on all four sides to add up alike, but they failed in their attempt and
gave it up as impossible. It will be seen that the pips in the top row,
the bottom row, and the left-hand side all add up 14, but the right-hand
side sums to 23. Now, what they were trying to do is quite possible. Can
you rearrange the ten cards in the same formation so that all four sides
shall add up alike? Of course they need not add up 14, but any number
you choose to select.

[Illustration]


382.--THE CROSS OF CARDS.

[Illustration]

In this case we use only nine cards--the ace to nine of diamonds. The
puzzle is to arrange them in the form of a cross, exactly in the way
shown in the illustration, so that the pips in the vertical bar and in
the horizontal bar add up alike. In the example given it will be found
that both directions add up 23. What I want to know is, how many
different ways are there of rearranging the cards in order to bring
about this result? It will be seen that, without affecting the solution,
we may exchange the 5 with the 6, the 5 with the 7, the 8 with the 3,
and so on. Also we may make the horizontal and the vertical bars change
places. But such obvious manipulations as these are not to be regarded
as different solutions. They are all mere variations of one fundamental
solution. Now, how many of these fundamentally different solutions are
there? The pips need not, of course, always add up 23.