https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_14&feed=atom&action=history2016 AMC 12A Problems/Problem 14 - Revision history2024-03-28T22:55:41ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_14&diff=75549&oldid=prevSwe1: Fixed redirect2016-02-05T16:29:28Z<p>Fixed redirect</p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:29, 5 February 2016</td>
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</table>Swe1https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_14&diff=75548&oldid=prevSwe1: Redirected to matching 10A problem2016-02-05T16:29:05Z<p>Redirected to matching 10A problem</p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 16:29, 5 February 2016</td>
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<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">==</del>Problem<del class="diffchange diffchange-inline">==</del></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">#REDIRECT [[2016_AMC_12A_Problems/</ins>Problem 18]<ins class="diffchange diffchange-inline">]</ins></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div> </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">Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?</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> </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"><math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math></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> </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 1===</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> </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">First of all, the adjacent faces have same sum (</del>18<del class="diffchange diffchange-inline">, because 1+2+3+4+5+6+7+8=36, 36/2=18), </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 class="diffchange diffchange-inline">consider the the <math>opposite \text{ } sides</math> (the two sides which are parallel but not in same face of the cube)</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 class="diffchange diffchange-inline">they must have same  sum value too.</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 class="diffchange diffchange-inline">Now think about the extreme condition 1 and 8 , if they are not sharing the same side, which means they would become end points of <math>opposite \text{ } sides</math>,</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 class="diffchange diffchange-inline">we should have 1+X=8+Y, but no solution for [2,7</del>]<del class="diffchange diffchange-inline">, contradiction. </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> </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">Now we know 1 and 8 must share the same side, which sum is 9, the <math>opposite \text{ } side</math> also must have sum of 9, same thing for the other two parallel sides.</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> </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">now we have 4 parallel sides 1-8, 2-7, 3-6, 4-5.</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 class="diffchange diffchange-inline">thinking about 4 end points number need to have sum of 18.</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 class="diffchange diffchange-inline">it is easy to notice only 1-7-6-4 vs 8-2-3-5 would work.</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> </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">so if we fix one direction 1-8 (or 8-1) all other 3 parallel sides must lay in one particular direction.(1-8,7-2,6-3,4-5) or (8-1,2-7,3-6,5-4)</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> </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">Now, the problem is same as the problem to arrange 4 points in a 2-D square. which is 4!/4=<math>\boxed{\textbf{(C) }6.}</math></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> </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 ===</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> </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">Again, all faces sum to <math>18.</math> If <math>x,y,z</math> are the vertices next to one, then the remaining vertices are <math>17-x-y, 17-y-z, 17-x-z, x+y+z-16.</math> Now it remains to test possibilities. Note that we must have <math>x+y+z>17.</math> WLOG let <math>x<y<z.</math></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> </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"><math>3,7,8:</math> Does not work.</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 class="diffchange diffchange-inline"><math>4,6,8:</math> Works.</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 class="diffchange diffchange-inline"><math>5,6,7:</math> Does not work.</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 class="diffchange diffchange-inline"><math>5,6,8:</math> Works.</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 class="diffchange diffchange-inline"><math>5,7,8:</math> Does not work.</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 class="diffchange diffchange-inline"><math>6,7,8:</math> Works.</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> </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">So our answer is <math>3\cdot 2=\boxed{\textbf{(C) }6.}</math></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> </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 3 ===</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> </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">We know the sum of each face is <math>18.</math> If we look at an edge of the cube whose numbers sum to <math>x</math>, it must be possible to achieve the sum <math>18-x</math> in two distinct ways, looking at the two faces which contain the edge. If <math>8</math> and <math>6</math> were on the same face, it is possible to achieve the desired sum only with the numbers <math>1</math> and <math>3</math> since the values must be distinct. Similarly, if <math>8</math> and <math>7</math> were on the same face, the only way to get the sum is with <math>1</math> and <math>2</math>. This means that <math>6</math> and <math>7</math> are not on the same edge as <math>8</math>, or in other words they are diagonally across from it on the same face, or on the other end of the cube.</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> </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">Now we look at three cases, each yielding two solutions which are reflections of each other:</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> </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">1) <math>6</math> and <math>7</math> are diagonally opposite <math>8</math> on the same face.</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 class="diffchange diffchange-inline">2) <math>6</math> is diagonally across the cube from <math>8</math>, while <math>7</math> is diagonally across from <math>8</math> on the same face.</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 class="diffchange diffchange-inline">3) <math>7</math> is diagonally across the cube from <math>8</math>, while <math>6</math> is diagonally across from <math>8</math> on the same face.</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> </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">This means the answer is <math>3\cdot 2=\boxed{\textbf{(C) }6.}</math></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> </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">==See Also==</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 class="diffchange diffchange-inline">{{AMC12 box|year=2016|ab=A|num-b=13|num-a=15}}</del></div></td><td colspan="2"> </td></tr>
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</table>Swe1https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12A_Problems/Problem_14&diff=75485&oldid=prev1915933: Created page with "==Problem== Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the..."2016-02-04T21:16:48Z<p>Created page with "==Problem== Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the..."</p>
<p><b>New page</b></p><div>==Problem==<br />
<br />
Each vertex of a cube is to be labeled with an integer <math>1</math> through <math>8</math>, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?<br />
<br />
<math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math><br />
<br />
===Solution 1===<br />
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First of all, the adjacent faces have same sum (18, because 1+2+3+4+5+6+7+8=36, 36/2=18), <br />
consider the the <math>opposite \text{ } sides</math> (the two sides which are parallel but not in same face of the cube)<br />
they must have same sum value too.<br />
Now think about the extreme condition 1 and 8 , if they are not sharing the same side, which means they would become end points of <math>opposite \text{ } sides</math>,<br />
we should have 1+X=8+Y, but no solution for [2,7], contradiction. <br />
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Now we know 1 and 8 must share the same side, which sum is 9, the <math>opposite \text{ } side</math> also must have sum of 9, same thing for the other two parallel sides.<br />
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now we have 4 parallel sides 1-8, 2-7, 3-6, 4-5.<br />
thinking about 4 end points number need to have sum of 18.<br />
it is easy to notice only 1-7-6-4 vs 8-2-3-5 would work.<br />
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so if we fix one direction 1-8 (or 8-1) all other 3 parallel sides must lay in one particular direction.(1-8,7-2,6-3,4-5) or (8-1,2-7,3-6,5-4)<br />
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Now, the problem is same as the problem to arrange 4 points in a 2-D square. which is 4!/4=<math>\boxed{\textbf{(C) }6.}</math><br />
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=== Solution 2 ===<br />
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Again, all faces sum to <math>18.</math> If <math>x,y,z</math> are the vertices next to one, then the remaining vertices are <math>17-x-y, 17-y-z, 17-x-z, x+y+z-16.</math> Now it remains to test possibilities. Note that we must have <math>x+y+z>17.</math> WLOG let <math>x<y<z.</math><br />
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<math>3,7,8:</math> Does not work.<br />
<math>4,6,8:</math> Works.<br />
<math>5,6,7:</math> Does not work.<br />
<math>5,6,8:</math> Works.<br />
<math>5,7,8:</math> Does not work.<br />
<math>6,7,8:</math> Works.<br />
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So our answer is <math>3\cdot 2=\boxed{\textbf{(C) }6.}</math><br />
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=== Solution 3 ===<br />
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We know the sum of each face is <math>18.</math> If we look at an edge of the cube whose numbers sum to <math>x</math>, it must be possible to achieve the sum <math>18-x</math> in two distinct ways, looking at the two faces which contain the edge. If <math>8</math> and <math>6</math> were on the same face, it is possible to achieve the desired sum only with the numbers <math>1</math> and <math>3</math> since the values must be distinct. Similarly, if <math>8</math> and <math>7</math> were on the same face, the only way to get the sum is with <math>1</math> and <math>2</math>. This means that <math>6</math> and <math>7</math> are not on the same edge as <math>8</math>, or in other words they are diagonally across from it on the same face, or on the other end of the cube.<br />
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Now we look at three cases, each yielding two solutions which are reflections of each other:<br />
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1) <math>6</math> and <math>7</math> are diagonally opposite <math>8</math> on the same face.<br />
2) <math>6</math> is diagonally across the cube from <math>8</math>, while <math>7</math> is diagonally across from <math>8</math> on the same face.<br />
3) <math>7</math> is diagonally across the cube from <math>8</math>, while <math>6</math> is diagonally across from <math>8</math> on the same face.<br />
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This means the answer is <math>3\cdot 2=\boxed{\textbf{(C) }6.}</math><br />
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==See Also==<br />
{{AMC12 box|year=2016|ab=A|num-b=13|num-a=15}}<br />
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