compelled to continue with 1—2, 2—3, 3—4, 4—5, 5—6. There are three dominoes that can never be used at all. These are 0—5, 0—6, and 1—6. If we used a box of dominoes extending to 9—9, there would be forty different ways.
379.—THE FIVE DOMINOES.
There are just ten different ways of arranging the dominoes. Here is one of them:—
(2—0) (0—0) (0—1) (1—4) (4—0).
I will leave my readers to find the remaining nine for themselves.
380.— THE DOMINO FRAME PUZZLE.
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The illustration is a solution. It will be found that all four sides of the frame add up 44. The sum of the pips on all the dominoes is 168, and if we wish to make the sides sum to 44, we must take care that the four comers sum to 8, because these corners are counted twice, and 168 added to 8 will equal 4 times 44, which is necessary. There are many different solutions.
Even in the example given certain interchanges are possible to produce different arrangements. For example, on the left-hand side the string of dominoes from 2—2 down to 3—2 may be reversed, or from 2—6 to 3—2, or from 3—0 to 5—3. Also, on the right-hand side we may reverse from 4—3 to 1—4. These changes will not affect the correctness of the solution.
381.—THE CARD FRAME PUZZLE.
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 comer 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 comers. We need then only try different arrangements with the four corners always summing to 17, and we soon discover the following solution:—
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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 inter-changed, 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.
382.—THE CROSS OF CARDS.
There are eighteen fundamental arrangements, as follows, where I only give the numbers in the horizontal bar, since the remainder must naturally fall into their places.
| 5 | 6 | 1 | 7 | 4 | 2 | 4 | 5 | 6 | 8 |
| 3 | 5 | 1 | 6 | 8 | 3 | 4 | 5 | 6 | 7 |
| 3 | 4 | 1 | 7 | 8 | 1 | 4 | 7 | 6 | 8 |
| 2 | 5 | 1 | 7 | 8 | 2 | 3 | 7 | 6 | 8 |
| 2 | 5 | 3 | 6 | 8 | 2 | 4 | 7 | 5 | 8 |
| 1 | 5 | 3 | 7 | 8 | 3 | 4 | 9 | 5 | 6 |
| 2 | 4 | 3 | 7 | 8 | 2 | 4 | 9 | 5 | 7 |
| 1 | 4 | 5 | 7 | 8 | 1 | 4 | 9 | 6 | 7 |
| 2 | 3 | 5 | 7 | 8 | 2 | 3 | 9 | 6 | 7 |
It will be noticed that there must always be an odd number in the centre, that there are four ways each of adding up 23, 25, and 27, but only three ways each of summing to 24 and 26.
