Amusements in Mathematics HTML version

Remember that the dominoes must be correctly placed one against another as in the
In the illustration we have a frame constructed from the ten playing cards, ace to ten of
diamonds. The children who made it wanted the pips 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.
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.
An entertaining little puzzle with cards is to take the nine cards of a suit, from ace to nine
inclusive, and arrange them in the form of the letter "T," as shown in the illustration, so
that the pips in the horizontal line shall count the same as those in the column. In the
example given they add up twenty-three both ways. Now, it is quite easy to get a single
correct arrangement. The puzzle is to discover in just how many different ways it may be
done. Though the number is high, the solution is not really difficult if we attack the
puzzle in the right manner. The reverse way obtained by reflecting the illustration in a
mirror we will not count as different, but all other changes in the relative positions of the
cards will here count. How many different ways are there?
Here you pick out the nine cards, ace to nine of diamonds, and arrange them in the form
of a triangle, exactly as shown in the illustration, so that the pips add up the same on the
three sides. In the example given it will be seen that they sum to 20 on each side, but the
particular number is of no importance so long as it is the same on all three sides. The
puzzle is to find out in just how many different ways this can be done.