2, 3, 4, 5, in
succession will show that we may continue this "winding up" process for
ever; and as there will always be an unobstructed way (however long and
circuitous) from stations B and E to their respective main lines, it is
evident that the number of routes for line A alone is infinite. Therefore
the number of complete solutions must also be infinite, if railway lines,
like other lines, have no breadth; and indeterminate, unless we are told
the greatest number of parallel lines that it is possible to construct in
certain places. If some clear condition, restricting these "windings up,"
were given, there would be no great difficulty in giving the number of
solutions. With any reasonable limitation of the kind, the number would,
I calculate, be little short of two thousand, surprising though it may
appear.

81.--_The Eight Clowns._

This is a little novelty in magic squares. These squares may be formed
with numbers that are in arithmetical progression, or that are not in
such progression. If a square be formed of the former class, one place
may be left vacant, but only under particular conditions. In the case of
our puzzle there would be no difficulty in making the magic square with 9
missing; but with 1 missing (that is, using 2, 3, 4, 5, 6, 7, 8, and 9)
it is not possible. But a glance at the original illustration will show
that the numbers we have to deal with are not actually those just
mentioned. The clown that has a 9 on his body is portrayed just at the
moment when two balls which he is juggling are in mid-air. The positions
of these balls clearly convert his figure into the recurring decimal
.[.9]. Now, since the recurring decimal .[.9] is equal to 9/9, and
therefore to 1, it is evident that, although the clown who bears the
figure 1 is absent, the man who bears the figure 9 by this simple
artifice has for the occasion given his _figure_ the value of the
_number_ 1. The troupe can consequently be grouped in the following
manner:--

7 5
2 4 6
3 8 .[.9]

Every column, every row, and each of the two diagonals now add up to 12.
This is the correct solution to the puzzle.

82.--_The Wizard's Arithmetic._

This puzzle is both easy and difficult, for it is a very simple matter to
find one of the multipliers, which is 86. If we multiply 8 by 86, all we
need do is to place the 6 in front and the 8 behind in order to get the
correct answer, 688. But the second number is not to be fo