Part 31

No doubt some of my readers will smile at the statement that a man in a boat on smooth water can pull himself across with the tiller rope! But it is a fact. If the jester had fastened the end of his rope to the stern of the boat and then, while standing in the bows, had given a series of violent jerks, the boat would have been propelled forward. This has often been put to a practical test, and it is said that a speed of two or three miles an hour may be attained. See W. W. Rouse Ball's Mathematical Recreations.

53.--The Royal Gardens.

This puzzle must have struck many readers as being absolutely impossible. The jester said: "I had, of a truth, entered every one of the sixteen gardens once, and never more than once." If we follow the route shown in the accompanying diagram, we find that there is no difficulty in once entering all the gardens but one before reaching the last garden containing the exit B. The difficulty is to get into the garden with a star, because if we leave the B garden we are compelled to enter it a second time before escaping, and no garden may be entered twice. The trick consists in the fact that you may enter that starred garden without necessarily leaving the other. If, when the jester got to the gateway where the dotted line makes a sharp bend, his intention had been to hide in the starred garden, but after he had put one foot through the doorway, upon the star, he discovered it was a false alarm and withdrew, he could truly say: "I entered the starred garden, because I put my foot and part of my body in it; and I did not enter the other garden twice, because, after once going in I never left it until I made my exit at B." This is the only answer possible, and it was doubtless that which the jester intended.

See "The Languishing Maiden," in A. in M.

54.--Bridging the Ditch.

The solution to this puzzle is best explained by the illustration. If he had placed his eight planks, in the manner shown, across the angle of the ditch, he would have been able to cross without much trouble. The king's jester might thus have well overcome all his difficulties and got safely away, as he has told us that he succeeded in doing.

THE SQUIRE'S CHRISTMAS PUZZLE PARTY

HOW THE VARIOUS TRICKS WERE DONE

The record of one of Squire Davidge's annual "Puzzle Parties," made by the old gentleman's young lady relative, who had often spent a merry Christmas at Stoke Courcy Hall, does not contain the solutions of the mysteries. So I will give my own answers to the puzzles and try to make them as clear as possible to those who may be more or less novices in such matters.

55.--The Three Teacups.

Miss Charity Lockyer clearly must have had a trick up her sleeve, and I think it highly probable that it was conceived on the following lines. She proposed that ten lumps of sugar should be placed in three teacups, so that there should be an odd number of lumps in every cup. The illustration perhaps shows Miss Charity's answer, and the figures on the cups indicate the number of lumps that have been separately placed in them. By placing the cup that holds one lump inside the one that holds two lumps, it can be correctly stated that every cup contains an odd number of lumps. One cup holds seven lumps, another holds one lump, while the third cup holds three lumps. It is evident that if a cup contains another cup it also contains the contents of that second cup.

There are in all fifteen different solutions to this puzzle. Here they are:--

1 0 9 1 4 5 9 0 1 3 0 7 7 0 3 7 2 1 1 2 7 5 2 3 5 4 1 5 0 5 3 4 3 3 6 1 3 2 5 1 6 3 1 8 1

The first two numbers in a triplet represent respectively the number of lumps to be placed in the inner and outer of the two cups that are placed one inside the other. It will be noted that the outer cup of the pair may itself be empty.

56.--The Eleven Pennies.

It is rather evident that the trick in this puzzle was as follows:--From the eleven coins take five; then add four (to those already taken away) and you leave nine--in the second heap of those removed!

57.--The Christmas Geese.

Farmer Rouse sent exactly 101 geese to market. Jabez first sold Mr. Jasper Tyler half of the flock and half a goose over (that is, 50-1/2 + 1/2, or 51 geese, leaving 50); he then sold Farmer Avent a third of what remained and a third of a goose over (that is, 16-2/3 + 1/3, or 17 geese, leaving 33); he then sold Widow Foster a quarter of what remained and three-quarters of a goose over (that is, 8-1/4 + 3/4 or 9 geese, leaving 24); he next sold Ned Collier a fifth of what he had left and gave him a fifth of a goose "for the missus" (that is, 4-4/5 + 1/5 or 5 geese, leaving 19). He then took these 19 back to his master.

58.--The Chalked Numbers.

This little jest on the part of Major Trenchard is another trick puzzle, and the face of the roguish boy on the extreme right, with the figure 9 on his back, showed clearly that he was in the secret, whatever that secret might be. I have no doubt (bearing in mind the Major's hint as to the numbers being "properly regarded") that his answer was that depicted in the illustration, where boy No. 9 stands on his head and so converts his number into 6. This makes the total 36--an even number--and by making boys 3 and 4 change places with 7 and 8, we get 1278 and 5346, the figures of which, in each case, add up to 18. There are just three other ways in which the boys may be grouped: 1368--2457, 1467--2358, and 2367--1458.

59.--Tasting the Plum Puddings.

The diagram will show how this puzzle is to be solved. It is the only way within the conditions laid down. Starting at the pudding with holly at the top left-hand corner, we strike out all the puddings in twenty-one straight strokes, taste the steaming hot pudding at the end of the tenth stroke, and end at the second sprig of holly.

Here we have an example of a chess rook's path that is not re-entrant, but between two squares that are at the greatest possible distance from one another. For if it were desired to move, under the condition of visiting every square once and once only, from one corner square to the other corner square on the same diagonal, the feat is impossible.

There are a good many different routes for passing from one sprig of holly to the other in the smallest possible number of moves--twenty-one--but I have not counted them. I have recorded fourteen of these, and possibly there are more. Any one of these would serve our purpose, except for the condition that the tenth stroke shall end at the steaming hot pudding. This was introduced to stop a plurality of solutions--called by the maker of chess problems "cooks." I am not aware of more than one solution to this puzzle; but as I may not have recorded all the tours, I cannot make a positive statement on the point at the time of writing.

60.--Under the Mistletoe Bough.

Everybody was found to have kissed everybody else once under the mistletoe, with the following additions and exceptions: No male kissed a male; no man kissed a married woman except his own wife; all the bachelors and boys kissed all the maidens and girls twice; the widower did not kiss anybody, and the widows did not kiss each other. Every kiss was returned, and the double performance was to count as one kiss. In making a list of the company, we can leave out the widower altogether, because he took no part in the osculatory exercise.

7 Married couples 14 3 Widows 3 12 Bachelors and Boys 12 10 Maidens and Girls 10 Total 39 Persons

Now, if every one of these 39 persons kissed everybody else once, the number of kisses would be 741; and if the 12 bachelors and boys each kissed the 10 maidens and girls once again, we must add 120, making a total of 861 kisses. But as no married man kissed a married woman other than his own wife, we must deduct 42 kisses; as no male kissed another male, we must deduct 171 kisses; and as no widow kissed another widow, we must deduct 3 kisses. We have, therefore, to deduct 42+171+3=216 kisses from the above total of 861, and the result, 645, represents exactly the number of kisses that were actually given under the mistletoe bough.