1 2 3 4 5 6 7 8
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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