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?