Henry Ernest Dudeney/Modern Puzzles/Magic Square Problems

Henry Ernest Dudeney: Modern Puzzles: Magic Square Problems

$171$ - An Irregular Magic Square

Here we have a perfect magic square composed of the numbers $1$ to $16$ inclusive.

$\qquad \begin{array}{|c|c|c|c|} \hline 1 & 14 & 7 & 12 \\ \hline 15 & 4 & 9 & 6 \\ \hline 10 & 5 & 16 & 3 \\ \hline 8 & 11 & 2 & 13 \\ \hline \end{array}$

The rows, columns, and two long diagonals all add up to $34$.

Now, supposing you were forbidden to use the two numbers $2$ and $15$, but allowed, in their place, to repeat any two numbers already used,

how would you construct your square so that rows, columns, and diagonals should still add up to $34$?

Your success will depend on which two numbers you select as substitutes for $2$ and $15$.

$172$ - A Magic Square Delusion

Here is a magic square of the fifth order.

$\qquad \begin{array}{|c|c|c|c|c|} \hline 17 & 24 & 1 & 8 & 15 \\ \hline 23 & 5 & 7 & 14 & 16 \\ \hline 4 & 6 & 13 & 20 & 22 \\ \hline 10 & 12 & 19 & 21 & 3 \\ \hline 11 & 18 & 25 & 2 & 9 \\ \hline \end{array}$

I have found that a great many people who have not gone very profoundly into these things believe that the central number in all squares of this order must be $13$.

One correspondent who had devoted years to amusing himself with this particular square was astounded when I told him that any number from $1$ to $25$ might be in the centre.

I will show that this is so.

Try to form such a magic square with $1$ in the central cell.

$173$ - Difference Squares

Can you rearrange the nine digits in the square so that in all the eight directions the difference between one of the digits and the sum of the remaining two shall always be the same?

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

In the example shown it will be found that all the rows and columns give the difference $3$:

(thus $4 + 2 - 3$, and $1 + 9 - 7$, and $6 + 5 - 8$, etc.),

but the two diagonals are wrong, because $8 - \paren {4 + 1}$ and $6 - \paren {1 + 2}$ is not allowed:

the sum of the two must not be taken from the single digit, but the single digit from the sum.

How many solutions are there?

$174$ - Swastika Magic Square

It is a magic square, the rows, columns, and two diagonals all adding up to $65$,

and all prime numbers that occur between $1$ and $25$

(viz. $1$, $2$, $3$, $5$, $7$, $11$, $13$, $17$, $19$, $23$)

are to be found in the swastika except $11$.

"This number," he says, "in occult lore is ominous and is associated with the eleven Curses of Ebal,

so it is just as well it does not come into this potent charm of good fortune."

He is clearly under the impression that $11$ cannot be got into the swastika with the other primes.

But in this he is wrong, and the reader may like to try to reconstruct the square

so that the swastika contains all the prime numbers

and yet forms a correct magic square, for it is quite possible.

$175$ - Is it Very Easy?

Here is a simple magic square, the three columns, three rows, and two diagonals adding up to $72$.

$\qquad \begin{array}{|c|c|c|} \hline 27 & 20 & 25\\ \hline 22 & 24 & 26 \\ \hline 23 & 28 & 21 \\ \hline \end{array}$

The puzzle is to convert it into a multiplying magic square,

in which the numbers in all the eight lines if multiplied together give the same product in every case.

You are not allowed to change, or add to, any of the figures in a cell

or use any arithmetical sign whatever!

But you may shift the two figures within a cell.

Thus, you may write $27$ as $72$, if you like.