Henry Ernest Dudeney/Puzzles and Curious Problems/289 - Magic Square Trick/Solution

by : $289$

 * Magic Square Trick

Solution
The solution given by appears to be flawed:


 * $\begin{array} {|c|c|c|}

\hline \dfrac {\paren {8 \times 8} - 8} 8 + \dfrac 8 {8 + 8} & 1 + 1 + 1 + 1 + 1 & \dfrac {6 + 6} 6 \times \dfrac 6 {6 + 6} \\ \hline 3 - 3 & 5 & \dfrac {\paren {7 \times 7} + 7 + 7 + 7} 7 \\ \hline \dfrac {\paren {4 \times 4} + 4 + 4 + 4} 4 + \dfrac 4 {4 + 4} & \dfrac {9 + 9 + 9 + 9 + 9} 9 \ & 2 + \dfrac 2 {2 + 2} \\ \hline \end{array}$

These evaluate to:


 * $\begin{array} {|c|c|c|}

\hline 7 \tfrac 1 2 & 5 & 2 \tfrac 1 2 \\ \hline 0 & 5 & 10 \\ \hline 7 \tfrac 1 2 & 5 & 2 \tfrac 1 2 \\ \hline \end{array}$

which do indeed add to $15$ along all rows, columns and diagonals.

Nothing, of course, was said about these numbers needing all to be different.

Of course, the individual digits themselves form the traditional magic square of order $3$:


 * $\begin{array} {|c|c|c|}

\hline 8 & 1 & 6 \\ \hline 3 & 5 & 7 \\ \hline 4 & 9 & 2 \\ \hline \end{array}$

However, it is seen that the multiplicities of the digits of these entries themselves are:


 * $\begin{array} {|c|c|c|}

\hline 7 & 5 & 6 \\ \hline 2 & 1 & 6 \\ \hline 9 & 6 & 4 \\ \hline \end{array}$

which as can be seen are not all different.

And, trivially, if these all do not have to be different, then why are we wasting our time with this when they could all be given a multiplicity of $1$?