THE BASKETS OF PLUMS.
(
Magic Squares Problem.)
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.
Answer:
As the merchant told his man to distribute the contents of one of the
baskets of plums "among some children," it would not be permissible to
give the complete basketful to one child; and as it was also directed
that the man was to give "plums to every child, so that each should
receive an equal number," it would also not be allowed to select just as
many children as there were plums in a basket and give each child a
single plum. Consequently, if the number of plums in every basket was a
prime number, then the man would be correct in saying that the proposed
distribution was quite impossible. Our puzzle, therefore, resolves
itself into forming a magic square with nine different prime numbers.
A B
+-----+-----+-----+ +-----+-----+-----+
| | | | | | | |
| 7 | 61 | 43 | | 83 | 29 | 101 |
|_____|_____|_____| |_____|_____|_____|
| | | | | | | |
| 73 | 37 | 1 | | 89 | 71 | 53 |
|_____|_____|_____| |_____|_____|_____|
| | | | | | | |
| 31 | 13 | 67 | | 41 | 113 | 59 |
| | | | | | | |
+-----+-----+-----+ +-----+-----+-----+
C D
+-----+-----+-----+ +-----+-----+-----+
| | | | | | | |
| 103 | 79 | 37 | |1669 | 199 |1249 |
|_____|_____|_____| |_____|_____|_____|
| | | | | | | |
| 7 | 73 | 139 | | 619 |1039 |1459 |
|_____|_____|_____| |_____|_____|_____|
| | | | | | | |
| 109 | 67 | 43 | | 829 |1879 | 409 |
| | | | | | | |
+-----+-----+-----+ +-----+-----+-----+
In Diagram A we have a magic square in prime numbers, and it is the one
giving the smallest constant sum that is possible. As to the little trap
I mentioned, it is clear that Diagram A is barred out by the words
"every basket contained plums," for one plum is not plums. And as we
were referred to the baskets, "as shown in the illustration," it is
perfectly evident, without actually attempting to count the plums, that
there are at any rate more than 7 plums in every basket. Therefore C is
also, strictly speaking, barred. Numbers over 20 and under, say, 250
would certainly come well within the range of possibility, and a large
number of arrangements would come within these limits. Diagram B is one
of them. Of course we can allow for the false bottoms that are so
frequently used in the baskets of fruitsellers to make the basket appear
to contain more fruit than it really does.
Several correspondents assumed (on what grounds I cannot think) that in
the case of this problem the numbers cannot be in consecutive
arithmetical progression, so I give Diagram D to show that they were
mistaken. The numbers are 199, 409, 619, 829, 1,039, 1,249, 1,459,
1,669, and 1,879--all primes with a common difference of 210.
