Chapter 13

Ourearth'sfinalfate—enigmaeverdark45213 11 22 312 21

or,

OurdearRichard'stalebeginsatthesea.45213 11 22 31221

Our young friends may find amusement in forming lines for themselves as much superior to these as possible.

THE TEN TENS.

Take ten pieces of card, and upon each write any ten words; there is no restriction as to the initial letter of nine of the words, but the last word on each card must commence with certain letters which you must in your own mind associate with the numbers 1 to 10, so that by knowing the initial letter of the last word on each card, you can determine its number.

Here are ten cards, (call these theSelecting Cards,) which we give by way of example, though our readers will perhaps prefer having words of their own selection.

Jane.Ellen.George.James.Newton.Mary.Fanny.William.Clement.Davy.Matilda.Caroline.Frederick.Edward.Morse.Sarah.Isabel.Robert.Ralph.Fulton.Rosa.Flora.Edmund.Francis.Franklin.Elizabeth.Laura.John.Edwin.Arago.Harriet.Maria.Alfred.Walter.Spurzheim.Ann.Frances.Albert.Charles.Laplace.Emily.Edith.Henry.Samuel.Steers.Emma.Dorothea.Isaac.Theodore.Herschel.Sister.Rose.Friendship.Putnam.Clay.Brother.Violet.Happiness.Lafayette.Webster.Uncle.Lupin.Industry.Steuben.Calhoun.Aunt.Daisy.Ambition.Scott.Benton.Grandmother.Tulip.Energy.Taylor.Jefferson.Grandfather.Peony.Fidelity.Green.Adams.Nephew.Hyacinth.Affection.Harrison.Madison.Niece.Pink.Hope.Hamilton.Jackson.Cousin.Snowdrop.Justice.Wayne.Monroe.Father.Lily.Order.Washington.Napoleon.

For these the key words are, "Edith Flown," so that the letters

EDITHFLOWNStand for12345678910

For the success of the game, the key words and the numbers denoted by their letters, must be carefully concealed.

Take ten other cards, which call the "grouped cards," andupon one write down the first word from each of the selecting cards, being careful to write them in the same order. Let another card contain all the words which are second from the top, and so on till all the words have been grouped together. As an example, we give the 1st and 4th grouped cards.

1st.4th.Jane.Sarah.Ellen.Isabel.George.Robert.James.Ralph.Newton.Fulton.Sister.Aunt.Rose.Daisy.Friendship.Ambition.Putnam.Scott.Clay.Benton.

The object of the game is to guess which of the words from any of theselecting cardsany person may have fixed upon.

Let any one choose a card out of theselecting cards, and after he has fixed upon a word, give it back to you; when receiving it, carefully note the last word upon it, which will give you, by the aid of the key word, the number of the card; this you must keep secret, and you then give him all thegrouped cards, and request him to show you the cards which contain the words he fixed upon.

You can then announce the word; for the number of the word from the top on the grouped card is the same as the number of the selecting card, from which he made his choice.

Suppose he made his choice from the card which has Theodore for its last word—this is No. 4; when he shows you the grouped card, which he says contains the selected word, you will know that Ralph, the fourth from the top, is the name he fixed upon.

DIVIDING THE BEER.

During the siege of Sebastopol, when the troops were on "short allowance," a can of eight pints of porter was ordered to be equally divided between two messes; but having only a five pint can, and one which held three pints, it was found impossible to make this division, till one of theclever sappers suggested the following method; and, to understand it, we will put down the contents of each of the three cans at each stage of the process; commencing with

8-pt.5-pt.3-pt.The 8-pint can full, and the others empty,8001.Filled the 5-pint can3502.Filled the 3-pint can from the 5-pint3233.Pour the contents of the 3-pint into the 8-pint6204.Transfer the 2 pints from the 5-pint to the 3-pint6025.Filled the 5-pint from the 8-pint1526.Fill up the 3-pint from the 5-pint1437.Poured the 3 pints into the 8-pint; completing the feat440

This was a dexterous expedient of the worthy sapper, the only objections to it being the time the thirty men had to wait, and the resulting flat condition of the beer.

THE DIFFICULT CASE OF WINE.

A gentleman had a bottle containing 12 pints of wine, 6 of which he was desirous of giving to a friend; but he had nothing to measure it, except two other bottles, one of 7 pints, and the other of 5. How did he contrive to put 6 pints into the 7-pint bottle?

12-pt.7-pt.6-pt.Before he commenced, the contents of the bottles were12001.He filled the 5-pint7052.Emptied the 5-pint into the 7-pint7503.Filled again the 5-pint from the 12-pint2554.Filled up the 7-pint from the 52735.Emptied the 7-pint into the 12-pint9036.Poured the 3 pints from the 5 into the 79307.Filled the 5-pint from the 12-pint4358.Filled up the 7-pint from the 5-pint4719.Emptied the 7-pint into the 12-pint110110.Poured 1 pint from the 5-pint into the 7-pint111011.Filled the 5-pint from the 12-pint61512.Poured the contents of the 5-pint into the 7-pint660

ANOTHER DECIMATION OF FRUIT.

On the next visit of the youth to his uncle, the latter produced thirty apples and ten oranges, and offered him thefavorite oranges, if his nephew could arrange them in an oval, so that by taking every twelfth the apples should remain. But this he could not accomplish, and the old gentleman, being well versed in the "Recreations in Science,"proceededto arrange them thus:

The places which the oranges here occupy can be easily remembered, being Nos. 7, 8, 11, 12, 21, 22, 24, 34, 36, 37.

THE WINE AND THE TABLES.

A certain hotel-keeper was dexterous in contrivances to produce a large appearance with small means. In the dining-room were three tables, between which he could divide 21 bottles, of which 7 only were full, 7 half full, and 7 apparently just emptied, and in such a manner that each table had the same number of bottles, and the same quantity of wine. He did this in two ways:

Table.Full Hf.full.Empty.Table.Full Hf.full.Empty.1232|13132232|23133313|3151

He also performed a similar exploit with 24 bottles, 8 full, 8 half-full, and 8 empty:

Table.Full Hf.full.Empty.Table.Full Hf.full.Empty.1323|12422322|22423242|3404

Also with 27 bottles, 9 full, 9 half-full, and 9 empty:

Table.Full Hf.full.Empty.Table.Full Hf.full.Empty.1252|11712333|24143414|3414

THE THREE TRAVELERS.

Three men met at a caravansary or inn, in Persia; and two of them brought their provision along with them, according to the custom of the country; but the third not having provided any, proposed to the others that they should eat together, and he would pay the value of his proportion. This being agreed to, A produced 5 loaves, and B 3 loaves,all of which the travelers ate together, and C paid 8 pieces of money as the value of his share, with which the others were satisfied, but quarreled about the division of it. Upon this the matter was referred to the judge, who decided impartially. What was his decision?

At first sight it would seem that the money should be divided according to the bread furnished; but we must consider that, as the 3 ate 8 loaves, each one ate 2-2/3 loaves of the bread he furnished. This from 5 would leave 2-1/3 loaves furnished the stranger by A; and 3 - 2-2/3 = 1/3 furnished by B, hence 2-1/3 to 1/3 = 7 to 1, is the ratio in which the money is to be divided. If you imagine A and B to furnish, and C to consume all, then the division will be according to amounts furnished.

WHICH COUNTER HAS BEEN THOUGHT OF OUT OF SIXTEEN?

Take sixteen pieces of card, and number them 1 to 16. Arrange them in two rows, as at A B.

ABCBDMEBFNGBH191922294229621031044610861105311511661113141184127128851275312751313113413614143148147151551531581616716716

Desire a person to think of one of the numbers, and to tell you in which row it is. Suppose he fixes on 6; he will tell you that the row A contains the number he thought of.

Take up the row A, and arrange the numbers on each side of the row B, as shown at C D, so that the first number of the row A may be the first of the row C, the second of A be the first of D, the third of A be the second of C, and so on.

Ask in which of the rows, C or D, is the number thought of: in the case supposed it is in D.

Take up the rows C D, and put one underneath the other, as at M, taking care that the half-row in which is the number thought of, shall be above the other.

Divide it again into two rows, as at E F, on each side of B, in the same way as before. Ask again in which row it is: it is now in E.

Place one row under the other, as at N, and divide again into two rows, which will now be as G H.

You will be informed that the number is in row H, and you may then announce it to be the top number of that row.

The number thought of will always beat the top of one of the rows after three transpositions. If there were 32 counters it would be at the top after four transpositions.

MAGIC SQUARES.

The name "Magic Square" is given to a square divided into several smaller squares, in which numbers are placed in such a manner that every column of numbers, whether vertical, horizontal, or from corner to corner, shall amount to the same sum.

They are divided into three principal classes: 1st, Those which have an odd number of squares in each band; 2d, Those which have an even number of squares in each band, this even number being divisible exactly by 4; 3d, Where the even number of squares in each band cannot be divided by 4 without a fraction.

ODD MAGIC SQUARES.

Squares of this kind are formed thus. Imagine an exterior line of squares above the magic square you wish to form, and another exterior line on the right hand of it These two imaginary lines are shown in the figure.

Then attend to the two following rules:

1st. In placing the numbers in the squares we must go in an ascending oblique direction from left to right; any number which, by pursuing this direction, would fall into the exterior line, must be carried along that line of squares, whether vertical or horizontal, to the last square. Thus, 1 having been placed in the center of the top line, (see the first table on p. 228,) 2 would fall into the exterior square above the fourth vertical line; it must be therefore carried down to the lowest square of that line; then, ascending obliquely 3 falls into the square, but four falls out of it, to the end of a horizontal line, and it must be carried along that line to the extreme left, and there placed. Resuming our oblique ascension to the right, we place 5, where the reader sees it, and would place 6 in the middle of the top band, but finding it occupied by 1, we look for direction to the2d Rule, which prescribes that, when in ascending obliquely, we come to a square already occupied, we must place the number, which according to the first rule should go into that occupied square, directly under the last number placed. Thus, in ascending with 4, 5, 6, the 6 must be placed directly under the 5, because the square next to 5 in an oblique direction is "engaged."

Magic squares of this class, however large in the number of compartments, can be easily filled up by attending to these two rules.

We give opposite, a seven-placed square.

There are various other kinds of magic squares; but explanations of them would be too lengthy for our work.

The invention of these contrivances has been traced back to the early ages of science, and talismanic properties were attributed to them. Modern philosophers have amused themselves in bringing them to perfection, and none has contributed so much as "the model of practical wisdom," Dr. Franklin.

THE SQUARE OF GOTHAM.

The wise men of Gotham, famous for their eccentric blunders, once undertook the management of a school; they arranged their establishment in the form of a square divided into 9 rooms. The playground occupied the center, and 24 scholars the rooms around it, 3 being in each. In spite of the strictness of discipline, it was suspected that the boys were in the habit of playing truant, and it was determined to set a strict watch. To assure themselves that all the boys were on the premises, they visited the rooms, and found three in each, or 9 in each row. Four boys then went out, and the wise men soon after visited the rooms, and finding 9 in each row, thought all was right. The four boys then came back, accompanied by four strangers; and the Gothamites, on their third round, finding still 9 in each row, entertained no suspicion of what had taken place. Then 4 more "chums" were admitted; but the clever men, on examining the establishment a fourth time, still found 9 in each row, and so came to an opinion that their previous suspicions had been unfounded. How was all this possible?

The following figures represent the contents of each room at the four different visits; the first, at the commencement of the watch; the second, when four had gone out; thethird, when these 4, accompanied by another 4, had returned; and the fourth, when 4 more had joined them.

On each change the boys arranged themselves in the rooms in such a manner that, when the corner rooms were counted as a part of two rows, each entire row of three rooms contained the same number of boys. The illusion of the wise men was due to their mistake in counting each corner room twice.

THE MATHEMATICAL BLACKSMITH.

A blacksmith had a stone weighing 40 lbs. A mason coming into the shop, hammer in hand, struck it and broke it into four pieces. "There," says the smith, "you have ruined my weight." "No," says the mason, "I have made it better, for whereas you could before weigh but 40 lbs. with it, now you can weigh every pound from 1 to 40." Required size of the pieces?

Ans.1, 3, 9, 27; for in any geometrical series proceeding in a triple ratio, each term is 1 more than twice the sum of all the preceding, and the above series might proceed to any extent. In using the weights, they must be put in one or both scales as may be necessary: as to weigh 2, put 1 in one scale, and 3 in the other.

CURIOUS PROPERTIES OF SOME FIGURES.

Select any two numbers you please, and you will find that one of the two, their amount when added together, or their difference, is always 3, or a number divisible by 3.

Thus, if the numbers are 3 and 8, the first number is 3; let the numbers be 1 and 2, their sum is 3; let them be 4 and 7, the difference is 3. Again, 15 and 22, the first number is divisible by 3: 17 and 26, their difference is divisible by 3, &c.

All the odd numbers above 3, that can only be divided by 1, can be divided by 6, by the addition or subtraction ofa unit. For instance, 13 can only be divided by 1; but after deducting 1, the remainder can be divided by 6; for example, 5 + 1 = 6 ; 7 - 1 = 6; 17 + 1 = 18; 19 - 1 = 18; 25 - 1 = 24, and so on.

If you multiply 5 by itself, and the quotient again by itself, and the second quotient by itself, the last figure of each quotient will always be 5. Thus 5 × 5 = 25; 25 × 25 = 125; 125 × 125 = 625, &c. Again, if you proceed in the same manner with the figure 6, the last figure will constantly be 6; thus, 6 × 6 = 36; 36 × 36 = 216; 216 × 216 = 1,296, and so on.

To multiply by 2 is the same as to multiply by 10 and divide by 5.

Any number of figures you may wish to multiply by 5, will give the same result if divided by 2—a much quicker operation than the former; but you must remember to annex a cipher to the answer where there is no remainder, and where there is a remainder, annex a 5 to the answer. Thus, multiply 464 by 5, the answer will be 2320; divide the same number by 2, and you have 232, and as there is no remainder you add a cipher. Now, take 357, and multiply by 5—the answer is 1785. On dividing 357 by 2, there is 178, and a remainder; you therefore place 5 at the right of the line, and the result is again 1785.

There is something more curious in the properties of the number 9. Any number multiplied by 9 produces a sum of figures which, added together, continually makes 9. For example, all the first multiples of 9, as 18, 27, 36, 45, 54, 63, 72, 81, sum up 9 each. Each of them multiplied by any number whatever produces a similar result; as 8 times 81 are 648, these added together make 18, 1 and 8 are 9. Multiply 648 by itself, the product is 419,904—the sum of these digits is 27, 2 and 7 are 9. The rule is invariable. Take any number whatever and multiply it by 9; or any multiple of 9, and the sum will consist of figures which, added together, continually number 9. As 17 × 18 = 306, 6 and 3 are 9; 117 × 27 = 3,159, the figures sum up 18, 8 and 1 are 9; 4591 × 72 = 330,552, the figures sum up 18, 8 and 1 are 9. Again, 87,363 × 54 = 4,717,422; added together the product is 27, or 2 and 7 are 9, and so always. If any row of two or more figures be reversed and subtracted from itself, the figures composing the remainder, will, when added horizontally, be a multiple of nine:

42886326246881623—————18 – 9 × 2.198 – 9 × 2.1638 – 9 × 2.

If a multiplicand be formed of the digits in their regular order, omitting the 8, a multiplier may be found by a rule, which will give a product, each figure of which shall be the same. Thus if 12345679 be given, and it be required to find a multiplier which shall give the product all in 2, that multiplier will be 18: if in 3, the multiplier will be 27: if all 4, it will be 36—and so forth.

12345679 12345679 12345679182736———— ———— ————98765432 86419753 7407407412345679 24691358 37037037———— ———— ————222222222 333333333 444444444

The rule by which the multiplier is discovered (but which we do not attempt to explain) is this: Multiply the last figure (the 9) of the multiplicand by the figure of which you wish the product to be composed, and that number will be the required multiplier. Thus, when it was required to have the product composed of 2, the 2 multiplied by 9 gives 18, the multiplier: 3 multiplied by 9 gives 27, the multiplier to give the product in 3; &c.

If a figure, with a number of ciphers attached to it, be divided by 9, the quotient will be composed of one figure only, namely, the first figure of the dividend, as—

9)600,0009)40,000—————————66,666–64,444–4

{ 9)549If any sum of figures can be divided by 9 as, {———{61

the amount of these figures, when added together, can be divided by 9:—thus, 5, 4, 9, added together, make 18, which is divisible by 9. If the sum 549 is multiplied by any figure, the product can also be divided by 9, as—

}{3549}{26}{9———}And the amount of the figures of{49)3294}the product can also be divided by{—————}9, thus,{2)18366}{——}{9

To multiply by 9, add a cipher, and deduct the sum that is to be multiplied: thus,

43,260}{4,3264,326}Produces the same result as{9——}{——38,934}{38,934

In the same manner, to multiply by 99, add two ciphers; by 999, three ciphers, &c. These properties of the figure 9 will enable the young arithmetician to perform an amusing trick, quite sufficient to excite the wonder of the uninitiated.

Any series of numbers that can be divided by 9, as 365, 472,821,754, &c., being shown, a person may be requested to multiply secretly either of these series by any figure he pleases, to strike out one number of the quotient, and to let you know the figures which remain, in any order he likes; you will then, by the assistance of the knowledge of the above properties of 9, easily declare the number which has been erased. Thus, suppose 365,472 are the numbers chosen, and the multiplier is six; if then, 8 is stuck out, the numbers returned to you will be

}2}1365472}96}2———}3219232}2}—}19

The amount of these numbers is 19; but 19, divided by 9, leaves a remainder of 1; you, therefore, want 8 to complete another 9: 8, then, is the number erased.

The component figures of the product made by the multiplication of every digit into the number 9, when added together, make NINE.

The order of these component figures is reversed after the said number has been multiplied by 5.

The component figures of the amount of the multipliers (viz. 45,) when added together, make NINE.

The amount by the several products, or multiples of 9 (viz. 405,) when divided by 9, gives for a quotient, 45; that is, 4 + 5 = NINE.

The amount of the first product (viz. 9,) when added to the other product, whose respective component figures make 9, is 81; which is the square of NINE.

The said number 81, when added to the above mentioned amount of the several products, or multiples of 9 (viz. 405) makes 486, which, if divided by 9, gives for a quotient 54: that is, 5 + 4 = NINE.

It is also observable, that the number of changes that may be rung on nine bells is 362,880; which figures, added together, make 27; that is, 2 + 7 = NINE.

And the quotient of 362,880, divided by 9, will be 40,320; that is 4 + 0 + 3 + 2 + 0 = NINE.

If number 37 be multiplied by any of the progressive numbers arising from the multiplication of 3 with any of the units, the figures in the quotient will be similar, and the result may be known beforehand by merely inspecting the progressive numbers, thus, 3, 6, 9, 12, 15, 18, 21, 24, 27, &c., are the progressive numbers formed by 3 multiplied by the units 1 to 9; and the result of the multiplication of any of these numbers with 37 may be seen in the following examples:—37 × 3 = 111; 37 × 6 = 222; 37 × 12 = 444; 37 × 24 = 888; by which it appears that the numbers of which the quotient is formed are the same as the units by which number 3 was multiplied to obtain the respective progressive numbers. Thus—3 multiplied by 2 is equal to 6, and 37 multiplied by 9 is equal to 222; so, again, 4 multiplied by 3 produces 12, and 37 multiplied by 12 is equal to 444, and so on.

THE INDUSTRIOUS FROG.

There was a well 30 feet deep, and at the bottom a frog anxious to get out. He got up 3 feet per day, but regularly fell back 2 feet at night. Required the number of days necessary to enable him to get out?

The frog appears to have cleared one foot per day, and at the end of 27 days, he would be 27 feet up, or within 3 feet of the top, and the next day he would get out. He would therefore be 28 days getting out.

THE COUNCIL OF TEN.

Ten cards or counters, numbered from one to ten, or the first ten playing cards of any suit disposed in a circular form may be employed with great convenience for performing this feat. The accompanying figure shows the cards thus arranged, number one, or the ace, designated by A, and the ten by K.

Having placed the cards in the above order, desire a bystander to think of a card or number, and when he has done so, to touch any other card or number. Request him then to add to the number of the card touched the number of the cards employed, which in this case is ten. Then desire him to count the sum in an order contrary to that of the natural numbers, beginning at the card he touched, and assigning it the number of the card he thought of. By counting in this manner, he will end at the number or card he thought of, and consequently you will immediately know it.

Thus, for example, suppose the person had thought of 3 C, and touched 6 F; then, if 10 be added to 6, the sum will be 16; and if that number be counted from F, the number touched, towards E D B C A, and so on, in the retrograde order, counting F three, the number thought of, E five, D six, and so round to sixteen, that number will terminate at C, showing that the person thought of 3, the number which corresponds to C.

A greater or less number of cards or counters may be employed at pleasure; but in every instance the whole number of cards must be added to the number of the card touched.

THE TWO TRAVELERS.

Two travelers trudged along the road together,Talking, as Yankees do, about the weather;When, lo! beside their path the foremost spiesThree casks, and loud exclaims "A prize, a prize!"One large, two small, but all of various size.This way and that they gazed, and all around,Each wondering if an owner might be found:But not a soul was there—the coast was clear,So to the barrels they at once drew near,And both agree whatever may be thereIn friendly partnership they'll fairly share.Two they find empty, but the other full,And straightway from his pocket one doth pullA large clasp-knife. A heavy stone lay handy,And thus in time they found their prize was brandy.'Tis tasted and approved: their lips they smack,And each pronounces 'tis the famed Cognac."Won't we have many a jolly night, my boy!May no ill luck our present hopes destroy!"'Twas fortunate one knew the mathematics,And had a smattering of hydrostatics;Then measured he the casks, and said, "I seeThis is eight gallons, those are five and three."The question then was how they might divideThe brandy, so that each should be suppliedWith just four gallons, neither less nor more.With eight, and five, and three they puzzle sore,Filled up the five—filled up the three, in vain;At length a happy thought came o'er the brainOf one: 'twas done, and each went home content,And their good dames declared 'twas excellent.With those three casks they made division true;I found the puzzle out, say, friend, can you?

Two travelers trudged along the road together,

Talking, as Yankees do, about the weather;

When, lo! beside their path the foremost spies

Three casks, and loud exclaims "A prize, a prize!"

One large, two small, but all of various size.

This way and that they gazed, and all around,

Each wondering if an owner might be found:

But not a soul was there—the coast was clear,

So to the barrels they at once drew near,

And both agree whatever may be there

In friendly partnership they'll fairly share.

Two they find empty, but the other full,

And straightway from his pocket one doth pull

A large clasp-knife. A heavy stone lay handy,

And thus in time they found their prize was brandy.

'Tis tasted and approved: their lips they smack,

And each pronounces 'tis the famed Cognac.

"Won't we have many a jolly night, my boy!

May no ill luck our present hopes destroy!"

'Twas fortunate one knew the mathematics,

And had a smattering of hydrostatics;

Then measured he the casks, and said, "I see

This is eight gallons, those are five and three."

The question then was how they might divide

The brandy, so that each should be supplied

With just four gallons, neither less nor more.

With eight, and five, and three they puzzle sore,

Filled up the five—filled up the three, in vain;

At length a happy thought came o'er the brain

Of one: 'twas done, and each went home content,

And their good dames declared 'twas excellent.

With those three casks they made division true;

I found the puzzle out, say, friend, can you?

The five-gallon barrel was filled first, and from that the three-gallon barrel, thus leaving two gallons in the five-gallon barrel; the three-gallon barrel was then emptied into the eight-gallon barrel, and the two gallons poured from the five-gallon barrel into the empty three-gallon barrel; the five-gallon barrel was then filled, and one gallon poured into the three-gallon barrel, therefore leaving four gallons in the five-gallon barrel, one gallon in the eight-gallon barrel, and three gallons in the three-gallon barrel, which was then emptied into the eight-gallon barrel. Thus each person had four gallons of brandy in the eight and five-gallon barrels respectively.

ARITHMETICAL PUZZLE.

If from 6 you take 9, and from 9 you take 10; and if 50 from 40 be taken, there will just half a dozen remain.

ANSWER.

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