379.—­THE FIVE DOMINOES.

[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?

380.—­THE DOMINO FRAME PUZZLE.

[Illustration]

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.

381.—­THE CARD FRAME PUZZLE.

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.

[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.