Regular octagon abcdefgh has its center at z . 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 aze, bzf, czg, and djh are all equal. In how many ways can this be done?
Regular octagon abcdefgh has its center at z .
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 aze, bzf, czg, and dzh are all equal.
In how many ways can this be done?
\(\begin{array}{|r|c|c|c|c|c|c|c|c|c|} \hline \text{sum} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline 1 & - & 3 & 4 & 5 & 6 & 7 & 8 & \color{red} 9 & \color{red}10 \\ \hline 2 & & - & 5 & 6 & 7 & 8 & \color{red}9 & \color{red}10 & \color{red}11 \\ \hline 3 & & & - & 7 & 8 & \color{red}9 & \color{red}10 & \color{red}11 & 12 \\ \hline 4 & & & & - &\color{red} 9 & \color{red}10 & \color{red}11 & 12 & 13 \\ \hline 5 & & & & & - & \color{red}11 & 12 & 13 & 14 \\ \hline 6 & & & & & & - & 13 & 14 & 15 \\ \hline 7 & & & & & & & - & 15 & 16 \\ \hline 8 & & & & & & & & - & 17 \\ \hline 9 & & & & & & & & & - \\ \hline \end{array} \)
We need equal sums in four different ways: There are three possible sums: 9, 10, and 11.
\(\begin{array}{|rcll|} \hline \text{sum }~=9 \\ \hline 9 &=& 1+8 \\ 9 &=& 2+7 \\ 9 &=& 3+6 \\ 9 &=& 4+5 \\ \hline \text{There are }\\ ~ 4! = 24~ \text{ ways} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{sum }~=10 \\ \hline 10 &=& 1+9 \\ 10 &=& 2+8 \\ 10 &=& 3+7 \\ 10 &=& 4+6 \\ \hline \text{There are }\\ 4! = 24~ \text{ ways} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{sum }~=11 \\ \hline 11 &=& 2+9 \\ 11 &=& 3+8 \\ 11 &=& 4+7 \\ 11 &=& 5+6 \\ \hline \text{There are }\\ ~ 4! = 24~ \text{ ways} \\ \hline \end{array}\)
In total so far these are \(3\times 4!=3\times 24 = \mathbf{72} \text{ ways}\).
Swapping the two elements of a pair of numbers:
\(\begin{array}{|rcll|} \hline \text{One of the $\mathbf{72}$ ways} \\ \text{swapping the two elements of a pair of numbers} \\ \hline \mathbf{ab} & \mathbf{cd} & \mathbf{ef} & \mathbf{gh} \\ ab & cd & ef & hg \\ ab & cd & fe & gh \\ ab & cd & fe & hg \\ \hline ab & dc & ef & gh \\ ab & dc & ef & hg \\ ab & dc & fe & gh \\ ab & dc & fe & hg \\ \hline ba & cd & ef & gh \\ ba & cd & ef & hg \\ ba & cd & fe & gh \\ ba & cd & fe & hg \\ \hline ba & dc & ef & gh \\ ba & dc & ef & hg \\ ba & dc & fe & gh \\ ba & dc & fe & hg \\ \hline \text{There are }\\ ~ \mathbf{16}~ \text{ ways} \\ \hline \end{array}\)
This can be done in \(16\times 72 = \mathbf{1152}\) ways.