ARITHMETICAL PUZZLES.

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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."

1 14 15 4
8 11 10 5
12 7 6 9
13 2 3 16

Fig. 1.—A Solution of the "34" Puzzle.

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):—

1 2 3 4
5 6 7 8
9 10 11 12
13 14 15

Fig. 3.—"15" Puzzle—The Cubes in Order.

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
2. 15 3
3. 8 5
4. 11 6
5. 10 7
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.

2 9 4
7 5 3
6 1 8

Fig. 5.—The Magic Nine.

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.

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.

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.

There is a still more complicated puzzle of this class to be performed. It is called

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 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.

1 2 3
4 ABBOTT 5
6 7 8

Fig. 8.—The Twenty-four Monks.

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.

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.

A costermonger bought 120 oranges at two for a penny, and 120 more at three for a penny, and mixed the oranges all together in a basket. He sold them out, hoping to receive back his money again, at the rate of five for twopence; but on counting his money he found that he had sold the oranges for fourpence less than they had cost him. How this happened will be seen by following the accompanying figures. The first forty purchasers of the oranges would take 200 out of the 240 oranges, and taking it for granted that the fruit was equally mixed, would receive for their money 100 of the oranges originally bought at two a penny and 100 of those at three a penny, and would pay for them the sum of 6s. 8d. The forty remaining oranges would bring in, at the same rate, 1s. 4d. only, making 8s. in all. The cost of the fruit was, for the first 120, 5s., and for the second 120, 3s. 4d., or 8s. 4d. in all, making the loss of 4d. on the lot. To more fully explain the matter, we will suppose the oranges not mixed, but standing in separate baskets, from which, for each purchaser, the costermonger takes two of the two a penny oranges and three of the three a penny oranges, disposing of them in that way for twopence; it will then be clearly seen that the basket containing the three a penny oranges will be first exhausted, for the first forty purchasers, each having three oranges from one basket, will take all the 120 oranges purchased at three a penny, but will require only 80 oranges from the other basket, thus leaving 40 of the two a penny oranges to be sold at five for twopence, or a loss of fourpence on the last 40 sold.

THE PROGRESSION OF NUMBERS.

An illustration of the progression of numbers may be gathered from the description given of the American puzzle of "15," at the commencement of this section on Arithmetical Amusements. It is there stated that the different number of combinations or different arrangements of the fifteen cubes that can be made are 1,307,674,318,000. The reader may prove this for himself in the following manner:—The number of combinations that can be made with two cubes is 2, of three cubes 6, of four cubes 24, of five cubes 120, of six cubes 720, and so on, multiplying the result each time by one number higher than the previous result was multiplied by, until the amazing total quoted is reached; the arrangement of the cubes in rows and columns introducing additional variations of combinations. There are numerous instances on record in which it is stated that advantage has been taken of the known progression that ensues upon a repeated doubling of a given result. The Horse-dealer's Bargain is frequently quoted. A horse-dealer having a horse to dispose of, to which a gentleman had taken a great fancy, was asked to name any price he thought fit. Wishing at the first blush to appear generous, he offered to sell his horse, calculating its price according to the number of nails that were used to fasten on the four shoes, a farthing being allowed for the first nail, a halfpenny for the second, a penny for the third, twopence for the fourth, and so on. Upon examination it was found that it took six nails to fasten on each of the shoes, making in all twenty-four nails. The amount arrived at by repeatedly doubling the amount until the twenty-fourth nail had been allowed for was £8,738 2s. 8d.

The story of the Sovereign and the Sage gives a still more wonderful result. A king once, anxious to reward one of his subjects for valuable services performed to the State, asked in what way the subject would take his recompense. The king and the subject were both sixty-four years of age, and the wise man asked that he might be granted a kernel of wheat for the first year of their lives, two for the second, four for the third, eight for the fourth, sixteen for the fifth, and so on. By continuing the calculation until the result has been doubled for the sixty-fourth time, the astounding number of 9,223,372,036,854,775,808 will appear. It is generally conceded that the average number of wheat kernels in a pint is 9,216, which will give 18,432 for a quart, 73,728 for a peck, and 589,824 for a bushel, or 31,274,997,411,298 bushels of grain as the courtier's reward for his services, a larger amount than the whole world would produce in several years.

The Pin in the Hold of the "Great Eastern" Steamship is comparatively a modern calculation, based on this principle. It is calculated that 200 pins go to the ounce, and that if for the fifty-two weeks in the year one pin were dropped into the hold during the first week, two in the second, four in the third, and so on, that by the end of the year the weight of the whole would be no less than 628,292,358 tons of pins. As the Great Eastern steamship was built to carry 22,500 tons only, it follows that to carry all the pins there would be required 27,924 ships of the size of the Great Eastern.

As a last illustration of this subject we will instance the feat of counting a billion, which all boys know is a million millions. Allowing that so many as 200, which is an outside number, could be counted in a minute, it would, excluding the 366th day in leap years, take one person upwards of 9,512 years before the task would be completed. It is not, therefore, probable that any one person has yet counted a billion.

We next proceed to give a few of the rules showing

HOW A NUMBER THOUGHT OF OR OTHERWISE INDICATED MAY BE TOLD.

These rules and puzzles are numerous, and in practising them in company it is well to have several methods at command, in order that those of the company not in the secret may be the more mystified; and, indeed, those who only know one or two ways will themselves be astonished if they see others proceeding on principles differing from those with which they are familiar.

The Cancelled Figure.—Write down on a slate a series of numbers, the sum of each of which shall be 9: such, for example, as 18, 27, 36, 45, 144, 234, 612, 711, 252, 342, 261, 360, 432, 315, &c. &c. The greater the variety the better. Tell some person to fix on two of these numbers, and after adding them together, to strike out any one of the figures of the result, and then, upon his stating the sum of the remaining figure or figures, the figure struck out may be arrived at by ascertaining the difference between that sum and 9 or 18, according to whether the sum is less or more than 9. If the sum remaining be 9, the figure struck out will have been 9. Suppose, for instance, the numbers selected are 711 and 252, the total of which will be 963; if the figure struck out of that number be 6, the sum of the two remaining figures will be 12, or 6 less than 18. Again, take the numbers 18 and 27, making a total of 45; strike out the 5, and it will be seen that the difference between 4 and 9 is 5.

In the following methods any number may be thought of, and the subsequent calculations are to be mentally or otherwise made by him thinking of the number.

First Method.—Instruct that the number thought of be multiplied by 3, that 1 be added to the result, the result again being multiplied by 3, to which result the number first thought of has to be added; ask the result, strike off mentally the final figure, which will be a 3, and the figure or figures then left will represent the number first thought of. For example:—

The number thought of is 11
Multiplied by 3, it is 33
Add 1 34
Multiply by 3 102
Add the number (11) thought of 113

The result of which, when told, will show 11 to be the number thought of.

Second Method. Let any number be thought of, which we will again suppose to be 11
Instruct that it be doubled 22
Instruct that some stated even number be added (say 54) 76
Let the result be halved 38
Deduct the number first thought of (11) 27

The result will always be the half of the number that was instructed to be added.

Third Method. Multiply the number thought of by itself (say 11). 121
Take 1 from the number thought of, and multiply the result by itself (10+10) 100
Ask the difference between the two results 21

To this number the player, who is exhibiting his powers, must mentally add 1=22, and divide that number by 2, which gives 11, the number thought of.

Fourth Method. Add 1 to the number thought of (again 11) 12
Multiply by 3 36
Add 1 37
Add the number thought of (11) 48

Ask the result, from which mentally subtract 4, and divide the result by 4, which will again correctly give the original number.

Fifth Method. Let the number (11) thought of be doubled 22
Add 4 26
Multiply by 5 130
Add 12 142
Multiply by 10 1,420

Ask the result, from which mentally deduct 320, giving 1,100, from which strike off the noughts, and the result is again as before.

Sixth Method. Let 1 be deducted from the number (11) thought of 10
Multiply by 2 20
Add number first thought of 31

Ask the result, and to it mentally add 3=34, divide by 3, and the quotient of full numbers will be the number thought of. The above methods of guessing a number thought of will be about as many as any lad can remember.

MAGICAL ADDITION.

The following is a peculiar arrangement of the figures 1 to 9, so that by adding them together they amount to 100:—

15

36

47

___

98

2

___

100

To find the sum-total of three lines of figures upon the first line being shown, let any one write down a row of figures, and suppose they are 76854. Then take the paper, and, leaving space for two more rows of figures, say that the result of the addition of that row with two other rows can be given, the first of which rows may be written by any one present. Proceed by deducting 1 from the final right-hand figure, and place the figure 1 on the left-hand side; let the top row be folded over, and the paper handed back for the second row to be written, which we will suppose to be 34721; fill in the third row by making each figure in the second row up to 9 by writing 65278; the figures given according to the instructions will be the addition of the three rows.

76854

34721

65278

______

176853

______

The addition of five rows of figures may be told in a similar manner. Let any one present write down, as before, a row of figures, and then the calculator may undertake to tell the addition of that row with four other rows, two of which may be written by any person or persons present. This is attained by deducting 2 from the unit figure, and placing the figure 2 on the left-hand side. For example: again suppose the first row to be as before (76854); the result of the addition of that row and four others may be made to be 276852. After the first row has been written, fold it over, and request some one to write a second row of figures, which shall be supposed to be 34721; to this the magic calculator should write for the third row such figures as will make each of the above up to 9, namely, 65278; again, let a stranger write the fourth row, first turning over so as to conceal the second and third rows; whatever appears for the fourth row must be in each figure, as before, for the fifth row, made by the calculator to 9. Thus, if the fourth row be put down 80765, the necessary figures to add for the fifth row will be 19234, when the previously given total will be found to be correct.

76854

34721

65278

80765

19234

______

276852

______

In the last two examples of Magical Addition the only stipulations needed are that no subsequent row shall contain more figures than are contained in the first row, and that the first row shall end neither with a 1 nor a 0.

THE CLEVER LAWYER.

The following good story is very old:—A country attorney was once left executor to a will in which the testator bequeathed his stable of horses to be divided among three persons, in the proportions of half of the horses to A, a third of the horses to B, and a ninth of the horses to C. When the will was made 18 horses were in the stable, but subsequently, and before the death of the testator, one died, leaving but 17. The division according to the will now seemed impossible; but to prevent disputes among the legatees, the lawyer gave a horse out of his own stable, then divided the horses according to the will, and yet received his own back, and all were satisfied. It was done in the following manner:—

A received the half of 18, namely 9 horses.
B third 6
C ninth 2
17
The lawyer's horse returned 1
18

A NEW WAY OF MULTIPLYING BY 9.

Suppose it be required to multiply the following figures by 9, the result may be obtained in the following as well as in the ordinary way. In the first example the ordinary method has been pursued; the new way consists in adding a 0 on the right-hand side of the figures, and subtracting the number to be multiplied.

467543

9

_______

4207887

4675430

467543

_______

4207887

TO REWARD THE FAVOURITES, AND SHOW NO FAVOURITISM.

The proprietor of a ladies' school once received an invitation for one-half of her pupils to attend a flower show, but was a long time before she could decide how to pick out those who ought to be rewarded without hurting the feelings of those left behind. There were thirty pupils in the school, and fifteen were to be taken and the like number left at home. The following plan was the one hit upon:—The pupils were arranged in a row, four intended to go were placed first, five not intended next, and so on, as shown below, the letter A denoting those it was intended should partake of the offered pleasure, the letter B denoting those it was wished to leave out, and they were told when so arranged that the ninth girl, and each succeeding ninth, would be left at home.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
A A A A B B B B B A A B A A A B A B B A A B B B A B B A A B

The counting commenced with No. 1, and went round and round consecutively, each ninth pupil being dropped out, as designated to stay at home. It will be seen that those to be left were dropped out in the following order:—9th, 18th, 27th, 6th, 16th, 26th, 7th, 19th, 30th, 12th, 24th, 8th, 22nd, 5th, 23rd, thus leaving fifteen only.

THE DISHONEST SERVANTS.

Three gentlemen, with their servants, had to cross over a river in a boat in which two passengers only could be transported at one time. The servants were known to have planned to murder and rob one or more of the masters if two servants were left with one master or three servants with two masters. The question to be decided was how these six persons were to cross so that the boat could be returned, and yet so that the servants on either side of the river should not outnumber the masters. The following is one of the several ways in which the difficulty might have been overcome:—Two servants go over first, one returns; two servants go over again, one again returning with the boat; two of the masters next go over, and a master and one of the previously taken servants returns; then two of the masters again go over, and the servant already crossed takes the boat back, leaving the three masters safely crossed; the servants are left to come over in any manner they choose.

LORD DUNDREARY'S FINGER PUZZLE TO COUNT ELEVEN FINGERS ON THE TWO HANDS.

Begin on one hand, and count the ten fingers throughout. Begin next time at the finger last counted in the first round, counting this time backwards—ten, nine, eight, seven, six—then holding up the other hand, say "And five are eleven."

UNIFORM RESULTS OF MULTIPLICATION.

The digits 15873 multiplied by 7 give 111111
31746 7 222222
47619 7 333333
63492 7 444444
79365 7 555555
95238 7 666666
126984 7 888888
142857 7 999999
Of course, it would need the digits 111111 to make 777777

TO ASCERTAIN A SQUARE NUMBER AT A GLANCE.

Every boy knows that a square number is a number produced by the multiplication of any number into itself; thus 7, multiplied by itself, gives 49 as a result, 49 consequently is a square number, 7 being termed the square root from which it springs. In high numbers the extraction of the square root is an affair of time and trouble, and after all the necessary calculations have been made it may perhaps be found that the number is not a square number. This unnecessary trouble may be saved if the following instructions are remembered:—Every square number ends with one of the figures 1, 4, 5, 6, or 9, or with two ciphers preceded by one or other of those figures. Again, every square number is either equally divisible by 4, or when divided by 4 will have a remainder of 1; thus, as shown above, the square of 7 is 49, which divided by 4 gives us a quotient 12 and 1 over; 64 again is a square number, and it is exactly divisible by 4.

TO DISTINGUISH COINS BY ARITHMETICAL CALCULATION.

Request some person to place in one of his hands a bronze coin and in the other a silver one, and to let no one know which hand contains either particular coin. This may be ascertained by the following calculation:—The calculator should assign an even number, say 4, to the bronze coin, and an odd number, say 7, to the silver coin. The person holding the coins should be requested to multiply the number assigned to the coin held in his right hand by an even number, and that assigned to the coin held in the left hand by an odd number. Instruct that the products of the two calculations be added together, and if the whole sum be even the silver coin will have been placed in the right hand, and the bronze coin in the left. If the result be an odd number, the reverse arrangement will of course have been made.

We will conclude this section by stating shortly some of the

PROPERTIES OF NUMBERS.

By a careful study of these properties many amusing arithmetical puzzles and numerical combinations may be arrived at:—

1. Every odd number multiplied by an odd number produces an odd number.

2. Every odd number multiplied by an even number produces an even number.

3. Every even number multiplied by an odd number produces an even number.

4. An even number added to or subtracted from an even number, or an odd number to or from an odd number, produces an even number.

5. An odd number added to or subtracted from an even number produces an odd number.

6. The digits of the nine times multiplication table added together make either 9 or 18 (twice 9), thus:—

9 × 1 = 9
9 × 2 = 18 or 8 plus 1 = 9
9 × 3 = 27 or 7 2 = 9
9 × 4 = 36 or 6 3 = 9
And so on to
9 × 11 = 99 or 9 plus 9 = 18

Then so on again up to 9 times 24, each table making 9, with the exception of 9 times 22 =198=8+9+1=18. Indeed, the digits, added together, of the product of any number multiplied by 9, will be found to be 9 or a multiple thereof.

7. The digits 1 to 9 may be placed to form 362880 combinations; this number divided by 9 gives 40320; these figures added together make 9.

8. If two numbers are divisible by any one number, their sum and their difference will also be divisible by the same number.

9. If two numbers divisible by 9 be added together, the sum of the figures will be either 9 or a multiple of 9.

                                                                                                                                                                                                                                                                                                           

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