https://artofproblemsolving.com/wiki/index.php?title=1960_AHSME_Problems/Problem_25&feed=atom&action=history1960 AHSME Problems/Problem 25 - Revision history2024-03-29T12:58:11ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=1960_AHSME_Problems/Problem_25&diff=95268&oldid=prevRockmanex3: /* Solution */2018-06-18T05:21:03Z<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;"><div>If <math>a</math> and <math>b</math> are both even, then <math>a-b</math> is even.  If <math>a</math> and <math>b</math> are both odd, then <math>a-b</math> is even as well.  If <math>a</math> is odd and <math>b</math> is even (or vise versa), then <math>a+b+1</math> is even.  Therefore, in all cases, <math>8</math> can be divided into all numbers with the form <math>m^2-n^2</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>If <math>a</math> and <math>b</math> are both even, then <math>a-b</math> is even.  If <math>a</math> and <math>b</math> are both odd, then <math>a-b</math> is even as well.  If <math>a</math> is odd and <math>b</math> is even (or vise versa), then <math>a+b+1</math> is even.  Therefore, in all cases, <math>8</math> can be divided into all numbers with the form <math>m^2-n^2</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>This can be confirmed by setting <math>m=3</math> and <math>n=1</math>, making <math>m^2-n^2=9-1=8</math>.  Since <math>8</math> is not a multiple of <math>3</math> and is less than <math>16</math>, we can confirm that the answer is <math>\boxed{\textbf{(D)}}</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>This can be confirmed by setting <math>m=3</math> and <math>n=1</math>, making <math>m^2-n^2=9-1=8</math>.  Since <math>8</math> is not a multiple of <math>3</math> and is less than <math>16</math>, we can confirm that the answer is <math>\boxed{\textbf{(D)}}</math><ins class="diffchange diffchange-inline">.</ins></div></td></tr>
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</table>Rockmanex3https://artofproblemsolving.com/wiki/index.php?title=1960_AHSME_Problems/Problem_25&diff=94563&oldid=prevRockmanex3: /* See Also */2018-05-17T23:10:20Z<p><span dir="auto"><span class="autocomment">See Also</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:10, 17 May 2018</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>{{AHSME 40p box|year=1960|num-b=24|num-a=26}}</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>{{AHSME 40p box|year=1960|num-b=24|num-a=26}}</div></td></tr>
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</table>Rockmanex3https://artofproblemsolving.com/wiki/index.php?title=1960_AHSME_Problems/Problem_25&diff=94413&oldid=prevRockmanex3: Solution to Problem 252018-05-11T16:05:41Z<p>Solution to Problem 25</p>
<p><b>New page</b></p><div>==Problem==<br />
<br />
Let <math>m</math> and <math>n</math> be any two odd numbers, with <math>n</math> less than <math>m</math>. <br />
The largest integer which divides all possible numbers of the form <math>m^2-n^2</math> is:<br />
<br />
<math>\textbf{(A)}\ 2\qquad<br />
\textbf{(B)}\ 4\qquad<br />
\textbf{(C)}\ 6\qquad<br />
\textbf{(D)}\ 8\qquad<br />
\textbf{(E)}\ 16 </math><br />
<br />
==Solution==<br />
First, factor the [[difference of squares]].<br />
<cmath>(m+n)(m-n)</cmath><br />
Since <math>m</math> and <math>n</math> are odd numbers, let <math>m=2a+1</math> and <math>n=2b+1</math>, where <math>a</math> and <math>b</math> can be any integer.<br />
<cmath>(2a+2b+2)(2a-2b)</cmath><br />
Factor the resulting expression.<br />
<cmath>4(a+b+1)(a-b)</cmath><br />
If <math>a</math> and <math>b</math> are both even, then <math>a-b</math> is even. If <math>a</math> and <math>b</math> are both odd, then <math>a-b</math> is even as well. If <math>a</math> is odd and <math>b</math> is even (or vise versa), then <math>a+b+1</math> is even. Therefore, in all cases, <math>8</math> can be divided into all numbers with the form <math>m^2-n^2</math>.<br />
<br />
This can be confirmed by setting <math>m=3</math> and <math>n=1</math>, making <math>m^2-n^2=9-1=8</math>. Since <math>8</math> is not a multiple of <math>3</math> and is less than <math>16</math>, we can confirm that the answer is <math>\boxed{\textbf{(D)}}</math><br />
<br />
==See Also==<br />
{{AHSME 40p box|year=1960|num-b=24|num-a=26}}</div>Rockmanex3