[Illustration]
Here is a new little puzzle that is not difficult, but will probably be
found entertaining by my readers. It will be seen that the five dominoes
are so arranged in proper sequence (that is, with 1 against 1, 2 against
2, and so on), that the total number of pips on the two end dominoes is
five, and the sum of the pips on the three dominoes in the middle is
also five. There are just three other arrangements giving five for the
additions. They are: --
(1--0) (0--0) (0--2) (2--1) (1--3)
(4--0) (0--0) (0--2) (2--1) (1--0)
(2--0) (0--0) (0--1) (1--3) (3--0)
Now, how many similar arrangements are there of five dominoes that shall
give six instead of five in the two additions?