</del>7</math><del class="diffchange diffchange-inline">. This </del>is <del class="diffchange diffchange-inline">true because </del>when we take an alternating sum, each <del class="diffchange diffchange-inline">term of </del><math>S</math> <del class="diffchange diffchange-inline">has </del>the opposite sign of each corresponding <del class="diffchange diffchange-inline">term </del>of <math>S\cup <del class="diffchange diffchange-inline">7</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>Then the alternating sum of <math>S</math><ins class="diffchange diffchange-inline">, </ins>plus the alternating sum of <math>S <ins class="diffchange diffchange-inline">\cup \{7\}</ins></math><ins class="diffchange diffchange-inline">, </ins>is <ins class="diffchange diffchange-inline"><math></ins>7<ins class="diffchange diffchange-inline"></math></ins>. <ins class="diffchange diffchange-inline">This is because</ins>, <ins class="diffchange diffchange-inline">since </ins><math>7</math> is <ins class="diffchange diffchange-inline">the largest element, </ins>when we take an alternating sum, each <ins class="diffchange diffchange-inline">number in </ins><math>S</math> <ins class="diffchange diffchange-inline">ends up with </ins>the opposite sign of each corresponding <ins class="diffchange diffchange-inline">element </ins>of <math>S\cup \<ins class="diffchange diffchange-inline">{</ins>7\}</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> </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">Because there are <math>63</math> of these pairs, the sum of all possible subsets of our given set is <math>63*7</math>. However, we forgot to include the subset that only contains <math>7</math>, so our answer is <math>64</del>\<del class="diffchange diffchange-inline">cdot </del>7<del class="diffchange diffchange-inline">=</del>\<del class="diffchange diffchange-inline">boxed{448</del>}</math>.</div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'>−</td><td style="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 2 (Almost the same as Solution 1)==</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="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">Consider a given subset <math>T</math> of <math>S</math> that contains 7; then </del>there <del class="diffchange diffchange-inline">is a subset </del><math><del class="diffchange diffchange-inline">T'</del></math> <del class="diffchange diffchange-inline">which contains all the elements </del>of <del class="diffchange diffchange-inline"><math>T</math> except for 7, and only those. Since each element </del>of <math><del class="diffchange diffchange-inline">T'</math> has one element fewer preceding it than it does in <math>T</del></math>, <del class="diffchange diffchange-inline">their signs are opposite; so </del>the sum of <del class="diffchange diffchange-inline">the alternating sums </del>of <math><del class="diffchange diffchange-inline">T</del></math> <del class="diffchange diffchange-inline">and </del><math><del class="diffchange diffchange-inline">T'</math> is equal to </del>7<del class="diffchange diffchange-inline">. There are <math>2^6</del></math> <del class="diffchange diffchange-inline">subsets containing 7</del>, so <del class="diffchange diffchange-inline">our </del>answer is <math>7 <del class="diffchange diffchange-inline">* 2^6 </del>= \boxed{448}</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">Because </ins>there <ins class="diffchange diffchange-inline">are </ins><math><ins class="diffchange diffchange-inline">2^{6} - 1 = 63</ins></math> of <ins class="diffchange diffchange-inline">these pairs </ins>of <ins class="diffchange diffchange-inline">sets (subtracting </ins><math><ins class="diffchange diffchange-inline">1</ins></math> <ins class="diffchange diffchange-inline">to exclude the empty set)</ins>, the sum of <ins class="diffchange diffchange-inline">all possible subsets </ins>of <ins class="diffchange diffchange-inline">our given set is </ins><math><ins class="diffchange diffchange-inline">63 \cdot 7</ins></math><ins class="diffchange diffchange-inline">. However, we forgot to include the subset that only contains </ins><math>7</math>, so <ins class="diffchange diffchange-inline">the </ins>answer is <math><ins class="diffchange diffchange-inline">64 \cdot </ins>7=\boxed{448}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>==Solution <del class="diffchange diffchange-inline">3</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">=</ins>== Solution <ins class="diffchange diffchange-inline">2 (almost the same as Solution 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;"></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">Denote the desired total </del>of <del class="diffchange diffchange-inline">alternating sums of an </del><math><del class="diffchange diffchange-inline">n</del></math> <del class="diffchange diffchange-inline">element set with </del><math><del class="diffchange diffchange-inline">S_n</del></math><del class="diffchange diffchange-inline">. We are looking for </del><math><del class="diffchange diffchange-inline">S_7</del></math><del class="diffchange diffchange-inline">. Note that </del>all <del class="diffchange diffchange-inline">alternating sums </del>of <del class="diffchange diffchange-inline">an </del><math><del class="diffchange diffchange-inline">n</del></math> <del class="diffchange diffchange-inline">element set are also alternating sums of an </del><math><del class="diffchange diffchange-inline">n+1</del></math> element <del class="diffchange diffchange-inline">set. However, when we go from an </del><math><del class="diffchange diffchange-inline">n</del></math> <del class="diffchange diffchange-inline">to </del><math><del class="diffchange diffchange-inline">n+1</del></math> <del class="diffchange diffchange-inline">element set, for each subset with the new element</del>, <del class="diffchange diffchange-inline">we </del>are <del class="diffchange diffchange-inline">adding </del>the <del class="diffchange diffchange-inline">new element and subtracting one </del>of the alternating sums of <del class="diffchange diffchange-inline">the </del><math><del class="diffchange diffchange-inline">n</del></math> <del class="diffchange diffchange-inline">element set</del>. There are <math>2^<del class="diffchange diffchange-inline">n</del></math> subsets <del class="diffchange diffchange-inline">of an </del><math><del class="diffchange diffchange-inline">n+1</del></math> <del class="diffchange diffchange-inline">element set that includes the new element, giving us the following relationship</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">Consider a given subset <math>T</math> </ins>of <math><ins class="diffchange diffchange-inline">S</ins></math> <ins class="diffchange diffchange-inline">that contains </ins><math><ins class="diffchange diffchange-inline">7</ins></math><ins class="diffchange diffchange-inline">; then there is a subset </ins><math><ins class="diffchange diffchange-inline">T'</ins></math> <ins class="diffchange diffchange-inline">which contains </ins>all <ins class="diffchange diffchange-inline">the elements </ins>of <math><ins class="diffchange diffchange-inline">T</ins></math> <ins class="diffchange diffchange-inline">except for </ins><math><ins class="diffchange diffchange-inline">7</ins></math><ins class="diffchange diffchange-inline">, and only those elements . Since each </ins>element <ins class="diffchange diffchange-inline">of </ins><math><ins class="diffchange diffchange-inline">T'</ins></math> <ins class="diffchange diffchange-inline">has one fewer element preceding it than it does in </ins><math><ins class="diffchange diffchange-inline">T</ins></math>, <ins class="diffchange diffchange-inline">their signs </ins>are <ins class="diffchange diffchange-inline">opposite. Thus </ins>the <ins class="diffchange diffchange-inline">sum </ins>of the alternating sums of <math><ins class="diffchange diffchange-inline">T</math> and <math>T'</ins></math> <ins class="diffchange diffchange-inline">is equal to 7</ins>. There are <math>2^<ins class="diffchange diffchange-inline">6</ins></math> subsets <ins class="diffchange diffchange-inline">containing 7, so our answer is </ins><math><ins class="diffchange diffchange-inline">7 * 2^6 = \boxed{448}</ins></math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="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>S_{n+1} </del>= <del class="diffchange diffchange-inline">S_n + 2^n(n+1) - S_n </del>= <del class="diffchange diffchange-inline">2^n(n+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">= Solution 3 ===</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">For </del><math>n = 6</math>, <del class="diffchange diffchange-inline">this becomes </del><math>S_ 7 = 2^6(7) = \boxed{448}</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">Denote the desired total of all alternating sums of an <math>n</math>-element set as <math>S_n</math>. We are looking for <math>S_7</math>. Notice that all alternating sums of an <math>n</math>-element set are also alternating sums of an <math>n+1</math>-element set. However, when we go from an <math>n</math> to <math>n+1</math> element set, for each subset with the new element, we are adding the new element and subtracting one of the alternating sums of the <math>n</math>-element set. There are <math>2^n</math> subsets of an <math>n+1</math>-element set that includes the new element, giving us the relationship <math>S_{n+1} = S_n + 2^n(n+1) - S_n = 2^n(n+1)</math>. When </ins><math>n = 6</math>, <ins class="diffchange diffchange-inline">we therefore get </ins><math>S_ 7 = 2^6(7) = \boxed{448}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== 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>Sevenoptimushttps://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_13&diff=103011&oldid=prevSevenoptimus: Fixed the problem statement2019-02-15T23:56:32Z<p>Fixed the problem statement</p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:56, 15 February 2019</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Problem ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Problem ==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="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><!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>For <math>\{1, 2, 3, \ldots, n\}</math> and each of its non-empty subsets<del class="diffchange diffchange-inline">, an </del>alternating sum is defined as follows. Arrange the <del class="diffchange diffchange-inline">number </del>in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for <math>\{1, 2, <del class="diffchange diffchange-inline">4</del>, 6,9\}</math> is <math>9-6+<del class="diffchange diffchange-inline">4</del>-2+1=<del class="diffchange diffchange-inline">6</del></math> and for <math>\{5\}</math> it is simply <math>5</math>. Find the sum of all such alternating sums for <math>n=7</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude></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><!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>For <math>\{1, 2, 3, \ldots, n\}</math> and each of its non-empty subsets <ins class="diffchange diffchange-inline">a unique '''</ins>alternating <ins class="diffchange diffchange-inline">sum''' </ins>sum is defined as follows. Arrange the <ins class="diffchange diffchange-inline">numbers </ins>in the subset in decreasing order and then, beginning with the largest, alternately add and subtract successive numbers. For example, the alternating sum for <math>\{1, 2, <ins class="diffchange diffchange-inline">3</ins>, 6,9\}</math> is <math>9-6+<ins class="diffchange diffchange-inline">3</ins>-2+1=<ins class="diffchange diffchange-inline">5</ins></math> and for <math>\{5\}</math> it is simply <math>5</math>. Find the sum of all such alternating sums for <math>n=7</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude></div></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;"><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>
</table>Sevenoptimus