‘Twenty-one players participated.’

‘Suppose that m men played for Doomshire, and n men played for Gloomshire.

Then we have:

3n/5 + (100-m)/3 + (2m-2n)/3 = 51

i.e. 5m-n=265 and hence n=5 (m-53).

Now n must be more than 50; m must be greater than n; and (100-m) must be divisible by three.

Having regard to these considerations, there is only one solution of the above equation: n=55, m=64.

So 64 men played for Doomshire.’

‘Gambitt should aim at securing nine points.

There cannot be four others with a higher score than this; whereas, if Gambitt only scores 8½ points, scores of 9½-9-9-9 are possible.’

‘It is obvious from Mrs Orme’s comments that the required number is one of four digits, and that Ronnie has assumed the number to be

bacdwhere in fact it isabcd.Trial and error does the rest – facilitated, no doubt, by acquaintance with the theory of numbers.

The numbers are 5760 (24 pairs of factors) and 7560 (32 pairs of factors) and the Ormes’s number is 7560.’

‘There are 64 letters in the cipher, which, taken with the chess “problem” is a strong hint that the cipher depends on a chessboard arrangement. Write the letters, therefore, in the form of a chessboard, ignoring the division into “words”. Then draw on transparent paper a chessboard of similar size and mark the squares referred to in the chess “problem”. Lay this over the gridded arrangement of letters, and read off the first 16 letters of the message, lying under the marked squares. Now rotate the diagram through a right angle, and read off the next 16 letters, and so on until the whole message appears.

The message is thus shown to be: “Moti Lal preparing attack station with three thousand Shaitani first April”.’

‘The title and the form of the cryptogram suggest the knight’s move in chess. The first word is in the bottom right-hand corner, and the words are picked out according to the following table:

The question then reads:

“Summa aetatium Mariae et Annae est quattuor et quadraginta anni. Maria est bis tantum annorum nata quantum Anna fuit cum Maria fuit dimidio tantum annorum nata quantum Anna erit cum Anna erit ter tantum annorum nata quantum Maria fuit cum Maria fuit ter tantum annorum nata quantum Anna fuit.

Quot annos nata est Maria?”

This is the old age problem, invented by Sam Loyd:

“The combined ages of Mary and Ann are 44 years. Mary is twice as old as Ann was when Mary was half as old as Ann will be when Ann is three times as old as Mary was when Mary was three times as old as Ann.

How old is Mary?”

The answer, which can readily be obtained by simple equations, is that Mary is 27½ and Ann is 16½.’