Part 105

Here is the best answer, in eleven manipulations:--

Treasure down. Boy down--treasure up. Youth down--boy up. Treasure down. Man down--youth and treasure up. Treasure down. Boy down--treasure up. Treasure down. Youth down--boy up. Boy down--treasure up. Treasure down.

378.--DOMINOES IN PROGRESSION.

There are twenty-three different ways. You may start with any domino, except the 4--4 and those that bear a 5 or 6, though only certain initial dominoes may be played either way round. If you are given the common difference and the first domino is played, you have no option as to the other dominoes. Therefore all I need do is to give the initial domino for all the twenty-three ways, and state the common difference. This I will do as follows:--

With a common difference of 1, the first domino may be either of these: 0--0, 0--1, 1--0, 0--2, 1--1, 2--0, 0--3, 1--2, 2--1, 3--0, 0--4, 1--3, 2--2, 3--1, 1--4, 2--3, 3--2, 2--4, 3--3, 3--4. With a difference of 2, the first domino may be 0--0, 0--2, or 0--1. Take the last case of all as an example. Having played the 0--1, and the difference being 2, we are 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.

+---+-------+-------+-------+-------+-------+-------+-------+ | 2 | 2 | 5 | 5 | 6 | 6 | 6 | 6 | 1 | 1 | | | | | 4 | | - +-------+-------+-------+-------+-------+-------+---+---+ | 2 | | 4 | +---+ | - | | 2 | | 3 | | - | +---+ | 6 | | 3 | +---+ T H E | - | | 6 | | 3 | | - | +---+ | 3 | | 3 | +---+ | - | | 3 | | 1 | | - | D O M I N O F R A M E +---+ | | | 1 | +---+ | - | | | | 1 | | - | +---+ | 5 | | 1 | +---+ -S-O-L-U-T-I-O-N- | - | | 5 | | 4 | | - | +---+ | 3 | | 4 | +---+ | - | | 3 | | 6 | | - | +---+ | 2 | | 6 | +---+---+-------+-------+-------+-------+-------+-------+ - | | 2 | 1 | 1 | 5 | 5 | 5 | 5 | 4 | 4 | 4 | 4 | 2 | 2 | | | +-------+-------+-------+-------+-------+-------+-------+---+

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 corners 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 corner 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 corners. We need then only try different arrangements with the four corners always summing to 17, and we soon discover the following solution:--

+-------+-------+-------+ | 2 | 10 | 6 | +---+---+------ +---+---+ | | | | | 3 | | 7 | | | | | +---+ +---+ | | | | | 8 | | 1 | | | | | +---+---+-------+--+----+ | 5 | 9 | 4 | +-------+-------+-------+

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 interchanged, 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.

383.--THE "T" CARD PUZZLE.

If we remove the ace, the remaining cards may he divided into two groups (each adding up alike) in four ways; if we remove 3, there are three ways; if 5, there are four ways; if 7, there are three ways; and if we remove 9, there are four ways of making two equal groups. There are thus eighteen different ways of grouping, and if we take any one of these and keep the odd card (that I have called "removed") at the head of the column, then one set of numbers can be varied in order in twenty-four ways in the column and the other four twenty-four ways in the horizontal, or together they may be varied in 24 x 24 = 576 ways. And as there are eighteen such cases, we multiply this number by 18 and get 10,368, the correct number of ways of placing the cards. As this number includes the reflections, we must divide by 2, but we have also to remember that every horizontal row can change places with a vertical row, necessitating our multiplying by 2; so one operation cancels the other.

384.--CARD TRIANGLES.

The following arrangements of the cards show (1) the smallest possible sum, 17; and (2) the largest possible, 23.

1 7 9 6 4 2 4 8 3 6 3 7 5 2 9 5 1 8

It will be seen that the two cards in the middle of any side may always be interchanged without affecting the conditions. Thus there are eight ways of presenting every fundamental arrangement. The number of fundamentals is eighteen, as follows: two summing to 17, four summing to 19, six summing to 20, four summing to 21, and two summing to 23. These eighteen fundamentals, multiplied by eight (for the reason stated above), give 144 as the total number of different ways of placing the cards.

385.--"STRAND" PATIENCE.

The reader may find a solution quite easy in a little over 200 moves, but, surprising as it may at first appear, not more than 62 moves are required. Here is the play: By "4 C up" I mean a transfer of the 4 of clubs with all the cards that rest on it. 1 D on space, 2 S on space, 3 D on space, 2 S on 3 D, 1 H on 2 S, 2 C on space, 1 D on 2 C, 4 S on space, 3 H on 4 S (9 moves so far), 2 S up on 3 H (3 moves), 5 H and 5 D exchanged, and 4 C on 5 D (6 moves), 3 D on 4 C (1), 6 S (with 5 H) on space (3), 4 C up on 5 H (3), 2 C up on 3 D (3), 7 D on space (1), 6 C up on 7 D (3), 8 S on space (1), 7 H on 8 S (1), 8 C on 9 D (1), 7 H on 8 C (1), 8 S on 9 H (1), 7 H on 8 S (1), 7 D up on 8 C (5), 4 C up on 5 D (9), 6 S up on 7 H (3), 4 S up on 5 H (7) = 62 moves in all. This is my record; perhaps the reader can beat it.