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?
[380.—THE DOMINO FRAME PUZZLE.]
It will be seen in the illustration that the full set of twenty-eight dominoes is arranged in the form of a square frame, with 6 against 6, 2 against 2, blank against blank, and so on, as in the game. It will be found that the pips in the top row and left-hand column both add up 44. The pips in the other two sides sum to 59 and 32 respectively. The puzzle is to rearrange the dominoes in the same form so that all of the four sides shall sum to 44. Remember that the dominoes must be correctly placed one against another as in the game.