by Dudeney, Henry Ernest, 1857-1930
Available in 215 free installments
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If there had been only three couples, the island might have been dispensed with, but with four or more couples it is absolutely necessary in order to cross under the conditions laid down. It can be done in seventeen passages from land to land (though French mathematicians have declared in their books that in such circumstances twenty-four are needed), and it cannot be done in fewer. I will give one way. A, B, C, and D are the young men, and a, b, c, and d are the girls to whom they are respectively engaged. The three columns show the positions of the different individuals on the lawn, the island, and the opposite shore before starting and after each passage, while the asterisk indicates the position of the boat on every occasion.
Having found the fewest possible passages, we should consider two other points in deciding on the "quickest method": Which persons were the most expert in handling the oars, and which method entails the fewest possible delays in getting in and out of the boat? We have no data upon which to decide the first point, though it is probable that, as the boat belonged to the girls' household, they would be capable oarswomen. The other point, however, is important, and in the solution I have given (where the girls do 8-13ths of the rowing and A and D need not row at all) there are only sixteen gettings-in and sixteen gettings-out. A man and a girl are never in the boat together, and no man ever lands on the island. There are other methods that require several more exchanges of places.
377.? STEALING THE CASTLE TREASURE.? solution
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.? solution
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.
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379.?THE FIVE DOMINOES.? solution
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.? solution
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.? solution
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:?
