SQUARES OF PRIMES.
The problem of constructing magic squares with prime numbers only was first
discussed by myself in The Weekly Dispatch for 22nd July and 5th August
1900; but during the last three or four years it has received great attention
from American mathematicians. First, they have sought to form these squares with
the lowest possible constants. Thus, the first nine prime numbers, 1 to 23
inclusive, sum to 99, which (being divisible by 3) is theoretically a suitable
series; yet it has been demonstrated that the lowest possible constant is 111,
and the required series as follows: 1, 7, 13, 31, 37, 43, 61, 67, and 73.
Similarly, in the case of the fourth order, the lowest series of primes that are
"theoretically suitable" will not serve. But in every other order, up to the
12th inclusive, magic squares have been constructed with the lowest series of
primes theoretically possible. And the 12th is the lowest order in which a
straight series of prime numbers, unbroken, from 1 upwards has been made to
work. In other words, the first 144 odd prime numbers have actually been
arranged in magic form. The following summary is taken from The Monist
(Chicago) for October 1913:—
||Henry E. |
and C. D. Shuldham.
||H. A. Sayles.|
||C. D. Shuldham |
and J. N. Muncey.
||J. N. Muncey.|
For further details the reader should consult the article itself, by W. S.
Andrews and H. A. Sayles.
investigators have also performed notable feats in constructing associated and
bordered prime magics, and Mr. Shuldham has sent me a remarkable paper in which
he gives examples of Nasik squares constructed with primes for all orders from
the 4th to the 10th, with the exception of the 3rd (which is clearly impossible)
and the 9th, which, up to the time of writing, has baffled all attempts.
BASKETS OF PLUMS.
This is the form in which I first introduced the question of magic squares
with prime numbers. I will here warn the reader that there is a little trap.
A fruit merchant had nine baskets. Every basket contained plums (all sound
and ripe), and the number in every basket was different. When placed as shown in
the illustration they formed a magic square, so that if he took any three
baskets in a line in the eight possible directions there would always be the
same number of plums. This part of the puzzle is easy enough to understand. But
what follows seems at first sight a little queer.
The merchant told one of his men to distribute the contents of any basket he
chose among some children, giving plums to every child so that each should
receive an equal number. But the man found it quite impossible, no matter which
basket he selected and no matter how many children he included in the treat.
Show, by giving contents of the nine baskets, how this could come about.
MANDARIN'S "T" PUZZLE.
Before Mr. Beauchamp Cholmondely Marjoribanks set out on his tour in the Far
East, he prided himself on his knowledge of magic squares, a subject that he had
made his special hobby; but he soon discovered that he had never really touched
more than the fringe of the subject, and that the wily Chinee could beat him easily. I
present a little problem that one learned mandarin propounded to our traveller,
as depicted on the last page.
The Chinaman, after remarking that the construction of the ordinary magic
square of twenty-five cells is "too velly muchee easy," asked our countryman so
to place the numbers 1 to 25 in the square that every column, every row, and
each of the two diagonals should add up 65, with only prime numbers on the
shaded "T." Of course the prime numbers available are 1, 2, 3, 5, 7, 11, 13, 17,
19, and 23, so you are at liberty to select any nine of these that will serve
your purpose. Can you construct this curious little magic square?
MAGIC SQUARE OF COMPOSITES.
As we have just discussed the construction of magic squares with prime
numbers, the following forms an interesting companion problem. Make a magic
square with nine consecutive composite numbers—the smallest possible.
MAGIC KNIGHT'S TOUR.
Here is a problem that has never yet been solved, nor has its impossibility
been demonstrated. Play the knight once to every square of the chessboard in a
complete tour, numbering the squares in the order visited, so that when
completed the square shall be "magic," adding up to 260 in every column, every
row, and each of the two long diagonals. I shall give the best answer that I
have been able to obtain, in which there is a slight error in the diagonals
alone. Can a perfect solution be found? I am convinced that it cannot, but it is
only a "pious opinion."