Part 109

The illustration explains itself. It will be found that the pips in every column, row, and long diagonal add up 18, as required.

407.--TWO NEW MAGIC SQUARES.

Here are two solutions that fulfil the conditions:--

SUBTRACTING DIVIDING 11 4 14 13 36 8 54 27 16 7 1 2 216 12 1 2 6 5 3 12 6 3 4 72 9 19 8 15 9 18 24 108

The first, by subtracting, has a constant 8, and the associated pairs all have a difference of 4. The second square, by dividing, has a constant 9, and all the associated pairs produce 3 by division. These are two remarkable and instructive squares.

408.--MAGIC SQUARES OF TWO DEGREES.

The following is the square that I constructed. As it stands the constant is 260. If for every number you substitute, in its allotted place, its square, then the constant will be 11,180. Readers can write out for themselves the second degree square.

7 53 | 41 27 | 2 52 | 48 30 12 58 | 38 24 | 13 63 | 35 17 ------+-------+-------+------ 51 1 | 29 47 | 54 8 | 28 42 64 14 | 18 36 | 57 11 | 23 37 ------+-------+-------+------ 25 43 | 55 5 | 32 46 | 50 4 22 40 | 60 10 | 19 33 | 61 15 ------+-------+-------+------ 45 31 | 3 49 | 44 26 | 6 56 34 20 | 16 62 | 39 21 | 9 59

The main key to the solution is the pretty law that if eight numbers sum to 260 and their squares to 11,180, then the same will happen in the case of the eight numbers that are complementary to 65. Thus 1 + 18 + 23 + 26 + 31 + 48 + 56 + 57 = 260, and the sum of their squares is 11,180. Therefore 64 + 47 + 42 + 39 + 34 + 17 + 9 + 8 (obtained by subtracting each of the above numbers from 65) will sum to 260 and their squares to 11,180. Note that in every one of the sixteen smaller squares the two diagonals sum to 65. There are four columns and four rows with their complementary columns and rows. Let us pick out the numbers found in the 2nd, 1st, 4th, and 3rd rows and arrange them thus :--

1 8 28 29 42 47 51 54 2 7 27 30 41 48 52 53 3 6 26 31 44 45 49 56 4 5 25 32 43 46 50 55

Here each column contains four consecutive numbers cyclically arranged, four running in one direction and four in the other. The numbers in the 2nd, 5th, 3rd, and 8th columns of the square may be similarly grouped. The great difficulty lies in discovering the conditions governing these groups of numbers, the pairing of the complementaries in the squares of four and the formation of the diagonals. But when a correct solution is shown, as above, it discloses all the more important keys to the mystery. I am inclined to think this square of two degrees the most elegant thing that exists in magics. I believe such a magic square cannot be constructed in the case of any order lower than 8.

409.--THE BASKETS OF PLUMS.

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.

410.--THE MANDARIN'S "T" PUZZLE.

There are many different ways of arranging the numbers, and either the 2 or the 3 may be omitted from the "T" enclosure. The arrangement that I give is a "nasik" square. Out of the total of 28,800 nasik squares of the fifth order this is the only one (with its one reflection) that fulfils the "T" condition. This puzzle was suggested to me by Dr. C. Planck.

+-----+-----+-----+-----+-----+ | | | | | | | 19 | 23 | 11 | 5 | 7 | |____|___|___|___|___| | | | | | | | 1 | 10 | 17 | 24 | 13 | |___|___|___|___|___| | | | | | | | 22 | 14 | 3 | 6 | 20 | |___|___|___|___|___| | | | | | | | 8 | 16 | 25 | 12 | 4 | |___|___|___|___|____| | | | | | | | 15 | 2 | 9 | 18 | 21 | | | | | | | +-----+-----+-----+-----+-----+

411.--A MAGIC SQUARE OF COMPOSITES.

The problem really amounts to finding the smallest prime such that the next higher prime shall exceed it by 10 at least. If we write out a little list of primes, we shall not need to exceed 150 to discover what we require, for after 113 the next prime is 127. We can then form the square in the diagram, where every number is composite. This is the solution in the smallest numbers. We thus see that the answer is arrived at quite easily, in a square of the third order, by trial. But I propose to show how we may get an answer (not, it is true, the one in smallest numbers) without any tables or trials, but in a very direct and rapid manner.

+-----+-----+-----+ | | | | | 121 | 114 | 119 | |____|___|___| | | | | | 116 | 118 | 120 | |___|___|____| | | | | | 117 | 122 | 115 | | | | | +-----+-----+-----+

First write down any consecutive numbers, the smallest being greater than 1--say, 2, 3, 4, 5, 6, 7, 8, 9, 10. The only factors in these numbers are 2, 3, 5, and 7. We therefore multiply these four numbers together and add the product, 210, to each of the nine numbers. The result is the nine consecutive composite numbers, 212 to 220 inclusive, with which we can form the required square. Every number will necessarily be divisible by its difference from 210. It will be very obvious that by this method we may find as many consecutive composites as ever we please. Suppose, for example, we wish to form a magic square of sixteen such numbers; then the numbers 2 to 17 contain the factors 2, 3, 5, 7, 11, 13, and 17, which, multiplied together, make 510510 to be added to produce the sixteen numbers 510512 to 510527 inclusive, all of which are composite as before.

But, as I have said, these are not the answers in the smallest numbers: for if we add 523 to the numbers 1 to 16, we get sixteen consecutive composites; and if we add 1,327 to the numbers 1 to 25, we get twenty-five consecutive composites, in each case the smallest numbers possible. Yet if we required to form a magic square of a hundred such numbers, we should find it a big task by means of tables, though by the process I have shown it is quite a simple matter. Even to find thirty-six such numbers you will search the tables up to 10,000 without success, and the difficulty increases in an accelerating ratio with each square of a larger order.

412.--THE MAGIC KNIGHT'S TOUR.

+----+----+----+----+----+----+----+----+ | 46 | 55 | 44 | 19 | 58 | 9 | 22 | 7 | +----+----+----+----+----+----+----+----+ | 43 | 18 | 47 | 56 | 21 | 6 | 59 | 10 | +----+----+----+----+----+----+----+----+ | 54 | 45 | 20 | 41 | 12 | 57 | 8 | 23 | +----+----+----+----+----+----+----+----+ | 17 | 42 | 53 | 48 | 5 | 24 | 11 | 60 | +----+----+----+----+----+----+----+----+ | 52 | 3 | 32 | 13 | 40 | 61 | 34 | 25 | +----+----+----+----+----+----+----+----+ | 31 | 16 | 49 | 4 | 33 | 28 | 37 | 62 | +----+----+----+----+----+----+----+----+ | 2 | 51 | 14 | 29 | 64 | 39 | 26 | 35 | +----+----+----+----+----+----+----+----+ | 15 | 30 | 1 | 50 | 27 | 36 | 63 | 38 | +----+----+----+----+----+----+----+----+

Here each successive number (in numerical order) is a knight's move from the preceding number, and as 64 is a knight's move from 1, the tour is "re-entrant." All the columns and rows add up 260. Unfortunately, it is not a perfect magic square, because the diagonals are incorrect, one adding up 264 and the other 256--requiring only the transfer of 4 from one diagonal to the other. I think this is the best result that has ever been obtained (either re-entrant or not), and nobody can yet say whether a perfect solution is possible or impossible.