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

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Puzzles and Curious Problems by Henry Ernest Dudeney: $289$

Magic Square Trick
Place in the empty squares such figures (different in every case, and no two squares containing the same figure)
so that they shall add up to $15$ in as many straight directions as possible.

$\qquad \begin{array} {|c|c|c|} \hline \ \ & \ \ & \ \ \\ \hline \ \ & 5 & \ \ \\ \hline \ \ & \ \ & \ \ \\ \hline \end{array}$


Solution

The solution given by Dudeney appears to be flawed:

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

These evaluate to:

$\qquad \begin{array} {|c|c|c|} \hline 7 \tfrac 1 2 & 5 & 2 \tfrac 1 2 \\ \hline 0 & 5 & 10 \\ \hline 7 \tfrac 1 2 & 5 & 2 \tfrac 1 2 \\ \hline \end{array}$

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 $3$:

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


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

$\qquad \begin{array} {|c|c|c|} \hline 7 & 5 & 6 \\ \hline 2 & 1 & 6 \\ \hline 9 & 6 & 4 \\ \hline \end{array}$

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 $1$?


Historical Note

Martin Gardner reports on simpler constructions for some of the results, suggested by Victor Meally:

\(\ds 4 + 4 - \dfrac 4 {4 + 4}\) \(=\) \(\ds 7 \tfrac 1 2\)
\(\ds 7 + \dfrac {7 + 7 + 7} 7\) \(=\) \(\ds 10\)
\(\ds 8 - \dfrac 8 {8 + 8}\) \(=\) \(\ds 7 \tfrac 1 2\)

but none of these come close to fixing the flaw in the basic premise of the solution of this puzzle.


Sources

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