The sum of all the pips on the ten cards is 55. Suppose we are trying to
get 14 pips on every side. Then 4 times 14 is 56. But each of the four
corner cards is added in twice, so that 55 deducted from 56, or 1, must
represent the sum of the four corner cards. This is clearly impossible;
therefore 14 is also impossible. But suppose we came to trying 18. Then
4 times 18 is 72, and if we deduct 55 we get 17 as the sum of the
corners. We need then only try different arrangements with the four
corners always summing to 17, and we soon discover the following
solution:--
[Illustration:
+-------+-------+-------+
| 2 | 10 | 6 |
+---+---+------ +---+---+
| | | |
| 3 | | 7 |
| | | |
+---+ +---+
| | | |
| 8 | | 1 |
| | | |
+---+---+-------+--+----+
| 5 | 9 | 4 |
+-------+-------+-------+
]
The final trials are very limited in number, and must with a little
judgment either bring us to a correct solution or satisfy us that a
solution is impossible under the conditions we are attempting. The two
centre cards on the upright sides can, of course, always be
interchanged, but I do not call these different solutions. If you
reflect in a mirror you get another arrangement, which also is not
considered different. In the answer given, however, we may exchange the
5 with the 8 and the 4 with the 1. This is a different solution. There
are two solutions with 18, four with 19, two with 20, and two with
22--ten arrangements in all. Readers may like to find all these for
themselves.