Part 14
[Illustration: Fig. 16.--The Accommodating Square.]
THE MAGIC CROSS.
Take three pieces of cardboard of the shape of the figure numbered 1 in Fig. 17, A, and one piece each of the shapes of 2 and 3. The pieces may be of any size, but it is hardly necessary to say that relatively each one must correspond with the sizes and shapes indicated in the diagram. Fig. 17, B, shows the pieces when put together and forming the cross.
[Illustration: Fig. 17.--a, The Pieces. b, The Cross.]
TO TAKE A MAN'S WAISTCOAT OFF WITHOUT REMOVING HIS COAT.
This puzzle is almost good enough to be included among conjuring tricks, but as there is neither magic nor sleight of hand involved, there is no alternative but to place it here. The puzzle seems ridiculous and unreasonable, as in performing it neither the coat nor vest may be torn, cut, or damaged, nor may either arm be removed from the sleeve of the coat. The puzzle cannot always be performed, as it depends upon the size of the coat-sleeves allowed by the fashions of the day, though as a rule a coat with suitable sleeves will be found in most households. The person whose waistcoat has to be removed should be the wearer of a coat the sleeves of which are sufficiently large at the wrist to admit of the hand of the operator being passed up and through them. Any person undertaking to perform the puzzle in a drawing-room should first request some one of the company to remove his evening coat, and to replace it by a light spring overcoat; this being done, it will be easy to carry out the following instructions: The waistcoat should first be unbuttoned in the front, and then the buckle at the back must be unloosed. The operator, standing in front of the person operated upon, should then place his hands underneath the coat at the back, taking hold of the bottom of the waistcoat, at the same time requesting the wearer to extend his arms at full length over his head. Now raise the bottom part of the waistcoat over the head of the wearer (if the waistcoat be tight it will be necessary to force it a little, but this must not be minded so long as the waistcoat is not torn); the waistcoat then will have been brought to the front of the wearer, across his chest. Take the _right_ side bottom-end of the waistcoat, and put it into the arm-hole of the coat at the shoulder, at the same time putting the hand up the sleeve, seizing the end, and drawing it down the sleeve; this action will release one arm-hole of the garment to be removed. The next thing to be done is to pull the waistcoat back again out of the sleeve of the coat, and put the _same end_ of the waistcoat into the _left_ arm-hole of the coat, again putting the hand up the sleeve of the coat as before, and seizing the end of the garment. It may then be drawn quite through the sleeve, and the puzzle is accomplished.
TO BREAK A STONE WITH A BLOW OF THE FIST.
To do this two stones are required, each one of which should be from three to six inches in length, and about half as thick. Place one of the stones flat, firmly and immovably, upon the ground, and on it place one end of the other stone, raising the opposite end to an angle of something like forty-five degrees, and just over the centre of the lower stone, with which it must form a T, being kept in that position by a piece of twig or stick of the necessary length. The top or elevated stone should then be smartly struck at about the centre with the little-finger side of the hand; the stick, of course, will give way, and the bottom stone will be broken to pieces.
THE KEY, THE HEART, AND THE DART.
This is a very old-fashioned puzzle, and easy of accomplishment to those who know how to do it. The puzzle is either to arrange the three articles in an apparently inextricable manner, or, if they are so arranged, to separate them without damaging either, or bending the cardboard out of which they should be made.
Cut out of some tough and elastic cardboard a double-headed dart, a key, small at the ring end, and a heart, in which should be cut four angular slits, shaped as in Fig. 18. To arrange them together, the lowermost cut in the heart must be pressed out so that it will form a loop, through which the ring end of the key has to be drawn, and so that one end of the dart may also be passed through without breaking the cardboard. Then fold the dart in the middle, so that one of its heads shall accurately fit upon the other head; bring the loop of the heart back into its former position, drawing it out of the ring of the key, which should then glide down the shaft of the dart, and hang fast held by the head. To disentangle the articles, reverse the order of procedure.
[Illustration: Fig. 18.]
THE PRISONERS' RELEASE PUZZLE.
Take two pieces of string or tape, and round the wrists of two persons tie the string, as shown in Fig. 19. It adds to the amusement of the puzzle if one of the persons is a lady and the other a gentleman. The puzzle is for them to liberate themselves, or for any one else to release them without untying the string. To do this, B makes a loop of his string pass under either of A's manacles, slips it over A's hands, and both will be free. Reverse the proceeding, and the manacles are again as before.
[Illustration: Fig. 19.--The Prisoners' Release Puzzle.]
As a finish to the Mechanical Puzzles, we will give the key to the world-renowned
HAMPTON COURT MAZE.
Upon entering the maze, turn to the right; afterwards, whenever there is a choice between the left and right, turn to the left, and the centre will soon be reached. Reverse the process in coming out.
ARITHMETICAL PUZZLES.
Under this heading we propose to give some arithmetical puzzles, to speak of the power of different numbers, to show some of the curious combinations of which numbers are capable, and generally to give such examples as our space will admit to explain how the science of numbers may be made to do service for our amusement.
Among the most popular of number puzzles are the
AMERICAN PUZZLES "15" AND "34,"
which have been christened "Boss." The materials of the puzzles are very simple, a description that may indeed be applied to all the amusements dealt with in this section. The puzzles, as purchased, consist of a square box of sixteen small wooden cubes, numbered from 1 to 16. The box of cubes may be purchased in the streets for a very trifling sum, or it may be obtained in the toy-shops in a more elaborate form, but still at a small cost. The popularity of the game may be guessed from the statement made by a New York toy-dealer to the effect that in one day he disposed of no less than 230 gross of a cheap variety. In London, street toy-vendors by the score sold them all day long for weeks together when they were first introduced, and a leading toy-dealer in the fashionable neighbourhood of Regent Street says the number sold retail from his shop daily was enormous. Their popularity in other countries is equally great.
The puzzle is twofold, and is described in the following quaint and curt manner in the little boxes sold in the streets:--
_The Puzzle of Fifteen._--"Remove the 16 block. Put the pieces in the box irregularly, and arrange them to regular order by shoving."
_The Magic Sixteen, or the Puzzle of Thirty-four._--"Arrange the sixteen blocks so that the sum of the numbers added up in any straight line, either vertical, horizontal, or diagonal, will be 34."
[Illustration:
+----+----+----+----+ | 1 | 14 | 15 | 4 | +----+----+----+----+ | 8 | 11 | 10 | 5 | +----+----+----+----+ | 12 | 7 | 6 | 9 | +----+----+----+----+ | 13 | 2 | 3 | 16 | +----+----+----+----+
Fig. 1.--A Solution of the "34" Puzzle.]
[Illustration:
+----+----+----+----+ | 1 | 15 | 14 | 4 | +----+----+----+----+ | 12 | 6 | 7 | 9 | +----+----+----+----+ | 8 | 10 | 11 | 5 | +----+----+----+----+ | 13 | 3 | 2 | 16 | +----+----+----+----+
Fig. 2.--Another Solution of the "34" Puzzle.]
It would appear that the "15" puzzle has the merit of being entirely new, a claim to which the "34" puzzle has no sort of right, it being found in many books of old and recent date. It is believed that there are in all sixteen different ways of arranging the numbered blocks so that the sum of the numbers will be 34 in every direction; but two ways will suffice to quote here, and they are as shown in Figs. 1, 2. The fascination and popularity of "Boss," however, all centre around the "15" puzzle; it is the solution of that which is said to have sent some people mad, to have made more forsake their ordinary occupation, and which claims to have given to a still larger and ever growing number of human beings a new incentive to life. The puzzle is fairly stated above in the words, "Put the pieces in the box irregularly," &c. As a first attempt, however, place the pieces as arranged when the "34" puzzle has been solved, and the "15" puzzle may be easily accomplished after a little practice. To describe the various moves would be unnecessary, but the object first to be aimed at is to get the first row of cubes, viz., 1, 2, 3, 4, into their proper places, attention being next directed to getting the 12 cube into its place; that cube will have to be again moved before all the cubes have been consecutively arranged, but it should always be kept as near to its proper position as possible. The cubes, when arranged, should read as follows (Fig. 3):--
[Illustration:
+-----------------+ | 1 2 3 4 | | | | 5 6 7 8 | | | | 9 10 11 12 | | | | 13 14 15 | +-----------------+
Fig. 3.--"15" Puzzle--The Cubes in Order.]
[Illustration:
+-----------------+ | 1 2 3 4 | | | | 5 6 7 8 | | | | 9 10 11 12 | | | | 13 15 14 | +-----------------+
Fig. 4.--"15" Puzzle--The Cubes set for Solution.]
"Boss," or the real American puzzle of "15" is to place the numbered cubes, as shown in Fig. 4, in the box, and then to arrange them, by sliding and without lifting any one cube, so that they shall read consecutively. It may at once be said that the American puzzle has never yet been solved. But why? is asked by every one, and every one tries to solve it. Articles on the puzzle have appeared in many periodicals, but no one has had the hardihood to publish a solution of the American puzzle. An ingenious calculator has stated that the fifteen cubes may be arranged in the box in 1,307,674,318,000 different combinations, and that it would take one individual a whole year to work out 105,000 of these arrangements, if only one arrangement was worked out every five minutes. Let the reader calculate at what remote period the whole of the different orders could be tested to see whether the "15-14" combination could be overcome. It seems to have been decided that there are a certain number of the combinations that can be solved, and that there are a certain number that cannot, and that the number of each is equal. If, when the fifteen cubes are placed in the box, the number of transpositions required to place the cubes in proper consecutive order is even, the puzzle may be solved; but if the number of transpositions required is odd, the puzzle cannot be solved. For example: take the first solution of the "34" puzzle (Fig. 1), and it will be found that six transpositions are required to place the numbers in the proper order, viz.:--
1. Transpose 14 and 2 4. Transpose 11 and 6 2. " 15 " 3 5. " 10 " 7 3. " 8 " 5 6. " 12 " 9
The number of transpositions being even, the puzzle is soluble; with the "15-14" order, there being only one transposition necessary, or an odd number, the puzzle is insoluble. With this information and a little practice any player may tell at a glance when any combination of the figures is shown whether the puzzle is soluble or no.
After the above lengthy dissertation on these clever puzzles we will now proceed to minor topics which may be treated as arithmetical amusements.
THE MAGIC NINE, OR THE PUZZLE OF FIFTEEN.
To arrange the numbers 1 to 9 in three rows, so that the sum of each row added together horizontally, vertically, or diagonally shall be 15. Fig. 5 shows how the arrangement has to be made.
[Illustration:
+---+---+---+ | 2 | 9 | 4 | +---+---+---+ | 7 | 5 | 3 | +---+---+---+ | 6 | 1 | 8 | +---+---+---+
Fig. 5.--The Magic Nine.]
THE MAGIC THIRTY-SIX, OR PUZZLE OF ONE HUNDRED AND ELEVEN.
This puzzle is similar in principle to the preceding one, and consists in so arranging the numbers 1 to 36 in six rows that the sum of each row, added together horizontally or vertically, shall be the same (Fig. 6). The sum of the rows will be found to be 111.
[Illustration:
+-----+-----+-----+-----+-----+-----+ | 8 | 30 | 27 | 10 | 25 | 11 | | | | | | | | +-----+-----+-----+-----+-----+-----+ | 35 | 6 | 33 | 34 | 1 | 2 | | | | | | | | +-----+-----+-----+-----+-----+-----+ | 17 | 13 | 22 | 21 | 24 | 14 | | | | | | | | +-----+-----+-----+-----+-----+-----+ | 20 | 19 | 16 | 15 | 18 | 23 | | | | | | | | +-----+-----+-----+-----+-----+-----+ | 5 | 31 | 4 | 3 | 36 | 32 | | | | | | | | +-----+-----+-----+-----+-----+-----+ | 26 | 12 | 9 | 28 | 7 | 29 | | | | | | | | +-----+-----+-----+-----+-----+-----+
Fig. 6.--Magic Thirty-six.]
There is a still more complicated puzzle of this class to be performed. It is called
THE MAGIC HUNDRED, OR THE PUZZLE OF FIVE HUNDRED AND FIVE.
This consists in arranging the numbers from 1 to 100 in ten rows, and in such a way that the sum of the numbers counted, horizontally, vertically, or diagonally, shall be 505, neither more nor less. This puzzle may be set when the Magic Nine, the Magic Fifteen, and the Magic Thirty-six have been solved. The key is printed in Fig. 7. Upon a close examination of the key the solution of the puzzle from memory will soon become quite an easy matter. Observe the rows are numbered on the right hand side from 1 to 5, commencing both at the top and at the bottom. It will be seen that the rows numbered 1 contain the numbers 1 to 10 and 91 to 100; the rows numbered 2 contain the numbers 11 to 20 and 81 to 90; the third rows contain all the numbers from 21 to 30 and from 71 to 80; the fourth rows contain the numbers 31 to 39 and 60 to 70, excluding 61, but including 41; in the fifth rows the numbers run from 42 to 59, and have also the numbers 40 and 61. Furthermore, note the lettered columns, and it will be seen that the unit figures in columns A are noughts and ones, in columns B twos and nines, in columns C threes and eights, in columns D fours and sevens, and in columns E fives and sixes.
[Illustration:
A B C D E E D C B A +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 91 | 2 | 3 | 97 | 6 | 95 | 94 | 8 | 9 | 100 | | | | | | | | | | | | 1 +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 20 | 82 | 83 | 17 | 16 | 15 | 14 | 88 | 89 | 81 | | | | | | | | | | | | 2 +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 21 | 72 | 73 | 74 | 25 | 26 | 27 | 78 | 79 | 30 | | | | | | | | | | | | 3 +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 60 | 39 | 38 | 64 | 66 | 65 | 67 | 33 | 32 | 41 | | | | | | | | | | | | 4 +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 50 | 49 | 48 | 57 | 55 | 56 | 54 | 43 | 42 | 51 | | | | | | | | | | | | 5 +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 61 | 59 | 58 | 47 | 45 | 46 | 44 | 53 | 52 | 40 | | | | | | | | | | | | 5 +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 31 | 69 | 68 | 34 | 35 | 36 | 37 | 63 | 62 | 70 | | | | | | | | | | | | 4 +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 80 | 22 | 23 | 24 | 75 | 76 | 77 | 28 | 29 | 71 | | | | | | | | | | | | 3 +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 90 | 12 | 13 | 87 | 86 | 85 | 84 | 18 | 19 | 11 | | | | | | | | | | | | 2 +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+ | 1 | 99 | 98 | 4 | 96 | 5 | 7 | 93 | 92 | 10 | | | | | | | | | | | | 1 +-----+-----+-----+-----+-----+-----+-----+-----+-----+-----+
Fig. 7.--Plan of the Magic Hundred.]
THE TWENTY-FOUR MONKS.
During the middle ages there existed a monastery, in which lived twenty-four monks, presided over by a blind abbot. The cells of the monastery were planned as shown in the accompanying figure (Fig. 8), passages being arranged along two sides of each of the outer cells and all round the inner cell, in which the abbot took up his quarters. Three monks were allotted to each cell, making, of course, nine monks in each row of cells. The abbot, being lazy as well as blind, was very remiss in making his rounds, but provided he could count nine heads on each side of the monastery he retired into his own cloister, contented and satisfied that the monks were all within the building, and that no outsiders were keeping them company. The monks, however, taking advantage of their abbot's blindness and remissness, conspired to deceive him, a portion of their number sometimes going out and at other times receiving friends in their cells. They accomplished their deception, and it never happened that strangers were admitted when monks were out, yet there never were more nor less than nine persons upon each side of the building. Their first deception consisted in four of their number going out, upon which four monks took possession of each of the cells numbered 1, 3, 6, and 8, one monk only being left in each of the other cells; nine monks being thus on each side of the building. Upon returning, the four monks brought in four friends, when it was necessary to arrange the twenty-eight persons, two in each of the cells 1, 3, 6, and 8, and five in each of the others; still nine heads only were to be counted in either row. Emboldened by success, eight outsiders were introduced, and the thirty-two persons now were arranged one only in each of the cells 1, 3, 6, and 8, but seven in each of the other cells; again, according to the abbot's system of counting, all was well. In the next endeavour, the strangers all went away and took six monks with them, leaving but eighteen at home to represent twenty-four; these eighteen placed themselves five in each of the cells 1 and 8 and four in each of the cells 3 and 6; the remaining cells were empty, but the cells on each side of the building still contained nine monks. On returning, the six truants each brought two friends to pass the night, and the thirty-six retired to rest, nine in each of the cells 2, 4, 5, and 7; the remainder were empty, and the abbot was quite satisfied that the monks were alone in the monastery.
[Illustration:
+-----++------++-----+ | || || | | 1 || 2 || 3 | | || || | +=====++======++=====+ | || || | | 4 ||ABBOTT|| 5 | | || || | +=====++======++=====+ | || || | | 6 || 7 || 8 | | || || | +-----++------++-----+
Fig. 8.--The Twenty-four Monks.]
TO TAKE ONE FROM NINETEEN, SO THAT THE REMAINDER SHALL BE TWENTY.
See how it is done: XIX. (nineteen), by taking away the one that stands between the two tens (XX.), twenty will remain.
A similar catch is to write down nine figures, the sum of which is 45, from that number to take away 50, and to let the remainder be fifteen. The numerals should be added together thus: 1+2+3+4+5+6+7+8+9=45, or XLV., from which take away L. (50), and there will be left XV. (15).
THE FAMOUS FORTY-FIVE.
The number 45 can be divided into four such parts that if to the first 2 is added, from the second 2 is subtracted, the third is multiplied by 2, and the fourth divided by 2:--the total of the addition, the remainder of the subtraction, the product of the multiplication, and the quotient of the division will be the same.
The first part is 8, to which add 2, and the total will be 10 The second is 12, from which subtract 2, and the total will be 10 The third is 5, which multiply by 2, and the result will be 10 The fourth is 20, which divide by 2, and the result will be 10 -- 45
Again, 45 may be subtracted from 45 in such a manner as to leave 45 for a remainder. Arrange the following figures, add the rows together, and each row will be 45; subtract the bottom row from the top row, and the sum of the result added together will also be 45.
9+8+7+6+5+4+3+2+1=45 1+2+3+4+5+6+7+8+9=45 -------------------- 8+6+4+1+9+7+5+3+2=45
THE COSTERMONGER'S PUZZLE.