https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&feed=atom&action=history
2013 AIME I Problems/Problem 15 - Revision history
2024-03-29T09:37:25Z
Revision history for this page on the wiki
MediaWiki 1.31.1
https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=193570&oldid=prev
Soilmilk: /* Solution 2 */
2023-05-27T03:14:32Z
<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 03:14, 27 May 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l38" >Line 38:</td>
<td colspan="2" class="diff-lineno">Line 38:</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>So the answer is <math>17 + 15 + \cdots + 1 = \boxed{272}</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>So the answer is <math>17 + 15 + \cdots + 1 = \boxed{272}</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;"><div>~SoilMilk</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>~SoilMilk</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
</table>
Soilmilk
https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=193569&oldid=prev
Soilmilk: /* Solution 2 */
2023-05-27T03:14:22Z
<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 03:14, 27 May 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l37" >Line 37:</td>
<td colspan="2" class="diff-lineno">Line 37:</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>When <math>x=44</math>, <math>B=45\Rightarrow 1</math> triple.  </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>When <math>x=44</math>, <math>B=45\Rightarrow 1</math> triple.  </div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>So the answer is <math>17 + 15 + \cdots + 1 = \boxed{272}</math> ~SoilMilk</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>So the answer is <math>17 + 15 + \cdots + 1 = \boxed{272}</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>~SoilMilk</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{AIME box|year=2013|n=I|num-b=14|after=Last 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>{{AIME box|year=2013|n=I|num-b=14|after=Last Problem}}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
</table>
Soilmilk
https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=193568&oldid=prev
Soilmilk: /* Solution2 */
2023-05-27T03:14:07Z
<p><span dir="auto"><span class="autocomment">Solution2</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 03:14, 27 May 2023</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;"></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">Solution2</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 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 class="diffchange diffchange-inline">Let <math>(A, B, C)</math> = <math>(B-x, B, B+x)</math> and <math>(b, a, c) = (a-y, a, a+y)</math>. Now the 3 differences would be </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 class="diffchange diffchange-inline"><cmath>\begin{align}</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 class="diffchange diffchange-inline">\label{1} &A-a = B-x-a \\</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 class="diffchange diffchange-inline">\label{2} &B - b = B-a+y \\</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 class="diffchange diffchange-inline">\label{3} &C - c = B+x-a-y </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 class="diffchange diffchange-inline">\end{align}</cmath></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </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">Adding equations <math>(1)</math> and <math>(3)</math> would give <math>2B - 2a - y</math>. Then doubling equation <math>(2)</math> would give <math>2B - 2a + 2y</math>. The difference between them would be <math>3y</math>. Since <math>p|\{(1), (2), (3)\}</math>, then <math>p|3y</math>. Since <math>p</math> is prime, <math>p|3</math> or <math>p|y</math>. However, since <math>p > y</math>, we must have <math>p|3</math>, which means <math>p=3</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> </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">If <math>p=3</math>, the only possible values of <math>(b, a, c)</math> are <math>(0, 1, 2)</math>. Plugging this into our differences, we get </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 class="diffchange diffchange-inline"><cmath>\begin{align*}</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 class="diffchange diffchange-inline"> &A-a = B-x-1 \hspace{4cm}(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 class="diffchange diffchange-inline"> &B - b = B \hspace{5.35cm}(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 class="diffchange diffchange-inline"> &C - c = B+x-2 \hspace{4cm}(6)</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 class="diffchange diffchange-inline">\end{align*}</cmath></ins></div></td></tr>
<tr><td colspan="2"> </td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">The difference between <math>(4)</math> and <math>(5)</math> is <math>x+1</math>, which should be divisible by 3. So <math>x \equiv 2 \mod 3</math>. Also note that since <math>3|(5)</math>, <math>3|B</math>. Now we can try different values of <math>x</math> and <math>B</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> </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">When <math>x=2</math>, <math>B=3, 6, ..., 96 \Rightarrow 17</math> triples.</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">When <math>x=5</math>, <math>B=6, 9, ..., 93\Rightarrow 15</math> triples..</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">... and so on until</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">When <math>x=44</math>, <math>B=45\Rightarrow 1</math> triple. </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">So the answer is <math>17 + 15 + \cdots + 1 = \boxed{272}</math> ~SoilMilk</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>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{AIME box|year=2013|n=I|num-b=14|after=Last 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>{{AIME box|year=2013|n=I|num-b=14|after=Last Problem}}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
</table>
Soilmilk
https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=193567&oldid=prev
Soilmilk: /* Solution */
2023-05-27T02:33:13Z
<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 02:33, 27 May 2023</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l8" >Line 8:</td>
<td colspan="2" class="diff-lineno">Line 8:</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>
<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>From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p | B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})</math>. Now we return to condition <math>\text{(c)}</math>, which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing positive integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us <math>1</math> solution. If <math>B=6</math>, we get <math>2</math> solutions, <math>(4,6,8)</math> and <math>(1,6,11)</math>. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</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>From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p | B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})</math>. Now we return to condition <math>\text{(c)}</math>, which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing positive integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us <math>1</math> solution. If <math>B=6</math>, we get <math>2</math> solutions, <math>(4,6,8)</math> and <math>(1,6,11)</math>. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</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;"></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;">==Solution2==</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>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{AIME box|year=2013|n=I|num-b=14|after=Last 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>{{AIME box|year=2013|n=I|num-b=14|after=Last Problem}}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
</table>
Soilmilk
https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=159987&oldid=prev
Puffer13: /* Solution 1 */
2021-08-10T18:30:20Z
<p><span dir="auto"><span class="autocomment">Solution 1</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 18:30, 10 August 2021</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>(d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</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>(d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</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">1</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>==Solution==</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>From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p | B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})</math>. Now we return to condition <math>\text{(c)}</math>, which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing positive integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us <math>1</math> solution. If <math>B=6</math>, we get <math>2</math> solutions, <math>(4,6,8)</math> and <math>(1,6,11)</math>. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</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>From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p | B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})</math>. Now we return to condition <math>\text{(c)}</math>, which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing positive integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us <math>1</math> solution. If <math>B=6</math>, we get <math>2</math> solutions, <math>(4,6,8)</math> and <math>(1,6,11)</math>. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
</table>
Puffer13
https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=159986&oldid=prev
Puffer13: /* Solution 2 */
2021-08-10T18:30:04Z
<p><span dir="auto"><span class="autocomment">Solution 2</span></span></p>
<table class="diff diff-contentalign-left" data-mw="interface">
<col class="diff-marker" />
<col class="diff-content" />
<col class="diff-marker" />
<col class="diff-content" />
<tr class="diff-title" lang="en">
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">← Older revision</td>
<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 18:30, 10 August 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l8" >Line 8:</td>
<td colspan="2" class="diff-lineno">Line 8:</td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 1==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 1==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 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>From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p | B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})</math>. Now we return to condition <math>\text{(c)}</math>, which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing positive integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us <math>1</math> solution. If <math>B=6</math>, we get <math>2</math> solutions, <math>(4,6,8)</math> and <math>(1,6,11)</math>. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</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>From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p | B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})</math>. Now we return to condition <math>\text{(c)}</math>, which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing positive integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us <math>1</math> solution. If <math>B=6</math>, we get <math>2</math> solutions, <math>(4,6,8)</math> and <math>(1,6,11)</math>. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</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="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;">== Solution 2 ==</del></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 style="font-weight: bold; text-decoration: none;">Condition c gives us that <math>A \equiv a \pmod p</math>, etc. Condition d then tells us that C and c can be expressed as <math>2B - A</math> and <math>2a - b</math>, respectively. However, plugging what we got from condition c into this, we find that <math>3a \equiv 3b \pmod p</math>. From there, we branch off into two cases; either <math>p = 3</math>, or <math>a \equiv b \pmod p</math>. Realize then that the second case leads to a contradiction, due to condition b. Then, <math>p = 3</math> means that <math>(b,a,c)</math> must be <math>(0, 1, 2)</math>. The bash from there is pretty similar to what was done in Solution 1. We get <math>\boxed{272}</math>. - Spacesam</del></div></td><td colspan="2"> </td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{AIME box|year=2013|n=I|num-b=14|after=Last 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>{{AIME box|year=2013|n=I|num-b=14|after=Last Problem}}</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>{{MAA Notice}}</div></td></tr>
</table>
Puffer13
https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=143141&oldid=prev
Hashtagmath at 00:14, 25 January 2021
2021-01-25T00:14:03Z
<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 00:14, 25 January 2021</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>==Problem <del class="diffchange diffchange-inline">15</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>==Problem ==</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>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions  </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>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions  </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>(a) <math>0\le A<B<C\le99</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>(a) <math>0\le A<B<C\le99</math>,  </div></td></tr>
</table>
Hashtagmath
https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=135824&oldid=prev
Smileymittens112307: /* Solution 1 */
2020-10-26T01:07:52Z
<p><span dir="auto"><span class="autocomment">Solution 1</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 01:07, 26 October 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l7" >Line 7:</td>
<td colspan="2" class="diff-lineno">Line 7:</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 1==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>==Solution 1==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p|B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})</math>. Now we return to condition <math>\text{(c)}</math>, which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing positive integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us <math>1</math> solution. If <math>B=6</math>, we get <math>2</math> solutions, <math>(4,6,8)</math> and <math>(1,6,11)</math>. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</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>From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p | B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})</math>. Now we return to condition <math>\text{(c)}</math>, which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing positive integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us <math>1</math> solution. If <math>B=6</math>, we get <math>2</math> solutions, <math>(4,6,8)</math> and <math>(1,6,11)</math>. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution 2 ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution 2 ==</div></td></tr>
</table>
Smileymittens112307
https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=135823&oldid=prev
Smileymittens112307: /* Problem 15 */
2020-10-26T01:06:54Z
<p><span dir="auto"><span class="autocomment">Problem 15</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 01:06, 26 October 2020</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>==Problem 15==</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 15==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>Let <math>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions (a) <math>0\le A<B<C\le99</math>, (b) there exist integers <math>a</math>, <math>b</math>, and <math>c</math>, and prime <math>p</math> where <math>0\le b<a<c<p</math>, (c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and (d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</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>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions  </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>(a) <math>0\le A<B<C\le99</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>(b) there exist integers <math>a</math>, <math>b</math>, and <math>c</math>, and prime <math>p</math> where <math>0\le b<a<c<p</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>(c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and  </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>(d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</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 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>
Smileymittens112307
https://artofproblemsolving.com/wiki/index.php?title=2013_AIME_I_Problems/Problem_15&diff=135822&oldid=prev
Smileymittens112307: /* Solution */
2020-10-26T01:06:06Z
<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 01:06, 26 October 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l2" >Line 2:</td>
<td colspan="2" class="diff-lineno">Line 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;"><div>Let <math>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions (a) <math>0\le A<B<C\le99</math>, (b) there exist integers <math>a</math>, <math>b</math>, and <math>c</math>, and prime <math>p</math> where <math>0\le b<a<c<p</math>, (c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and (d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</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>N</math> be the number of ordered triples <math>(A,B,C)</math> of integers satisfying the conditions (a) <math>0\le A<B<C\le99</math>, (b) there exist integers <math>a</math>, <math>b</math>, and <math>c</math>, and prime <math>p</math> where <math>0\le b<a<c<p</math>, (c) <math>p</math> divides <math>A-a</math>, <math>B-b</math>, and <math>C-c</math>, and (d) each ordered triple <math>(A,B,C)</math> and each ordered triple <math>(b,a,c)</math> form arithmetic sequences. Find <math>N</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==</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>From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p|B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})</math>. Now we return to condition <math>\text{(c)}</math>, which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing positive integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us <math>1</math> solution. If <math>B=6</math>, we get <math>2</math> solutions, <math>(4,6,8)</math> and <math>(1,6,11)</math>. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</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>From condition (d), we have <math>(A,B,C)=(B-D,B,B+D)</math> and <math>(b,a,c)=(a-d,a,a+d)</math>. Condition <math>\text{(c)}</math> states that <math>p\mid B-D-a</math>, <math>p|B-a+d</math>, and <math>p\mid B+D-a-d</math>. We subtract the first two to get <math>p\mid-d-D</math>, and we do the same for the last two to get <math>p\mid 2d-D</math>. We subtract these two to get <math>p\mid 3d</math>. So <math>p\mid 3</math> or <math>p\mid d</math>. The second case is clearly impossible, because that would make <math>c=a+d>p</math>, violating condition <math>\text{(b)}</math>. So we have <math>p\mid 3</math>, meaning <math>p=3</math>. Condition <math>\text{(b)}</math> implies that <math>(b,a,c)=(0,1,2)</math> or <math>(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})</math>. Now we return to condition <math>\text{(c)}</math>, which now implies that <math>(A,B,C)\equiv(-2,0,2)\pmod{3}</math>. Now, we set <math>B=3k</math> for increasing positive integer values of <math>k</math>. <math>B=0</math> yields no solutions. <math>B=3</math> gives <math>(A,B,C)=(1,3,5)</math>, giving us <math>1</math> solution. If <math>B=6</math>, we get <math>2</math> solutions, <math>(4,6,8)</math> and <math>(1,6,11)</math>. Proceeding in the manner, we see that if <math>B=48</math>, we get 16 solutions. However, <math>B=51</math> still gives <math>16</math> solutions because <math>C_\text{max}=2B-1=101>100</math>. Likewise, <math>B=54</math> gives <math>15</math> solutions. This continues until <math>B=96</math> gives one solution. <math>B=99</math> gives no solution. Thus, <math>N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
</table>
Smileymittens112307