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.

[Illustration:  THE MANDARIN’S “T” PUZZLE.

+-----+-----+-----+-----+-----+
|     |     |     |     |     |
|  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.

[Illustration]

+-----+-----+-----+
|     |     |     |
| 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.