Page:Amusements in mathematics.djvu/258

246 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

C D

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 comse we can allow for the false bottoms that are so fre- quently used in the baskets of fruitseUers 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 num- bers are 199, 409, 619, 829, 1,039, 1,249, i,459, 1,669, and 1,879 — all primes with a common difference of 210.

410.—THE MANDARIN'S "T" PUZZLE. are many different ways of arranging the numbers, and either the 2 or the 3 may be omitted from the "T" enclosure. The ar- rangement 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.

THE MANDARIN S PUZZLE. 411.—A MAGIC SQUARE OF COMPOSITES. 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 1131 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.

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