THE CANTERBURY PUZZLES

shown, from the numbers 3 and 4, which give the generators—37,

7 ; 37, 33 ; 37, 40. These three pairs of numbers solve the

indeterminate equation, a

3

b - b

3

a = 341,880. If we can find another

pair of values, the thing is done. These values are 56, 55, which

generators give the last triangle. The next best answer that I have

found is derived from 5 and 6, which give the generators, 91, 11 ;

91, 85 ; 91, 96. The fourth pair of values is 63, 42.

The reader will understand from what I have written above that

there is no limit to the number of rational sided R.A.T.'s of equal

area that may be found in whole numbers.

108.— Plato and the Nines.

The following is the simple solution of the three nines

puzzle :—

9 + 9 To divide 18 by '9 (or nine-tenths) we, of course,

*9 multiply by 10 and divide by 9. The result is 20,

as required.

109.—Noughts and Crosses.

The solution is as follows : Between two players who thoroughly

understand the play every game should be drawn. Neither player

could ever win except through the blundering of his opponent. If

Nought (the first player) takes the centre, Cross must take a corner,

or Nought may beat him with certainty. If Nought takes a corner

on his first play, Cross must take the centre at once, or again be

beaten with certainty. If Nought leads with a side, both players

must be very careful to prevent a loss, as there are numerous pitfalls.

But Nought may safely lead anything and secure a draw, and he can

only win through Cross's blunders.

110.—Ovid's Game.

The solution here is : The first player can always win, pro-

vided he plays to the centre on his first move. But a good variation

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