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Difference between revisions of "2013 AMC 10B Problems/Problem 22"

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==Problem==
  
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The regular octagon <math>ABCDEFGH</math> has its center at <math>J</math>.  Each of the vertices and the center are to be associated with one of the digits <math>1</math> through <math>9</math>, with each digit used once, in such a way that the sums of the numbers on the lines <math>AJE</math>, <math>BJF</math>, <math>CJG</math>, and <math>DJH</math> are all equal.  In how many ways can this be done?
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<math> \textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2  \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8 </math>
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==Solution==
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First of all, note that <math>J</math> must be <math>1</math>, <math>5</math>, or <math>9</math> to preserve symmetry.  We also notice that <math>A+E = B+F = C+G = D+H</math>. 
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WLOG assume that <math>J = 1</math>.  Thus the pairs of vertices must be <math>9</math> and <math>2</math>, <math>8</math> and <math>3</math>, <math>7</math> and <math>4</math>, and <math>6</math> and <math>5</math>.  There are <math>4! = 24</math> ways to assign these to the vertices.  Furthermore, there are <math>2^{4} = 16</math> ways to switch them (i.e. do <math>2</math> <math>9</math> instead of <math>9</math> <math>2</math>). 
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Thus, there are <math>16(24) = 384</math> ways for each possible J value.  There are <math>3</math> possible J values that still preserve symmetry: <math>384(3) = \boxed{\textbf{(C) }1152}</math>

Revision as of 15:50, 21 February 2013

Problem

The regular octagon $ABCDEFGH$ has its center at $J$. Each of the vertices and the center are to be associated with one of the digits $1$ through $9$, with each digit used once, in such a way that the sums of the numbers on the lines $AJE$, $BJF$, $CJG$, and $DJH$ are all equal. In how many ways can this be done?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ 8$

Solution

First of all, note that $J$ must be $1$, $5$, or $9$ to preserve symmetry. We also notice that $A+E = B+F = C+G = D+H$.

WLOG assume that $J = 1$. Thus the pairs of vertices must be $9$ and $2$, $8$ and $3$, $7$ and $4$, and $6$ and $5$. There are $4! = 24$ ways to assign these to the vertices. Furthermore, there are $2^{4} = 16$ ways to switch them (i.e. do $2$ $9$ instead of $9$ $2$).

Thus, there are $16(24) = 384$ ways for each possible J value. There are $3$ possible J values that still preserve symmetry: $384(3) = \boxed{\textbf{(C) }1152}$

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