1 2 3 12 4 11 5 10 6 9 7 8 1 2 4 11 6 9 8 7 10 5 12 3 1 2 5 10 8 7 11 4 3 12 6 9 1 2 6 9 10 5 3 12 7 8 11 4 1 2 7 8 12 3 6 9 11 4 5 10

Here 1 is a repeater, and the cycle is 2, 3, 4, 5,... 12.  We thus get 5 groups of 11 lines each.

274.—­THE MOUSE-TRAP PUZZLE.

If we interchange cards 6 and 13 and begin our count at 14, we may take up all the twenty-one cards—­that is, make twenty-one “catches”—­in the following order:  6, 8, 13, 2, 10, 1, 11, 4, 14, 3, 5, 7, 21, 12, 15, 20, 9, 16, 18, 17, 19.  We may also exchange 10 and 14 and start at 16, or exchange 6 and 8 and start at 19.

275.—­THE SIXTEEN SHEEP.

The six diagrams on next page show solutions for the cases where we replace 2, 3, 4, 5, 6, and 7 hurdles.  The dark lines indicate the hurdles that have been replaced.  There are, of course, other ways of making the removals.

276.—­THE EIGHT VILLAS.

There are several ways of solving the puzzle, but there is very little difference between them.  The solver should, however, first of all bear in mind that in making his calculations he need only consider the four villas that stand at the corners, because the intermediate villas can never vary when the corners are known.  One way is to place the numbers nought to 9 one at a time in the top left-hand corner, and then consider each case in turn.

Now, if we place 9 in the corner as shown in the Diagram A, two of the corners cannot be occupied, while the corner that is diagonally opposite may be filled by 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 persons.  We thus see that there are 10

[Illustration: 

+---+---+  +-+-----+  +---+---+
|O OHO O|  |OHO O O|  |O OHO O|
|   H   |  | +     |  |   +=+ |
|O OHO O|  |OHO O O|  |O OHOHO|
+-+ +-+-+  +-+-----+  +---+ + |
|O|O O|O|  |O|O O O|  |O O O|O|
| +---+ |  | +-+-+ |  |   +-+ |
|O O O O|  |O O OHO|  |O O|O O|
+-------+  +-------+  +-------+
2          3          4

+-----+-+  +-+-----+  +-------+
|O O OHO|  |OHO O O|  |O O O O|
|   +=+ |  | +=+   |  | +=+=+=+
|O OHO O|  |OHOHO O|  |OHOHO O|
| +-+-+ +  | + +-+ |  + + +   |
|O|O O|O|  |O|O O|O|  |O|OHO O|
+=+   +=+  | +   +=+  +=+ +   |
|O O O O|  |OHO O O|  |O O|O O|
+-------+  +-+-----+  +---+---+
5          6          7
THE SIXTEEN SHEEP

]

solutions with a 9 in the corner.  If, however, we substitute 8, the two corners in the same row and column may contain 0, 0, or 1, 1, or 0, 1, or 1, 0.  In the case of B, ten different selections may be made for the fourth corner; but in each of the cases C, D, and E, only nine selections are possible, because we cannot use the 9.  Therefore with 8 in the top left-hand corner there are 10 + (3 x 9) = 37 different solutions.  If we then try 7 in the corner, the result will be 10 + 27 + 40, or 77 solutions.  With 6 we get 10 + 27 + 40 + 49 = 126; with 5, 10 + 27 + 40 + 49 + 54 = 180; with 4, the same as with 5,