Part 23
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 seeming impossibility 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:
1 ---- 2 ---- 3 ---- 4 ---- 5 ---- 6 ---- 7 ---- 8 ---- 9 ---- 10 ---- 11 ---- 12 ---- 13 ---- 14 ---- 15 ---- 16 ---- 17 ---- 18 ---- 19 ---- 20 ----
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 an even 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.
+-----+----+----+----+----+----+----+----+----+----+ | 10 | 92 | 93 | 7 | 5 | 96 | 4 | 98 | 99 | 1 | | 11 | 19 | 18 | 84 | 85 | 86 | 87 | 13 | 12 | 90 | | 71 | 29 | 28 | 77 | 76 | 75 | 24 | 23 | 22 | 80 | | 70 | 62 | 63 | 37 | 36 | 35 | 34 | 68 | 69 | 31 | | 41 | 52 | 53 | 44 | 46 | 45 | 47 | 58 | 59 | 60 | | 51 | 42 | 43 | 54 | 56 | 55 | 57 | 48 | 49 | 50 | | 40 | 32 | 33 | 67 | 65 | 66 | 64 | 38 | 39 | 61 | | 30 | 79 | 78 | 27 | 26 | 25 | 74 | 73 | 72 | 21 | | 81 | 89 | 88 | 14 | 15 | 16 | 17 | 83 | 82 | 20 | | 100 | 9 | 8 | 94 | 95 | 6 | 97 | 3 | 2 | 91 | +-----+----+----+----+----+----+----+----+----+----+
[Sidenote: 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?
+----+----+----+----+----+----+ +----+----+----+----+----+----+ | 3 | 5 | 7 | 9 | 11 | 1 | | 5 | 6 | 7 | 13 | 12 | 4 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 13 | 15 | 17 | 19 | 21 | 23 | | 14 | 15 | 20 | 21 |22 | 23 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 25 | 27 | 29 | 31 | 33 | 35 | | 28 | 29 | 30 | 31 | 36 | 37 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 37 | 39 | 41 | 43 | 45 | 47 | | 52 | 38 | 39 | 44 | 45 | 46 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 49 | 51 | 53 | 55 | 57 | 59 | | 47 | 53 | 54 | 55 | 60 | 13 | +----+----+----+----+----+----+ +----+----+----+----+----+----+
+----+----+----+----+----+----+ +----+----+----+----+----+----+ | 9 | 10 | 11 | 12 | 13 | 8 | | 3 | 6 | 7 | 10 | 11 | 2 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 14 | 15 | 24 | 25 | 26 | 27 | | 14 | 15 | 18 | 19 | 22 | 23 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 28 | 29 | 30 | 31 | 40 | 41 | | 26 | 27 | 30 | 31 | 34 | 35 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 42 | 43 | 44 | 45 | 46 | 47 | | 38 | 39 | 42 | 43 | 46 | 47 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 56 | 57 | 58 | 59 | 60 | 13 | | 50 | 51 | 54 | 55 | 58 | 59 | +----+----+----+----+----+----+ +----+----+----+----+----+----+
+----+----+----+----+----+----+ +----+----+----+----+----+----+ | 17 | 18 | 19 | 20 | 21 | 16 | | 33 | 34 | 35 | 36 | 37 | 32 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 22 | 23 | 24 | 25 | 26 | 27 | | 38 | 39 | 40 | 41 | 42 | 43 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 28 | 29 | 30 | 31 | 48 | 49 | | 44 | 45 | 46 | 47 | 48 | 49 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 50 | 51 | 52 | 53 | 54 | 55 | | 50 | 51 | 52 | 53 | 54 | 55 | +----+----+----+----+----+----+ +----+----+----+----+----+----+ | 56 | 57 | 58 | 59 | 30 | 60 | | 56 | 57 | 58 | 59 | 60 | 41 | +----+----+----+----+----+----+ +----+----+----+----+----+----+
Request the person to give you the cards containing the number, and then add the right hand upper corner figures together, 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 contain 73728; 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, 3½; and 2½; 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 sold _as 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:--
11 11 11 11 11 11 11 11 11 1 2 3 4 5 6 7 8 9 -- -- -- -- -- -- -- -- -- 11 22 33 44 55 66 77 88 99 -- -- -- -- -- -- -- -- --
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 the number 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 least common 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.
15 36 47 -- 98 2 --- 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 5½ miles.
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 is 8; to which add 2, the sum is 10 The 2nd is 12; subtract 2, the remainder is 10 The 3rd is 5; multiplied by 2, the product is 10 The 4th is 20; divided by 2, the quotient is 10 -- 45
Required to subtract 45 from 45, and leave 45 as a remainder?
Solution.--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
SUBTRACTION.
From 1 mile subtract 7 furlongs, 39 rods, 5 yards, 1 foot, 5 inches.
miles, furlongs, rods, yards, feet, inches. From 1 0 0 0 0 0 Take 0 7 39 5 1 5 -------------------------------------- 0 0 0 0 0 1
In this problem, instead of borrowing 1 foot, we borrow ½ a foot = 6 inches, from which we take 5 inches, and 1 remains; we then carry ½ to 1, and borrowing ½ a yard = 1½ feet, we have 1½ from 1½ = 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 add them 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.
76542-24 24 ------ 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.
86,214 42,680 57,319 62,854 37,145 ------ Qt. 268,212
67,856 47,218 52,781 ------ Qt. 167,855
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.
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.