Chapter I: II. III. IV (1)

___________ ___________ ___________ ___________
| | | | | | | | | | | | | | | |
| 3 | 3 | 3 | | 4 | 1 | 4 | | 2 | 5 | 2 | | 1 | 7 | 1 |
|___|___|___| | __|___|___| | __|___|___| | __|___|___|
| | | | | | | | | | | | | | | |
| 3 | | 3 | | 1 | | 1 | | 5 | | 5 | | 7 | | 7 |
|___|___|___| | __|___|___| | __|___|___| | __|___|___|
| | | | | | | | | | | | | | | |
| 3 | 2 | 3 | | 4 | 1 | 4 | | 2 | 5 | 2 | | 1 | 7 | 1 |
|___|___|___| | __|___|___| | __|___|___| | __|___|___|

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 of a 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:

42 886 326
24 688 1623
-- --- ----
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 12345679
18 27 36
-------- -------- -------
98765432 86419753 74074074
12345679 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)600,000 9)40,000
---------- ---------
66,666-6 4,444-4

{ 9)549
If 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--

} { 3
549 } { 2
6 } { 9
------ } And the amount of the figures of { 4
9)3294 } the product can also be divided by { ----
----- } 9, thus, { 2)18
366 } { ---
} { 9

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

43,260 } { 4,326
4,326 } 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
} 1
} 9
365472 } 2
6 } 3
------ } 2
219232 } --
} 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.

3
C
2 B D 4
1 A E 5
10 K F 6
9 I G 7
H
8

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 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.

From SIX From IX From XL
Take IX Take X Take L
--- -- --
S I X Remains.

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 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.

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 take away 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 is 77
To which Thomas adding his age 35
---
The sum is 112
Then by taking away the first figure 1, and adding
it to the figure 2, the sum is 13
Which add to John's age 22
---
Gives the age of Thomas 35

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, that HALF of whatever sum you request to be added during the working of the sum is THE 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 of 7
Double it 14
Add 10 to it 10
---
Halve it 2)24
---
Which will leave 12
Subtract the number thought of 7
---
THE REMAINDER will be 5

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 us suppose 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 2⅓ times. In this case, we must, as before, multiply 4 by 2⅓, which will give 9⅓; to which, if 4 be added, we shall have 13⅓, or 40/3; if 1, then, be taken from 2⅓, the remainder will be 1⅓, or 4/3, by which, if 40/3 be 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?