FromSIXFromIXFromXLTakeIXTakeXTakeL———SIXRemains.
THE MONEY GAME.
A person having in one hand a piece of gold, and in the other a piece of silver, you may tell in which hand he has the gold, and in which the silver, by the following method: Some value, represented by an even number, such as 8, must be assigned to the gold; and a value represented by an odd number, such as three, must be assigned to the silver; after which, desire the person to multiply the number in the right hand by any even number whatever, such as 2, and that in the left by an odd number, as 3; then bid him add together the two products, and if the whole sum be odd, the gold will be in the right hand, and the silver in the left; if the sum be even, the contrary will be the case.
To conceal the artifice better, it will be sufficient to ask whether the sum of the two products can be halved without a remainder; for in that case the total will be even, and in the contrary case odd.
It may be readily seen, that the pieces, instead of being in the two hands of the same person, may be supposed to be in the hands of two persons, one of whom has the even number, or piece of gold, and the other the odd number, or piece of silver. The same operations may then be performed in regard to these two persons, as are performed in regard to the two hands of the same person, calling the one privately the right, and the other the left.
THE PHILOSOPHER'S PUPILS.
To find a number of which the half, fourth, and seventh added to three shall be equal to itself.
This was a favorite problem among the ancient Grecian arithmeticians, who stated the question in the following manner: "Tell us, illustrious Pythagoras, how many pupils frequent thy school?" "One half," replied the philosopher,"study mathematics, one fourth natural philosophy, one seventh observe silence, and there are three females besides."
The answer is, 28: 14 + 7 + 4 + 3 = 28.
TO DISCOVER A SQUARE NUMBER.
A square number is a number produced by the multiplication of any number into itself; thus, 4 multiplied by 4 is equal to 16, and 16 is consequently a square number, 4 being the square root from which it springs. The extraction of the square root of any number takes some time; and after all your labor you may perhaps find that the number is not a square number. To save this trouble, it is worth knowing that every square number ends either with a 1, 4, 5, 6, or 9, or with two cyphers, preceded by one of these numbers.
Another property of a square number is, that if it be divided by 4, the remainder, if any, will be 1—thus, the square of 5 is 25, and 25 divided by 4 leaves a remainder of 1; and again, 16, being a square number, can be divided by 4 without leaving a remainder.
THE SHEEP-FOLD.
A farmer had a pen made of 50 hurdles, capable of holding 100 sheep only; supposing he wanted to make it sufficiently large to hold double that number, how many additional hurdles would he have occasion for?
Answer.—Two. There were 24 hurdles on each side of the pen; a hurdle at the top, and another at the bottom; so that, by moving one of the sides a little back, and placing an additional hurdle at the top and bottom, the size of the pen would be exactly doubled.
COUNTRYWOMAN AND EGGS.
A countrywoman carried eggs to a garrison, where she had three guards to pass, sold to the first guard half the number she had, and half an egg more; to the second, the half of what remained, and half an egg besides; and to the third guard she sold the half of the remainder, and half another egg. When she arrived at the market-place, she had three dozen still to sell; how was this possible, without breaking any of the eggs? It would seem at the first view that this is impossible, for how can half an egg be sold without breaking any of the eggs? The possibility of this seemingimpossibility will be evident, when it is considered, that by taking the greater half of an odd number, we take the exact half + 1/2. When the countrywoman passed the first guard, she had 295 eggs; by selling to that guard 148, which is the half + 1/2, she had 147 remaining; to the second guard she disposed of 74, which is the major half of 147; and, of of course, after selling 37 out of 74 to the last guard, she had still three dozen remaining.
HOW TO RUB OUT TWENTY CHALKS AT FIVE TIMES, RUBBING OUT EVERY TIME AN ODD ONE.
To do this trick, you must make twenty chalks, or long strokes, upon a board, as in the margin:
Then begin and count backwards, as 20, 19, 18, 17, rub out these four; then proceed saying, 16, 15, 14, 13, rub out these four; and begin again, 12, 11, 10, 9, and rub out these; and proceed again, 8, 7, 6, 5, then rub out these; and lastly say, 4, 3, 2, 1, when these four are rubbed out. The whole twenty are rubbed out at five times, and every time an odd one, that is, 17th, 13th, 9th, 5th, and 1st.
This is a trick which, if once seen, may be easily retained; and the puzzle at first is, it not occurring immediately to the mind to begin to rub them out backwards. It is as simple as any thing possibly can be.
THE IMPOSSIBLE TRIANGLE.
The longest side of a triangle is 100 rods; and each of the other sides 50. Required the value of the grass at $5 per acre.
This is a catch question, as a triangle cannot be formed unless any two of the lines are longer than the third.
ODD OR EVEN.
Every odd number multiplied by an odd number produces an odd number; every odd number multiplied by an even number produces an even number; and every even number multiplied by an even number also produces an even number. So, again, an even number added to an even number, and an odd number added to an odd number, produce aneven number; while an odd and even number added together produce an odd number.
If any one holds an odd number of counters in one hand, and an even number in the other, it is not difficult to discover in which hand the odd or even number is. Desire the party to multiply the number in the right hand by an even number, and that in the left hand by an odd number, then to add the two sums together, and tell you the last figure of the product; if it is even, the odd number will be in the right hand; and if odd, in the left hand; thus, supposing there are 5 counters in the right hand, and 4 in the left hand, multiply 5 by 2, and 4 by 3, thus: 5 × 2 = 10, 4 × 3 = 12, and then adding 10 to 12, you have 10 + 12 = 22, the last figure of which, 2, is even, and the odd number will consequently be in the right hand.
THE FIGURES, UP TO 100, ARRANGED SO AS TO MAKE 505 IN EACH COLUMN, WHEN COUNTED IN TEN COLUMNS PERPENDICULARLY, AND THE SAME WHEN COUNTED IN TEN FILES HORIZONTALLY.
THE FIGURES, UP TO 100, ARRANGED SO AS TO MAKE 505 IN EACH COLUMN, WHEN COUNTED IN TEN COLUMNS PERPENDICULARLY, AND THE SAME WHEN COUNTED IN TEN FILES HORIZONTALLY.
109293759649899111191884858687131290712928777675242322807062633736353468693141525344464547585960514243545655574849504032336765666438396130797827262574737221818988141516178382201009894956973291
Each of these files, when added up, makes 505.
Each of these ten columns, when added up, makes 505.
THE OLD WOMAN AND HER EGGS.
At a time when eggs were scarce, an old woman who possessed some remarkably good-laying hens, wishing to oblige her neighbors, sent her daughter round with a basket of eggs to three of them; at the first house, which was the squire's, she left half the number of eggs she had and half a one over; at the second she left half of what remained and half an egg over; and at the third she again left half of the remainder, and half a one over; she returned with one egg in her basket, not having broken any. Required—the number she set out with.Ans.15 eggs.
THE MATHEMATICAL FORTUNE TELLER.
Procure six cards, and having ruled them the same as the following diagrams, write in the figures neatly and legibly.
It is required to tell the number thought by any person, the numbers being contained in the cards, and such numbers not to exceed 60. How is this done?
Request the person to give you the cards containing the number, and then add the right hand upper corner figurestogether, which will give the correct answer. For example: suppose 10 is the number thought of, the cards with 2 and 8 in the corners will be given, which makes the answer 10, and so on with the others.
THE DICE GUESSED UNSEEN.
A pair of dice being thrown, to find the number of points on each die without seeing them. Tell the person who cast the dice to double the number of points upon one of them, and add 5 to it; then to multiply the sum produced by 5, and to add to the product the number of points upon the other die. This being done, desire him to tell you the amount, and, having thrown out 25, the remainder will be a number consisting of two figures, the first of which, to the left, is the number of points on the first die, and the second figure, to the right, the number on the other. Thus:
Suppose the number of points of the first die which comes up to be 2, and that of the other 3; then, if to four, the double of the points of the first, there be added 5, and the sum produced, 9, be multiplied by 5, the product will be 45; to which, if 3, the number of points on the other die, be added, 48 will be produced, from which, if 25 be subtracted, 23 will remain; the first figure of which is 2, the number of points on the first die, and the second figure 3, the number on the other.
THE SOVEREIGN AND THE SAGE.
A sovereign being desirous to confer a liberal reward on one of his courtiers, who had performed some very important service, desired him to ask whatever he thought proper, assuring him it should be granted. The courtier, who was well acquainted with the science of numbers, only requested that the monarch would give him a quantity of wheat equal to that which would arise from one grain doubled sixty-three times successively. The value of the reward was immense; for it will be found by calculation that the sixty-fourth term of the double progression divided by 1, 2, 4, 8, 16, 32, &c., is 9223372036854775808. But the sum of all the terms of a double progression, beginning with 1, may be obtained by doubling the last term, and subtracting from it 1. The number of the grains of wheat, therefore, in the present case, will be 18446744073709551615. Now, if a pint contain 9216 grains of wheat, a gallon will contain73728; and, as eight gallons make one bushel, if we divide the above result by eight times 73728 we shall have 31274997411295 for the number of the bushels of wheat equal to the above number of grains, a quantity greater than what the whole surface of the earth could produce in several years, and which in value would exceed all the riches, perhaps, on the globe.
THE KNOWING SHEPHERD.
A shepherd was going to market with some sheep, when he met a man who said to him, "Good morning, friend, with your score." "No," said the shepherd, "I have not a score; but if I had as many more, half as many more, and two sheep and a half, I should have just a score." How many sheep had he?
He had 7 sheep: as many more 7; half as many more, 31⁄2; and 21⁄2; making in all 20.
THE CERTAIN GAME.
Two persons agree to take, alternately, numbers less than a given number, for example, 11, and to add them together till one of them has reached a certain sum, such as 100. By what means can one of them infallibly attain to that number before the other?
The whole artifice in this consists in immediately making choice of the numbers 1, 12, 23, 34, and so on, or of a series which continually increases by 11, up to 100. Let us suppose that the first person, who knows the game, makes choice of 1; it is evident that his adversary, as he must count less than 11, can at most reach 11, by adding 10 to it. The first will then take 1, which will make 12; and whatever number the second may add the first will certainly win, provided he continually add the number which forms the complement of that of his adversary to 11; that is to say, if the latter take 8, he must take 3; if 9, he must take 2; and so on. By following this method he will infallibly attain to 89, and it will then be impossible for the second to prevent him from getting first to 100; for whatever number the second takes he can attain only to 99; after which the first may say—"and 1 makes 100." If the second take 1 after 89, it would make 90, and his adversary would finish by saying—"and 10 make 100." Between two persons who are equally acquainted with the game, he who begins must necessarily win.
THE ASTONISHED FARMER.
A and B took each 30 pigs to market, A sold his at 3 for a dollar, B at 2 for a dollar, and together they received $25. A afterwards took 60 alone, which he soldas before, at 5 for $2, and received, but $24; what became of the other dollar?
This is rather a catch question, the insinuation that the first lot were sold at the rate of five for $2, being only true in part. They commence selling at that rate, but after making ten sales, A's pigs are exhausted, and they have received $20: B still has 10 which he sells at "2 for a dollar" and of course receives $5; whereas had he sold them at the rate of 5 for $2, he would have received but $4. Hence the difficulty is easily settled.
MAGICAL CENTURY.
If the number 11 be multiplied by any one of the nine digits, the two figures of the product will always be alike, as appears in the following example:—
111111111111111111123456789—————————112233445566778899—————————
Now, if another person and yourself have fifty counters a-piece, and agree never to stake more than ten at a time, you may tell him that if he permit you to stake first, you always complete the even century before him.
In order to succeed, you must first stake 1, and remembering the order of the above series, constantly add to what he stakes as many as will make one more than the numbers 11, 22, 33, &c., of which it is composed, till you come to 89, after which your opponent cannot possibly reach the even century himself, or prevent you from reaching it.
If your opponent has no knowledge of numbers, you may stake any other number first, under 10, provided you subsequently take care to secure one of the last terms, 56, 67, 78, &c.; or you may even let him stake first, if you take care afterward to secure one of these numbers.
This exercise may be performed with other numbers; but, in order to succeed, you must divide the number to be attained by a number which is a unit greater than what you can stake each time, and the remainder will then be thenumber you must first stake. Suppose, for example, the number to be attained be 52 (making use of a pack of cards instead of counters), and that you are never to add more than 6; then, dividing 52 by 7, the remainder, which is 3, will be the number which you must first stake; and whatever your opponent stakes, you must add as much to it as will make it equal to 7, the number by which you divided, and so in continuation.
THE UNLUCKY HATTER.
A blackleg passing through a town in Ohio, bought a hat for $8 and gave in payment a $50 bill. The hatter called on a merchant near by, who changed the note for him, and the blackleg having received his $42 change went his way. The next day the merchant discovered the note to be a counterfeit, and called upon the hatter, who was compelled forthwith to borrow $50 of another friend to redeem it with; but on turning to search for the blackleg he had left town, so that the note was useless on the hatter's hands. The question is, what did he lose—was it $50 besides the hat, or was it $50 including the hat?
This question is generally given with names and circumstances as a real transaction, and if the company knows such persons so much the better, as it serves to withdraw attention from the question; and in almost every case the first impression is, that the hatter lost $50 besides the hat, though it is evident he was paid for the hat, and had he kept the $8 he needed only to have borrowed $42 additional to redeem the note.
THE BASKET OF NUTS.
A person remarked that when he counted over his basket of nuts, two by two, three by three, four by four, five by five, or six by six, there was one remaining; but when he counted them by sevens, there was no remainder. How many had he?
The least common multiple of 2, 3, 4, 5, and 6 being 60, it is evident, that if 61 were divisible by 7, it would answer the conditions of the question. This not being the case, however, let 60 × 2 + 1, 60 × 3 + 1, 60 × 4 + 1, &c., be tried successively, and it will be found that 301 = 60 × 5 + 1, is divisible by 7; and consequently this number answers the conditions of the question. If to this we add 420, the leastcommon multiple of 2, 3, 4, 5, 6 and 7, the sum 721 will be another answer; and by adding perpetually 420, we may find as many answers as we please.
THE UNITED DIGITS.
Arrange the figures 1 to 9 in such order that, by adding them together, they amount to 100.
153647––982——100
153647––982——100
DECEMBER AND MAY.
An old man married a young woman; their united ages amounted to C. The man's age multiplied by 4 and divided by 9, gives the woman's age. What were their respective ages?
ANSWER.—The man's age, 60 years 12 weeks; the woman's age, 30 years 40 weeks.
THE TWO DROVERS.
Two drovers, A and B, meeting on the road, began discoursing about the number of sheep they each had. Says B to A, "Pray give me one of your sheep and I will have as many as you." "Nay," replied A, "but give me one of your sheep and I will have as many again as you." Required to know the number of sheep they each had?
A had seven and B had five sheep.
THE BASKET AND STONES.
If a hundred stones be placed in a straight line, at the distance of a yard from each other, the first being at the same distance from a basket, how many yards must the person walk who engages to pick them up, one by one, and put them into the basket? It is evident that, to pick up the first stone, and put it into the basket, the person must walk two yards; for the second, he must walk four; for the third, six: and so on increasing by two, to the hundredth.
The number of yards, therefore, which the person must walk will be equal to the sum of the progression, 2, 4, 6, &c., the last term of which is 200 (22). But the sum of the progression is equal to 202, the sum of the two extremes, multiplied by 50, or half the number of terms: that is to say, 10,100 yards, which makes more than 51⁄2miles.
THE FAMOUS FORTY-FIVE.
How can number 45 be divided into four such parts that, if to the first part you add 2, from the second part you subtract 2, the third part you multiply by 2, and the fourth part you divide by 2, the sum of the addition, the remainder of the subtraction, the product of the multiplication, and the quotient of the division be all equal?
The 1st is8;to which add2,the sum is10The 2nd is12;subtract2,the remainder is10The 3rd is5;multiplied by2,the product is10The 4th is20;divided by2,the quotient is1045
Required to subtract 45 from 45, and leave 45 as a remainder?
Solution.— 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 451 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45————————————————8 + 6 + 4 + 1 + 9 + 7 + 5 + 3 + 2 = 45
SUBTRACTION.
From 1 mile subtract 7 furlongs, 39 rods, 5 yards, 1 foot, 5 inches.
milesfurlongs,rods,yards,feet,inches.From100000Take0739515000001
In this problem, instead of borrowing 1 foot, we borrow1/2a foot = 6 inches, from which we take 5 inches, and 1 remains; we then carry1/2to 1, and borrowing1/2a yard = 11/2feet, we have 11/2from 11/2= 0, and afterwards proceed as usual.
THE EXPUNGED FIGURE.
In the first place desire a person to write down secretly, in a line, any number of figures he may choose, and addthem together as units; having done this, tell him to subtract that sum from the line of figures originally set down; then desire him to strike out any figure he pleases, and add the remaining figures in the line together as units, (as in the first instance,) and inform you of the result, when you will tell him the figure he has struck out.
76542-2424———76518
Suppose, for example, the figures put down are 76542; these, added together, as units, make a total of 24: deduct 24 from the first line, and 76518 remain; if 5, the center figure be struck out, the total will be 22. If 8, the first figure be struck out, 19 will be the total.
In order to ascertain which figure has been struck out, you make a mental sum one multiple of 9 higher than the total given. If 22 be given as the total, then 3 times 9 are 27, and 22 from 27 show that 5 was struck out. If 19 be given, that sum deducted from 27 shows 8.
Should the total be equal multiplies of 9, as 18, 27, 36, then 9 has been expunged.
With very little practice any person may perform this with rapidity, it is therefore needless to give any further examples. The only way in which a person can fail in solving this riddle is, when either the number 9 or a cipher is struck out, as it then becomes impossible to tell which of the two it is, the sum of the figure in the line being an even number of nines in both cases.
THE MYSTERIOUS ADDITION.
It is required to name the quotient of five or three lines of figures—each line consisting of five or more figures—only seeing the first line before the other lines are even put down. Any person may write down the first line of figures for you. How do you find the quotient?
EXAMPLE.—When the first line of figures is set down, subtract 2 from the last right-hand figure, and place it before the first figure of the line, and that is the quotient for five lines. For example, suppose the figures given are 86,214, the quotient will be 286,212. You may allow any person to put down the two first and the fourth lines, but you must always set down the third and fifth lines, and in doing so, always make up 9 with the line above, as in the following example:
Therefore in the annexed diagram you will see that you have made 9 in the third and fifth lines with the lines above them. If the person desire to put down the figures should set down a 1 or 0 for the last figure, you must say we will have another figure, and another, and so on until he sets down something above 1 or 2.
86,21442,68057,31962,85437,145———Qt. 268,212
In solving the puzzle with three lines, you subtract 1 from the last figure, and place it before the first figure, and make up the third line yourself to 9. For example: 67,856 is given, and the quotient will be 167,855, as shown in the above diagram.
67,85647,21852,781———Qt. 167,855
TO TELL AT WHAT HOUR A PERSON INTENDS TO RISE.
Let the person set the hand of the dial of a watch at any hour he pleases, and tell you what hour that is; and to the number of that hour you add in your mind 12; then tell him to count privately the number of that amount upon the dial, beginning with the next hour to that on which he proposes to rise, and counting backwards, first reckoning the number of the hour at which he has placed the hand. For example:
Suppose the hour at which he intends to rise be 8, and that he has placed the hand at 5; you will add 12 to 5, and tell him to count 17 on the dial, first reckoning 5, the hour at which the index stands, and counting backwards from the hour at which he intends to rise; and the number 17 will necessarily end at 8, which shows that to be the hour he chose.
TO FIND THE DIFFERENCE BETWEEN TWO NUMBERS, THE GREATEST OF WHICH IS UNKNOWN.
Take as many nines as there are figures in the smallest number, and subtract that sum from the number of nines. Let another person add the difference to the largest number, and talking away the first figure of the amount add it to the last figure, and that sum will be the difference of the two numbers.
For example: John, who is 22, tells Thomas, who is older, that he can discover the difference of their ages; he therefore privately deducts 22 from 99 (his age consisting of two figures, he of course takes two nines); the difference, which is 77, he tells Thomas to add to his age, and to takeaway the first figure from the amount, and add it to the last figure and that will be the difference of their ages; thus,
The difference between John's age and 99 is77To which Thomas adding his age35——The sum is112Then by taking away the first figure 1, andadding it to the figure 2, the sum is13Which add to John's age22——Gives the age of Thomas35
THE REMAINDER.
A very pleasing way to arrive at an arithmetical sum, without the use of either slate or pencil, is to ask a person to think of a figure, then to double it, then add a certain figure to it, now halve the whole sum, and finally to subtract from that the figure first thought of. You are then to tell the thinker what is the remainder.
The key to this lock of figures is, thatHALFof whatever sum you request to be added during the working of the sum isTHE REMAINDER. In the example given, five is the half of ten, the number requested to be added. Any amount may be added, but the operation is simplified by giving only even numbers, as they will divide without fractions.
Example.
Think of7Double it14Add 10 to it10——Halve it2)24——Which will leave12Subtract the number thought of7——THEREMAINDERwill be5
A PERSON HAVING AN EQUAL NUMBER OF COUNTERS, OR PIECES OF MONEY, IN EACH HAND, TO FIND HOW MANY HE HAS ALTOGETHER.
Request the person to convey any number, as 4, for example, from the one hand to the other, and then ask how many times the less number is contained in the greater. Let ussuppose that he says the one is the triple of the other; and, in this case, multiply 4, the number of the counters conveyed, by 3, and add to the product the same number, which will make 16. Lastly, take 1 from 3, and if 16 be divided by the remainder 2, the quotient will be the number contained in each hand, and consequently the whole number is 16.
This curious problem deserves another example. Let us again suppose that 4 counters are passed from one hand to the other, and the less number is contained in the greater 21/3times. In this case, we must, as before, multiply 4 by 21/3, which will give 91/3; to which, if 4 be added, we shall have 131/3, or40/3; if 1, then, be taken from 21/3, the remainder will be 11/3, or4/3, by which, if40/3be divided, the quotient 10 will be the number of counters in each hand.
THE THREE JEALOUS HUSBANDS.
Three jealous husbands, A, B, and C, with their wives, being ready to pass by night over a river, find at the water side a boat which can carry but two at a time, and for want of a waterman they are compelled to row themselves over the river at several times. The question is how those six persons shall pass, two at a time, so that none of the three wives may be found in the company of one or two men, unless her husband be present?
This may be effected in two or three ways; the following may be as good as any: Let A and wife go over—let A return—let B's and C's wives go over—A's wife returns—B and C go over—B and wife return, A and B go over—C's wife returns, and A's and B's wives go over—then C comes back for his wife. Simple as this question may appear, it is found in the works of Alcuin, who flourished a thousand years ago, hundreds of years before the art of printing was invented.
THE FALSE SCALES.
A cheese being put into one of the scales of a false balance, was found to weigh 16 lbs., and when put into the other only 9 lbs. What is the true weight?
The true weight is a mean proportional between the two false ones, and is found by extracting the square root of their product. Thus 16 × 9 = 144; and square root 144 = 12 lbs., the weight required.
THE APPLE WOMAN.
A poor woman, carrying a basket of apples, was met by three boys, the first of whom bought half of what she had, and then gave her back 10; the second boy bought a third of what remained, and gave her back 2; and the third bought half of what she had now left, and returned her 1; after which she found she had 12 apples remaining. What number had she at first?
From the 12 remaining, deduct 1, and 11 is the number she sold the last boy, which was half she had; her number at that time, therefore, was 22. From 22 deduct two, and the remaining 20 was 2/3 of her prior stock, which was therefore 30. From 30 deduct 10, and the remainder 20 is half her original stock; consequently she had at first 40 apples.
THE GRACES AND MUSES.
The three Graces, carrying each an equal number of oranges, were met by the nine Muses, who asked for some of them; and each Grace having given to each Muse the same number, it was then found that they had all equal shares. How many had the Graces at first?
The least number that will answer this question is twelve; for if we suppose that each Grace gave one to each Muse, the latter would each have three, and there would remain three for each Grace. (Any multiple of 12 will answer the conditions of the question.)
THE JESUITICAL TEACHER.
A teacher, having fifteen young ladies under her care, wished them to take a walk each day of the week. They were to walk in five divisions of three ladies each, but no two ladies were to be allowed to walk together twice during the week. How could they be arranged to suit the above conditions?
QUAINT QUESTIONS.
What is the difference between twenty four quart bottles, and four and twenty quart bottles?
Ans.—56 quarts difference.
What three figures, multiplied by 4, will make precisely 5?
Ans.—11/4, or 1·25.
What is the difference between six dozen dozen, and half-a-dozen dozen?
Ans.—792: Six dozen dozen being 864, and half-a-dozen dozen, 72.
Place three sixes together, so as to make seven.
Ans.—66/6.
Add one to nine and make it twenty.
Ans.IX—cross the1, it makes XX.
Place four fives so as to make six and a half.Ans.—55/5·5.
A room with eight corners had a cat in each corner, seven cats before each cat, and a cat on every cat's tail. What was the total number of cats?Ans.—Eight cats.
Prove that seven is the half of twelve.Ans.—Place the Roman figures on a piece of paper, and draw a line through the middle of it, the upper will be VII.
THE FOX, GOOSE AND CORN.
A countryman having a Fox, a Goose, and a peck of Corn, came to a river, where it so happened that he could carry but one over at a time. Now as no two were to be left together that might destroy each other, he was at his wit's end, for says he "Though the corn can't eat the goose, nor the goose eat the fox; yet the fox can eat the goose, and the goose eat the corn." How shall he carry them over, that they shall not destroy each other?
Let him first take over the Goose, leaving the Fox and Corn; then let him take over the Fox and bring the Goose back; then take over the Corn; and lastly take over the Goose again.
MULTIPLYING MONEY BY MONEY.
Amongst the various questions that are given for the purpose of puzzling the unwary arithmetician, the multiplication of money by money is one of the most curious: take for instance the following problems:
Multiply £99 19s.113/4d.by £99 19s.113/4d.Multiply £11 11s.11d.by £11 11s.11d.
To the uninitiated they usually appear easy of solution but the various modes of working them out, and the different results obtained, prove that there is something absurd and wrong in the questions themselves. Some reduce all to farthings, and after multiplying one term by the other, return the product into pounds, shillings, and pence. Others convert them into decimals; whilst some work the problem in the style of duodecimals.
Having sufficiently puzzled the tyros, the querist remarks: "The problem itself is absurd, it is incapable of solution; for what is the nature of the product of pounds, shillings, and pence multiplied by pounds, shillings and pence? We know that a yard multiplied by a yard is a square yard, but who can tell what is a penny multiplied by a penny, or a penny by a pound?"
Now all this is quite correct, provided the question is limited, as above to the product of pounds, shillings, and pence, into pounds, shillings, and pence;butsuppose the problem were put in this form—If a capital of £1 produces by compound interest, in a certain time, £99 19s.113/4d., how much would be produced by a capital of £99 19s.113/4d.?It is evident that, to answer this, we must multiply £99 19s.113/4d.by £99 19s.113/4d.: these are in fact the second and, third terms of an ordinary "rule of three;" and though one of the terms is a "concrete" quantity of pounds, shillings, and pence, the other must be regarded as an "abstract" mathematical quantity, being 99 and a fraction, of which the number of farthings in a pound is the denominator 960, and the number of farthings in the third term is the numerator, 959; or, instead of this, the shillings and pence might be converted into decimals of a pound, or into aliquot parts. The product of multiplying £99 19s.113/4d.by 99959/960is £9,999 15s.101/3840d.; the quickest way of doing this, is to multiply by 100, and to subtract from the product the 960th part of the multiplicand.
In the other question proposed, the product of £11 11s.11d.into £11 11s.11d., or 11143/240, is £134 9s.3493/240d.
Number and value are distinct abstract ideas, and cannot, without committing a logical absurdity, be confused. To multiply is to repeat a certain number oftimes, and it is obviouslyimpossible to bringvalueinto the question. Value is arbitrary; number is fixed. Put it in this way, and the absurdity is evident: One pound is equivalent to 20 shillings, or 240 pence, or 960 farthings. In value there is no difference whatever; but what an enormous difference between multiplying by 1, 20, 240, or 960!
THE UNFAIR DIVISION.
A gentleman rented a farm, and contracted to give to his landlord2/5of the produce;butprior to the time of dividing the corn, the tenant used 45 bushels. When the general division was made, it was proposed to give to the landlord 18 bushels from the heap, in lieu of his share of the 45 bushels which the tenant had used, and then to begin and divide the remainder as though none had been used. Would this method have been correct?
The landlord would lose 71/5bushels by such an arrangement, as the rent would entitle him to2/5of the 18. The tenant should give him 18 bushels from his own share after the division is completed, otherwise the landlord would receive but2/7of the first 63 bushels.
A POPULAR FALLACY.
It is often suggested from the pulpit and elsewhere, that enough persons have lived and died in the world to cover its whole surface with bodies; and even two or three strata deep. Is this probable?
Say the earth has existed 6000 years, the population always having been 800,000,000, and the average life of man 30 years; this being the utmost that could be claimed. Allow then the State of Virginia to contain 70,000 square miles, and each grave to occupy a space of 6 feet by 2; the territory of the State would contain 162,624,000,000; while the mighty army of the dead would number only 160,000,000,000; leaving 2,624,000,000 graves yet unoccupied. How wide of truth then is the position often set forth so positively!