OR, CURIOUS PROBLEMS IN ARITHMETIC.
As the principal object of this volume is to enable the
young reader to learn something in his sports, and to understand
what he is doing, we shall, before proceeding to the
curious tricks and feats connected with the science of numbers,
present him with some arithmetical aphorisms, upon
which most of the following examples are founded.
APHORISMS OF NUMBER.
1. If two even numbers be added together, or subtracted
from each other, their sum or difference will be an even
number.
2. If two uneven numbers be added or subtracted, their
sum or difference will be an even number.
3. The sum or difference of an even and an uneven number
added or subtracted, will be an uneven number.
4. The product of two even numbers will be an even
number, and the product of two uneven numbers will be an
uneven number.
5. The product of an even and uneven number will be an
even number.
6. If two different numbers be divisible by any one number,
their sum and their difference will also be divisible by
that number.
7. If several different numbers, divided by 3, be added
or multiplied together, their sum and their product will also
be divisible by 3.
8. If two numbers, divisible by 9, be added together, the
sum of the figures in the amount will be either 9, or a number
divisible by 9.
9. If any number be multiplied by 9, or by any other
number divisible by 9, the amount of the figures of the product
will be either 9, or a number divisible by 9.
10. In every arithmetical progression, if the first and last
term be each multiplied by the number of terms, and the
sum of the two products be divided by 2, the quotient will
be the sum of the series.
11. In every geometric progression, if any two terms be
multiplied together, their product will be equal to that term,
which answers to the sum of these two indices. Thus, in
the series—
If the third and fourth terms 8 and 16 be multiplied together,
the product 128 will be the seventh term of the
series. In like manner, if the fifth term be multiplied into
itself, the product will be the tenth term, and if that sum be
multiplied into itself, the product will be the twentieth term.
Therefore, to find the last, or any other term of a geometric
series, it is not necessary to continue the series beyond a
few of the first terms.
Previous to the numerical recreations, we shall here describe
certain mechanical methods of performing arithmetical
calculations, such as are not only in themselves entertaining,
but will be found more or less useful to the young
reader.
PALPABLE ARITHMETIC.
The blind mathematician, Dr. Saunderson, adopted a very
ingenious device for performing arithmetical operations by
the sense of touch.
Small cubes of wood were provided, and in one face of
each, nine holes were pierced, thus:
| 1 |
2 |
3 |
|
o |
o |
o |
| 4 |
5 |
6 |
|
o |
o |
o |
| 7 |
8 |
9 |
|
o |
o |
o |
These holes represented the nine digits, as in the figure, and
to denote any figure, a small peg was inserted into the hole
corresponding to it. If the number consisted of several
figures, more cubes were used, one for each. A cipher was
represented by a peg of different shape from that of the
others, and inserted in the central hole.
To perform any arithmetical process, a square board was
provided, divided by ridges into recesses of the same width
as the cubes, and by this the cubes were retained in the required
horizontal and perpendicular lines. Suppose it was
necessary to add together the numbers 763, 124, 859, the
cubes and pegs would be arranged thus:
THE ABACUS.
This instrument is used for teaching numeration, and the
first principles of arithmetic.
Upon a frame are placed wires, parallel
to one another, and at equal distances.
Ten small balls are strung
upon each wire, being placed as in
the margin. The right wire denotes
units, the next tens, and so on, the 7th
wire being the place of millions. In
using the abacus, all the balls are first
ranged at one end, and a number of
them are then moved to the other end
of each wire, to correspond to the
figures required. The example given
in the margin is 15,781, the height of
Mount Blanc.
NAPIER'S RODS.
The object of this contrivance is to render arithmetical
multiplication more easy, and to secure its correctness; it
was much used by astronomers before the invention of logarithms.
To appreciate the merits of this invention, we must consider
the process of multiplication as usually performed.
Suppose we had to multiply 8,679 by 8:
We first multiply 9 by 8 = 72, and putting down 2 as the
first figure in the product, carry the 7 to add to the next
product of 7 by 8 = 56; this gives us 63, the 3 being put
down as the second figure; 6 is carried to add to the product
of 6 by 8, and so on.
A blunder may be made in each part of this process; for
1st, we might reckon 8 times 9 as some other number than
72; 2d, after multiplying the 7 by the 8, we might add to
the resulting 56 some other figure than the 7, which we
carried; 3d, we may add 56 to 7 inaccurately,
making some other sum of it
than the right one, 63. Errors in a
long multiplication problem are usually
made in one of these three ways, and
to prevent such errors, Lord Napier[10]
introduced this useful contrivance.
Thin strips of card, wood, or bone, 9
times as long as they are broad, are
each divided into 9 equal squares, a
figure is printed or written on the top
square, and in each of the squares underneath
is the product of multiplying
that figure by 2, 3, 4, &c., up to 9.
To use these in multiplication, select
the strips, the top figures of which
make the number to be multiplied. For example:
To multiply 8,679 by 8, look at the eighth line of squares
from the top, and on that line will be found the product of
each of the integers 8, 6, 7, 9, when multiplied by 8. We have
then to write down the 2 as the first figure of the product,
add 7 and 6 together = 13; write 3 as the next figure, carry
1 to add to the sum of 8 and 5, and so on.
The reason for dividing the figures in each square by a
diagonal line, and for placing the left-hand figure higher
than the right is, that the eye may be thus assisted in adding
the carried figure of one slip to the unit of the next.
To provide for the occurrence of more than one of the same
figures in the multiplicand, there should be several slips or
rods for each of the digits.
In practice the rods are placed on a flat piece of wood,
with two ridges at right angles, by which they are preserved
in a proper position.
This instrument can be made useful in "divisions," by
making by means of it a table of the product of the divisor,
multiplied by each of the numbers 1 to 9.
THE ARITHMETICAL BOOMERANG.
The boomerang is an instrument of peculiar form, used
by the natives of New South Wales, for the purpose of
killing wild fowl and other small animals. If projected forwards,
it at first proceeds in a straight line, but afterwards
rises in the air, and after performing sundry peculiar gyrations,
returns in the direction of the place where it was
thrown.
The term is applied to those arithmetical processes by
which you can divine a number thought of by another. You
throw forwards the number by means of addition and multiplication,
and then, by means of subtraction and division,
you bring it back to the original starting point, making it
proceed in a track so circuitous as to evade the superficial
notice of the tyro.
TO FIND A NUMBER THOUGHT OF.
First Method.
This is an arithmetical trick which, to those who are unacquainted
with it, seems very surprising; but, when explained,
it is very simple. For instance, ask a person to
think of any number under 10. When he says he has done
so, desire him to treble that number. Then ask him whether
the sum of the number he has thought of (now multiplied
by 3) be odd or even; if odd, tell him to add 1 to make the
sum even. He is next to halve the sum, and then treble
that half. Again ask whether the amount be odd or even.
If odd, add 1 (as before) to make it even, and then halve it.
Now ask how many nines are contained in the remainder.
The secret is, to bear in mind whether the first sum be odd
or even; if odd, retain 1 in the memory; if odd a second
time, retain 2 more (making in all 3 to be retained in the
memory;) to which add 4 for every nine contained in the remainder.
For example, No. 7 is odd the first and also the second
time; and the remainder (17) contains one nine; so that
1, added to 2, make 3, and 3, added to 4, make 7, the number
thought of. No. 1 is odd the first time (retain 1), and
even the second (of which no notice is taken), but the remainder
is not equal to nine. No. 2 is even the first and odd
the second time (retain 2), but the remainder contains no
nine. No. 3 is odd the first and the second time, still there
is no nine in the remainder. No. 4 is even both times, and
contains one nine. No. 5 is odd the first time and the remainder
contains one nine. No. 6 is odd the second time,
and contains one nine in the remainder. No. 8 is even both
times, and the remainder contains two nines. No notice
need be taken of any overplus of a remainder, after being
divided by nine.
The following are illustrations of the result with each
number:
Second Method.
EXAMPLE.
| Let a person think of a number, say |
6 |
| 1. Let him multiply it by 3 |
18 |
| 2. Add 1 |
19 |
| 3. Multiply by 3 |
57 |
| 4. Add to this the number thought of |
63 |
Let him inform you what is the number produced; it will
always end with 3. Strike off the 3, and inform him that he
thought of 6.
Third Method.
EXAMPLE.
| Suppose the number thought of to be |
6 |
| 1. Let him double it |
12 |
| 2. Add 4 |
16 |
| 3. Multiply by 6 |
80 |
| 4. Add 12 |
92 |
| 5. Multiply by 10 |
920 |
Let him inform you what is the number produced. You
must in every case subtract 320; the remainder is, in this
example, 600; strike off the two ciphers, and announce 6
as the number thought of.
Fourth Method.
Desire a person to think of a number, say 6. He must
then proceed—
EXAMPLE.
| 1. To multiply this number by itself |
36 |
| 2. So take 1 from the number thought of |
5 |
| 3. To multiply this by itself |
25 |
| 4. To tell you the difference between this product and the former |
11 |
| You must then add 1 to it |
12 |
| And halve this number |
6 |
Which will be the number thought of.
Fifth Method.
Desire a person to think of a number, say 6. He must
then proceed as follows:
EXAMPLE.
| 1. Add 1 to it |
7 |
| 2. Multiply by 3 |
21 |
| 3. Add 1 again |
22 |
| 4. Add the number thought of |
28 |
Let him tell you the figures produced (28):
| 5. You then subtract 4 from it |
24 |
| 6. And divide by 4 |
6 |
Which you can say is the number thought of.
Sixth Method.
EXAMPLE.
| Suppose the number thought of |
6 |
| 1. Let him double it |
12 |
| 2. Desire him to add to this any number you tell him, say 4 |
16 |
| 3. To halve it |
8 |
You can then tell him that if he will subtract from this
the number he thought of, the remainder will be, in the case
supposed, 2.
Note.—The remainder is always half of the number you
tell him to add.
TO DISCOVER TWO OR MORE NUMBERS THAT A PERSON HAS THOUGHT OF.
1st Case.—Where each of the numbers is less than 10.
Suppose the numbers thought of were 2, 3, 5.
EXAMPLE.
| 1. Desire him to double the first number making |
4 |
| 2. To add 1 to it |
5 |
| 3. To multiply by 5 |
25 |
| 4. To add the second number |
28 |
There being a third number, repeat this process—
| 5. To double it |
56 |
| 6. To add 1 to it |
57 |
| 7. To multiply by 5 |
285 |
| 8. To add the third number |
290 |
And to proceed in the same manner for as many numbers as
were thought of. Let him tell you the last sum produced
(in this case 290). Then, if there were two numbers thought
of, you must subtract 5; if three, 55; if four, 555. You
must here subtract 55, leaving a remainder of 235, which
are the numbers thought of, 2, 3 and 5.
2d Case.—Where one or more of the numbers are 10, or
more than 10, and where there is an odd number of numbers
thought of.
Suppose he fixes upon five numbers, viz. 4, 6, 9, 15, 16.
He must add together the numbers as follows, and tell
you the various sums:
| 1. |
The sum of the 1st and 2d |
10 |
| 2. |
The sum of the 2d and 3d |
15 |
| 3. |
The sum of the 3d and 4th |
24 |
| 4. |
The sum of the 4th and 5th |
31 |
| 5. |
The sum of the 1st and last |
20 |
You must then add together the 1st, 3d and 5th sums, viz.
10 + 24 + 20 = 54, and the 2d and 4th, 15 + 31 = 46; take one
from the other, leaving 8. The half of this is the 1st number,
4; if you take this from the sum of the 1st and 2d you
will have the 2d number, 6; this taken from the sum of the
2d and 3d will give you the 3d, 9; and so on for the other
numbers.
3d Case.—Where one or more of the numbers are 10, or
more than 10, and where an even number of numbers has
been thought of.
Suppose he fixes on six numbers, viz. 2, 6, 7, 15, 16, 18.
He must add together the numbers as follows, and tell you
the sum in each case:—
| 1. |
The sum of the 1st and 2d |
8 |
| 2. |
The sum of the 2d and 3d |
13 |
| 3. |
The sum of the 3d and 4th |
22 |
| 4. |
The sum of the 4th and 5th |
31 |
| 5. |
The sum of the 5th and 6th |
34 |
| 6. |
The sum of the 2d and last |
24 |
You must then add together the 2d, 4th and 6th sums, 13 +
31 + 24 = 68, and the 3d and 5th sums, 22 + 34 = 56. Subtract
one from the other, leaving 12; the 2d number will be
6, the half of this; take the 2d from the sum of the 1st and
2d you will get the 1st; take the 2nd from the sum of the
2d and 3d, and you will have the 3d, and so on.
HOW MANY COUNTERS HAVE I IN MY HANDS?
A person having an equal number of counters in each
hand, it is required to find how many he has altogether.
Suppose he has 16 counters, or 8 in each hand. Desire
him to transfer from one hand to the other a certain number
of them, and to tell you the number so transferred. Suppose
it be 4, the hands now contain 4 and 12. Ask him how many
times the smaller number is contained in the larger; in this
case it is 3 times. You must then multiply the number
transferred, 4, by the 3, making 12, and add the 4, making
16; then divide 16 by the 3 minus 1; this will bring 8, the
number in each hand.
In most cases fractions will occur in the process: when
10 counters are in each hand, and if 4 be transferred, the
hands will contain 6 and 14.
He will divide 14 by 6 and inform you that the quotient is
22/6 or 21/3.
You multiply 4 by 21/3, which is 91/3.
Add 4 to this, making 131/3, equal to 40/3.
Subtract 1 from 21/3, leaving 11/3 or 4/3.
Divide 40/3 by 4/3 giving 10, the number in each hand.
THE MYSTERIOUS HALVINGS.
To tell the number a person has thought of.
One of the company must fix upon any one of the numbers
from 1 to 15; this he keeps secret, as well as the numbers
produced by the succeeding operations:
| Suppose he fixes on |
8 |
| He must add 1 to it, making |
9 |
| Triple it |
27 |
| Halve it[11]—1st halving—(larger half) |
14 |
| Triple it |
42 |
| Halve it—2d halving |
21 |
| Triple it |
63 |
| Halve it—3d halving—(larger half) |
32 |
| Triple it |
96 |
| Halve it—4th halving. |
48 |
He need not inform you that 48 is the figure produced, but
he must let you know in which of the four halvings he was
obliged to take a "larger half;" having ascertained this
point, you discover the number fixed upon in the following
manner. Carry in your mind, or on a slip of paper, the following
list of names in which the letter a occurs in one or
more of the three syllables of all except the last.
The three syllables are intended to represent the 1st, 2d,
and 3d halvings, and the occurrence of the letter A corresponds
to the occurrence of a "larger half" in one or more
of these three halvings. Having been informed where the
larger half was taken, refer to the word which has A in the
corresponding syllable, and against it stand two numbers,
one of which was the number thought of; and of these two,
the right hand number is the correct one if a larger half was
taken in the 4th stage, and the left hand one if the, 4th halving
was exact.
In the example given, a larger half occurred in the 1st
and 3d stage; this points us to Car-ro-way, and the halving
in the 4th stage being exact, shows us that 8 was the number
fixed upon.
| |
If the 4th halving is exact. |
If a larger half occurs in the 4th halving. |
| WAsh-ing-ton |
4 |
12 |
| LA-fAy-ette |
2 |
10 |
| CAr-row-wAy |
8 |
0 |
| MAn-hAt-tAn |
6 |
14 |
| Ger-mA-ny |
13 |
5 |
| Tel-e-grAph |
3 |
11 |
| Bo-nA-pArte |
1 |
9 |
| Long-fel-low |
15 |
7 |
It will be observed that there is always a difference of 8
between the numbers of the columns, so that it is necessary
to recollect only one of them. Perhaps some of our readers
who wish to be adepts in this game, would prefer recollecting
the above table if put in this form:
2-3 1-2 3 1-2-3 1-3 2 none
——————————————
1 2 3 4 8 13 15
where the upper line denotes the cases in which the "larger
half" was taken, and the lower line the numbers of the
left hand column above given.
Another Method.
The person having chosen any number from one to fifteen,
he is to add twenty-one to that number, and triple the
amount. Then,
1st. He is to take half of that triple, and triple that half.
2nd. To take the half of the last triple, and triple that
half.
3rd. To take the half of the last triple.
4th. To take the half of the last half.
In this operation there are four distinct cases or stages
where the half is to be taken. The three first are denoted
by one of the eight following Latin words, each word being
composed of three syllables, and the syllables containing the
letter i corresponding in numerical order with the cases
where the half cannot be taken without a fraction; consequently,
in those cases the person who makes the deduction
is to add one to the number to be divided. The fourth case
shows which of the two numbers corresponding to each
word has been chosen. For if the fourth half can be taken
without adding one, the number chosen is in the first, or
left-hand column; but if not, it is in the second column to
the right.
| The words. |
The numbers denoted. |
| Mi-ser-is |
8 |
0 |
| Ob-tin-git |
1 |
9 |
| Ni-mi-um |
2 |
10 |
| No-tar-i |
3 |
11 |
| In-fer-nos |
4 |
12 |
| Or-di-nes |
13 |
5 |
| Ti-mi-di |
6 |
14 |
| Te-ne-ant |
15 |
7 |
Example.—Suppose the number chosen to be nine, to
which is to be added one, making ten, and which last, being
tripled, gives thirty. Then:
| 1st case. |
The half of the triple is |
15 |
| |
which tripled, makes |
45 |
| 2nd case. |
The half of that triple, 1 being added to make an even number, is |
23 |
| |
and that tripled, makes |
69 |
| 3rd case. |
The half of the last triple, 1 being added, is |
35 |
| 4th case. |
The half of the last half, 1 being again added, is |
18 |
Here we see, that in the second and third case, one had
to be added, and, looking at the table, we find that the only
corresponding word having an i in its second and third
syllables is Ob-tin-git, which represents the figures one and
nine. Then, as one had to be added in the fourth case, we
know by the rule, that the figure in the second column, 9,
is the one required. Observe, that if no addition be required
at any of the four stages, the number thought of
will be fifteen; and if one addition only be required at the
fourth stage, the number will be seven.
WHO WEARS THE RING?
This is an elegant application of the principles involved
in discovering a number fixed upon. The number of persons
participating in the game should not exceed nine.
One of them puts a ring on one of his fingers, and it is
your object to discover—1st. The wearer of the ring. 2d.
The hand. 3d. The finger. 4th. The joint.
The company being seated in order the persons must be
numbered 1, 2, 3, &c.; the thumb must be termed the first
finger, the fore finger being the second; the joint nearest
the extremity must be called the first joint; the right hand
is one, and the left hand two.
These preliminaries having been arranged, leave the
room in order that the ring may be placed unobserved by
you. We will suppose that the third person has the ring
on the right hand, third finger, and first joint; your object
is to discover the figures 3131.
Desire one of the company to perform secretly the following
arithmetical operations:
| 1. |
Double the number of the person who has the ring; in the case supposed, this will produce |
6 |
| 2. |
Add 5 |
11 |
| 3. |
Multiply by 5 |
55 |
| 4. |
Add 10 |
65 |
| 5. |
Add the number denoting the hand |
66 |
| 6. |
Multiply by 10 |
660 |
| 7. |
Add the number of the finger |
663 |
| 8. |
Multiply by 10 |
6630 |
| 9. |
Add the number of the joint |
6631 |
| 10. |
Add 35 |
6666 |
He must apprise you of the figures now produced, 6666;
you will then in all cases subtract from it 3535; in the
present instance there will remain 3131, denoting the person
No. 3, the hand No. 1, the finger No. 3, and the joint
No. 1.
PROBABILITIES.[12]
When we look around us at results happening daily, of
the causes of which we are ignorant, we are led to regard
them as isolated incidents subject to no law or rule; but
could we see and understand the secret workings and connection
existing between cause and effect, we might frequently
discover that all works by rule. As it is, we may
readily mark the boundaries, within which events must
happen in very many instances; and do much to estimate
their probability. We speak of Chance, as something without
plan or design, but taking in a large range, our calculations
will approximate closely to the truth. When we
throw a copper into the air, the chances of "heads or tails,"
as the boys say, are equal, and though one or the other may
occur most frequently for a few throws, in a large number,
say a thousand, the results will be about equally divided.
In this case the sides of the coin must be equal in weight,
else it will be like the grumbler's bread and butter:
"I never had a piece of bread,
Particularly good and wide,
But fell upon the sanded floor,
And always on the buttered side."
Had he put on less butter, perhaps the sides would have
been more equal in weight, and the probability of the buttered
side being uppermost would have been increased.
Disturbing causes unknown to us, may often shape the
result; but in the absence of these, we may pretty accurately
estimate our chances.
We see accidents from fire and flood, happening at times
and points least expected; but the insurer has learned by
observation to estimate probabilities, and by taking a wide
range of country and a period of years, he does a comparatively
safe business. Death takes the young and the old;
but the life insurer has conned the bills of mortality and
studied the ages of those who have died, until he can estimate
at once the probability of duration of life, and determine
what he can afford to pay for an annuity contingent
on life, or engage for a present sum, or an annual sum paid
for life, to pay the heirs at the death of the insured. In
one instance his estimate may fall short, and in another
exceed, but the average will be about right.
So, too, the man who deals in lotteries and games of chance,
knows the data and calculates carefully the probabilities,
and though "luck" may sometimes be against him, his estimates
of probabilities are based on mathematical principles,
and he is secure in being ultimately the gaining party.
How these chances are calculated, depends on the data
in each case, and it is not within the range of our present
plan to attempt more than giving a general idea of the
subject; and this with any one of ordinary prudence, will
be sufficient to prevent all intermeddling with lotteries and
every other species of gambling. The probabilities are always
against the casual operator, even if all be conducted fairly;
what then must they be when fraud and dishonesty are
superadded? It is downright swindling!
In lottery schemes generally, fifteen per cent. is reserved
as profit, but this is a small part of what may be secured;
yet even this amounts to a great deal. If a man were to
draw a prize nominally of $100,000, fifteen thousand would
be deducted at once, and he would be entitled to only $85,000.
It is true that in his good fortune he would not probably regard
the abatement, but that does not change the principle.
VARIATIONS.
It is obvious that if we have a number of single things
arranged in any order, we may change the arrangement
into a variety of forms, and in doing so, we may take all
together, or we may take only part at once. For instance,
we may arrange the six vowels, a e, i, o, u, y, in a great
number of ways, as a e i o u y, a i e o u y, e a i o u y,
&c., &c.; or we may form them into groups, as ae, io, uy, ai,
eu, oy, &c.; or, we may take three, four, five, or, as above,
all at a time; and it is reasonable to suppose that the number
of possible changes may, in all cases, be calculated.
When all are taken together, the operation is called
Permutation; but if a part only be taken, it is called either
a Variation or a Combination; a e, i o, u y, are distinct combinations,
and are also considered one of the variations of
two of which those six letters are susceptible; e a, o i, y u,
are three other variations, but they are the same combinations;
for a change of order will constitute a new variation
but not a new combination; hence the number of variations
will always exceed the number of combinations.
The doctrine of variations and combinations forms the
basis of many forms of lotteries, and of other calculations
used in practical life.
COMBINATIONS AND PERMUTATIONS.
"Combinations" are the different ways in which a certain
number of things can be selected out of a larger number,
when taken 1 at a time, 2 at a time, or any other number
each time, but without regard to the order in which the
selected numbers can be arranged among themselves. The
latter is the province of "Permutation," which refers to the
different ways in which a number can be selected out of
one that is larger, and, in addition to this, to the different
ways of grouping these selected numbers.
Thus 4 things can be taken 2 at a time in 6 different ways;
for instance, the letters a, b, c, d, can be taken 2 at a time
thus, a and b, a and c, a and d, b and c, b and d, c and d;
if we regard the order of the selected letters we shall find
that these 4 letters are capable of 12 different permutations,
as ab, ba, ac, ca, ad, da, bc, cb, bd, db, cd, dc.
If we selected 3 letters at a time we could make 4 different
selections, and 24 different changes of grouping.
The rule to compute the number of these different ways
is very simple, but sometimes involves a multitude of
figures.
To determine the number of permutations, commence
with unity, and multiply by the successive terms of the
natural series 1, 2, 3, &c., until the highest multiplier shall
express the number of individual things. The last product
will indicate the number of possible changes.
Example 1. How many changes can be made in the arrangement
of 5 grains of corn, all of different colors, laid in
a row?
Solution. 1 × 2 × 3 × 4 × 5 = 120, Ans.
This may seem improbable, the number being so great,
but if there were but a single grain more, the possible
changes would be 720; and another would extend the limit
to 5040; and so onward in a constantly increasing ratio.
The reason, however, will be obvious on a little scrutiny. If
there were but one thing, as a, it would admit of but one
position; but if two, as a b, it would admit of two positions,
ab, ba. If three things, as a b c, then they will admit of
1 × 2 × 3 = 6 changes, for the last two will admit of two
variations, as a b c, a c b, and each of the three may successively
be placed first, and two changes made to each of
the others, so that 3 × 2 = 6, the number of possible changes.
In the same way we may show that if there be four individual
things, each one will be first in each of the six
changes which the other three will undergo, and consequently,
there will be 24 changes in all. In this way we
might show that when there are 5 individual things, there
will be 5 times as many changes as when there were but 4;
and when 6, there will be 6 times as many changes as when
there are only 5; and so on ad infinitum, according to the
same law.
Example 2. In how many ways may a family of 10 persons
seat themselves differently at dinner? Ans. 3,628,800.
When we consider that this would require a period of
9935-55/487 years, the mind is lost in astonishment. The story
of the man who bought a horse at a farthing for the first
nail in his shoe, a penny for the second, &c., is thrown into
the shade; and we incline to doubt whether there is not
some mistake; and yet on just such chances as one to all
these, do gamblers constantly risk their money!
Example 3. I have written the letters contained in the
word N I M R O D on 6 cards; being one letter on each,
and having thrown them confusedly into a hat, I am offered
$10 to draw the cards successively, so as to spell the name
correctly. What is my chance of success worth? Ans. 1 7/18
cents.
Example 4. In order to form a lottery scheme, I have put
into the wheel as many cards as I can put 4 letters of the
word Charleston on, without having the same letters in the
same order upon any two cards. I offer $100 to him who
draws the card having on it the first four letters of the said
word in their natural order (Char). What is the chance
of drawing a prize worth?
There are 10 letters in the word, and the combination is
of the 4th class; and, according to the mode of determining
combinations with repetitions, we find the whole number
of combinations of the 4th class which the word admits
of is 210. Then he has one chance in 210 of drawing the
letters Char, in some order. The number of permutations of
four individual things is 1 × 2 × 3 × 4 = 24, and 210 × 24 = 5040
his chance of drawing them in the right order, and $100
divided by 5040 gives Ans. 1 52/63 cents.
Suppose that the numbers from 1 to 78, inclusive, be
placed upon 78 cards, and the cards placed in a wheel by
which they are thoroughly mixed; and then 13 cards be
successively drawn out, by a person who has no means of
choosing, and the numbers on them registered. Suppose
also that tickets have been issued, containing each three of
the 78 numbers, but no two having all the same numbers,
and that he who holds the ticket having on it the first three
drawn numbers in their regular order, shall be entitled to
$100,000; what would the probability of drawing such a
ticket be worth?
Note.—It is usual also, to give smaller prizes to the holders
of tickets having the numbers in any order, or having
any two or one of the drawn numbers. Lotteries may be
arranged on a great diversity of plans, and in each the probability
of drawing prizes will vary.
A speaks the truth 3 times in 4; B 4 times in 5, and C 6
times in 7. What is the probability of an event which A
and B assert, and C denies? Ans. 140/143
Suppose a coin be thrown up, having two faces; what is
the probability that the obverse (heads) side will fall upward,
and what the reverse?
Here there are only two possible cases, and one favors each
of the contingencies the probability of each will be 1/(1 + 1) = 1/2;
there being no reason why one side should fall uppermost
rather than the other.
What would be the probability of either side presenting
upwards twice in two throws?
Here we have 4 possible cases, viz.:
Obverse and reverse;
Obverse both times;
Reverse and obverse;
Reverse both times.
Of the 4 possibilities there is only one which favors the
turning up of the obverse twice in succession, and the
same is true of the reverse, hence the probability of either
is only 1/4.
In like manner we might show that the probability of the
obverse presenting upwards three times in succession will
be 1/8, or 1/2 × 1/2 × 1/2; the general principle being to multiply
successively together the independent probabilities of an
event for the fraction expressing the chance of all the events
happening.
THE VISITORS TO THE CRYSTAL PALACE.
In a family consisting of 8 young people, it was agreed
that 3 at a time should visit the Crystal Palace, and that
the visit should be repeated each day as long as a different
trio could be selected. In how many days were the possible
combinations of 3 out of 8 completed?
We must multiply 8 × 7 × 6, and also 3 × 2 × 1, and divide
the product of the former, 336, by the product of the latter,
6; the result is 56, the number of visits, a different three
going each time. So much gratified were they with the results
of their agreement, that they wished to be allowed
another series of visits, to be continued as many days as
they could group 3 together in different order when starting.
If Paterfamilias had granted such permission he would have
had to wait 56 multiplied by 3 × 2 × 1, or 336 days, before
this "new series" of visits would have come to a finis.
HOW MANY CHANGES CAN BE GIVEN TO 7 NOTES OF A PIANO?
That is to say, in how many ways can 7 keys be struck
in succession, so that there shall be some difference in the
order of the notes each time?
The result of multiplying
7 × 6 × 5 × 4 × 3 × 2 × 1
is 5,040, the number of changes.
THE ARITHMETICAL TRIANGLE.
This name has been given to a contrivance said to have
originated with the famous Pascal, or to have been perfected
by him.
1
2 1
3 3 1
4 6 4 1
5 10 10 5 1
6 15 20 15 6 1
7 21 35 35 21 7 1
8 28 56 70 56 28 8 1
&c. &c.
This peculiar series of numbers is thus formed: Write
down the numbers 1, 2, 3, &c., as far as you please, in a vertical
row. On the right hand of 2 place 1, add them together,
and place 3 under the 1; then 3 added to 3 = 6, which place
under the 3; 4 and 6 are 10, which place under the 6, and
so on as far as you wish. This is the second vertical row,
and the third is formed from the second in a similar way.
This triangle has the property of informing us, without
the trouble of calculation, how many combinations can be
made, taking any number at a time out of a larger number.
Suppose the question were that just given; how many
selections can be made of 3 at a time out of 8? On the
horizontal row commencing with 8, look for the third number;
this is 56, which is the answer.
HOW MANY DIFFERENT DEALS CAN BE MADE WITH 13 CARDS OUT OF 52?
To discover this we must make a continued multiplication
of 52 × 51 × 50 × 49 × 48 × 47 × 46 × 45 × 44 × 43 × 42 × 41 × 40,
being 13 terms for the 13 cards, also a continued multiplication
of 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, and
having found the two products, we must divide one by the
other, and the quotient is the number of different deals out
of 52 cards. This "sum," that looks so formidable with
natural figures, is a very short one by logarithms.
THE THREE GRACES.
Three articles, or three names inscribed on cards, having
been distributed between three persons, you are to tell
which article or card each person has.
Designate the three persons in your own mind, as 1st, 2d,
and 3d, and the three articles, A, E, I. Provide 24 counters,
and give 1 to the first person, 2 to the 2d, 3 to the 3d.
Place the remaining 18 on the table. Request that the
three persons will distribute among themselves the three
articles, and that, having done so, the person who has the
one which you have secretly denoted by A, will take as many
counters as he may have already; the holder of E must take
twice as many as he may have; and the holder of I must
take four times as many. Then leave the room, in order
that the distribution of articles and of counters may be made
unobserved by you. We will suppose that the three articles
are three cards, on which are the words Clara, Rosa, Emily,
which you will yourself secretly denote by the letters A, E, I.
Suppose also that in the division the first person has Emily
(I), the second has Clara (A), and the third has Rosa (E),
then the 1st will take four times as many counters as he has
(1), and will therefore take 4; the 2d will take as many as
he has (2), and will therefore take 2; the 3d will take 6,
being twice as many as he has (3). On the table will be
left 6 counters. The distribution having been made, you
will return and observe the number of counters on the table,
from which you can find who is the holder of each card by
the following method.
It is plain that if the cards held by the 1st and 2d can be
told, that held by the 3d will be known. It will be found
that only six numbers can remain, viz. 1, 2, 3, 5, 6, 7; never
4, and never more than 7. Now the 6 combinations of a, e,
and i, here given, represent the articles held by the 1st and
2d persons.
| 1 |
2 |
3 |
4 |
5 |
6 |
7 |
| ae |
ea |
ai |
— |
ei |
ia |
ie |
In the case supposed, 6 counters being on the table, the
combination ia indicates that the first person has the card
you have called I (Emily), the 2d has A (Clara), so that the
3d has E (Rosa).
In order to recollect the combinations of A, E, and I, it will
be best to keep in memory some 7 words which form a sentence,
and which contain these vowels in the order just given.
Our young friends can amuse themselves in forming a
sentence for themselves, but as examples we supply three.
| 1 |
2 |
3 |
4 |
5 |
6 |
7 |
| ae |
ea |
ai |
— |
ei |
ia |
ie |
| James |
easy |
admires |
now |
reigning |
with a |
bride. |
| Anger, |
fear, |
pain |
may |
be hid |
with a |
smile. |
| Graceful |
Emma, |
charming |
she |
reigns |
in all |
circles. |
Or, if they prefer Latin, they can use the pentameter made
up by the inventor of this beautiful pastime:
| 1 |
2 |
3 |
5 |
6 |
7 |
| Salve |
certa |
animæ |
semita |
vita |
quies. |
ANOTHER METHOD.
The performer must mentally distinguish the articles by
the letters A, B, C, and the persons as 1st, 2d, and 3d. The
persons having made their choice, give 12 counters to the
1st, 24 to the 2d, and 36 to the 3d. Then request the 1st
person to add together the half of the counters of the person
who has chosen A, the 3d of the person who has chosen B,
and the 4th of those of the person who has chosen C, and
then ask the sum, which must be either 23, 24, 25, 27, 28,
or 29, as in the following table:
| First. |
Second. |
Third. |
|
| 12 |
24 |
36 |
|
| A |
B |
C |
23 |
| A |
C |
B |
24 |
| B |
A |
C |
25 |
| C |
A |
B |
27 |
| B |
C |
A |
28 |
| C |
B |
A |
29 |
This table shows that if the sum be 25, for example, the
1st person must have chosen B, the 2d A, and the 3d C; or
if it be 28, the 1st must have chosen B, the 2d C, and the
3d A.
ANOTHER METHOD.
Three things having been divided between three persons,
you are to determine the holder of each.
Call the persons in your own mind 1st, 2d, 3d.
Give to the 1st a card on which you have written the
number 12; to the 2d the number 24; to the 3d 36.
The three things you must denote as A, E, I.
To simplify it you may have three cards with a name upon
each, of which the initial letters are A, E, I, as Anna, Emma,
Isabel.
Request your friends to divide between them the three
articles, and then to add together certain parts of the numbers
on their cards, as follows:
Whoever has A must supply one half of the number on
his card;
Whoever has E must supply one third;
Whoever has I must supply one fourth;
This half, third and fourth having been added together,
the sum must be announced to you on your return; and from
this number you can tell who has A, who has E, and who has I.
| If the No. is |
the 1st has |
the 2d has |
the 3d has |
| 23 |
A |
E |
I |
| 24 |
A |
I |
E |
| 25 |
E |
A |
I |
| 27 |
I |
A |
E |
| 28 |
E |
I |
A |
| 29 |
I |
E |
A |
The sum which will be given to you can be one of six
only. There are only six ways in which the articles can be
divided, and there is a definite number for each of them.
The number 26 can never occur, and to recollect the six
which do occur, and which you perceive are consecutive,
you need take note only of what the 1st and 2d persons have.
| 23 |
24 |
25 |
26 |
27 |
28 |
29 |
| ae |
ai |
ea |
— |
ia |
ei |
ie |
If you make up a line of good (or bad) English, having
the vowels in the order here given, you will find it will aid
you in their recollection. We give one as a specimen:
| ae |
ai |
ea |
— |
ia |
ei |
ie |
| Brave |
dashing |
sea, |
like a |
giant |
revives |
itself. |
THE FORTUNATE NINTH.
A sharp youth, fresh from school, having gone to visit a
good-natured uncle, the latter placed on a table fifteen fine
oranges and fifteen apples, and desired his young friend to
take half. He, not liking the apples, was about to take the
fifteen oranges; but this monopoly of the best fruit being
objected to, the old gentleman told him to range all the
fruit in a circle, and to take every ninth. The clever fellow
ranged them in such a way as that, by taking away
every ninth, all the apples were left on the table, and all the
oranges were transferred to his capacious pockets. How did
he arrange them?
He placed them as in the margin, A representing
apples, and O oranges; and it
will be found that, by commencing at the
four apples, and going round and round
the circle, taking away every ninth, all
the oranges will be removed, and all the
apples will remain.
| If we let the vowels |
a |
e |
i |
o |
u |
| denote the figures |
1 |
2 |
3 |
4 |
5, |
the arrangement of the figures 4, 5, 2, 1, &c., can be easily
recollected by the following line:
| Our |
earth's |
final |
fate— |
enigma |
ever |
dark |
| 45 |
21 |
3 1 |
1 2 |
2 3 1 |
2 2 |
1 |
or,
| Our |
dear |
Richard's |
tale |
begins |
at |
the |
sea. |
| 45 |
21 |
3 1 |
1 2 |
2 3 |
1 |
2 |
21 |
Our young friends may find amusement in forming lines for
themselves as much superior to these as possible.
THE TEN TENS.
Take ten pieces of card, and upon each write any ten
words; there is no restriction as to the initial letter of nine
of the words, but the last word on each card must commence
with certain letters which you must in your own
mind associate with the numbers 1 to 10, so that by knowing
the initial letter of the last word on each card, you can
determine its number.
Here are ten cards, (call these the Selecting Cards,) which
we give by way of example, though our readers will perhaps
prefer having words of their own selection.
| Jane. |
Ellen. |
George. |
James. |
Newton. |
| Mary. |
Fanny. |
William. |
Clement. |
Davy. |
| Matilda. |
Caroline. |
Frederick. |
Edward. |
Morse. |
| Sarah. |
Isabel. |
Robert. |
Ralph. |
Fulton. |
| Rosa. |
Flora. |
Edmund. |
Francis. |
Franklin. |
| Elizabeth. |
Laura. |
John. |
Edwin. |
Arago. |
| Harriet. |
Maria. |
Alfred. |
Walter. |
Spurzheim. |
| Ann. |
Frances. |
Albert. |
Charles. |
Laplace. |
| Emily. |
Edith. |
Henry. |
Samuel. |
Steers. |
| Emma. |
Dorothea. |
Isaac. |
Theodore. |
Herschel. |
| |
|
|
|
|
| Sister. |
Rose. |
Friendship. |
Putnam. |
Clay. |
| Brother. |
Violet. |
Happiness. |
Lafayette. |
Webster. |
| Uncle. |
Lupin. |
Industry. |
Steuben. |
Calhoun. |
| Aunt. |
Daisy. |
Ambition. |
Scott. |
Benton. |
| Grandmother. |
Tulip. |
Energy. |
Taylor. |
Jefferson. |
| Grandfather. |
Peony. |
Fidelity. |
Green. |
Adams. |
| Nephew. |
Hyacinth. |
Affection. |
Harrison. |
Madison. |
| Niece. |
Pink. |
Hope. |
Hamilton. |
Jackson. |
| Cousin. |
Snowdrop. |
Justice. |
Wayne. |
Monroe. |
| Father. |
Lily. |
Order. |
Washington. |
Napoleon. |
For these the key words are, "Edith Flown," so that the
letters
| |
E |
D |
I |
T |
H |
F |
L |
O |
W |
N |
| Stand for |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
For the success of the game, the key words and the numbers
denoted by their letters, must be carefully concealed.
Take ten other cards, which call the "grouped cards," and
upon one write down the first word from each of the selecting
cards, being careful to write them in the same order.
Let another card contain all the words which are second
from the top, and so on till all the words have been grouped
together. As an example, we give the 1st and 4th grouped
cards.
| 1st. |
4th. |
| Jane. |
Sarah. |
| Ellen. |
Isabel. |
| George. |
Robert. |
| James. |
Ralph. |
| Newton. |
Fulton. |
| Sister. |
Aunt. |
| Rose. |
Daisy. |
| Friendship. |
Ambition. |
| Putnam. |
Scott. |
| Clay. |
Benton. |
The object of the game is to guess which of the words
from any of the selecting cards any person may have fixed
upon.
Let any one choose a card out of the selecting cards, and
after he has fixed upon a word, give it back to you; when
receiving it, carefully note the last word upon it, which will
give you, by the aid of the key word, the number of the
card; this you must keep secret, and you then give him all
the grouped cards, and request him to show you the cards
which contain the words he fixed upon.
You can then announce the word; for the number of the
word from the top on the grouped card is the same as the
number of the selecting card, from which he made his
choice.
Suppose he made his choice from the card which has
Theodore for its last word—this is No. 4; when he shows
you the grouped card, which he says contains the selected
word, you will know that Ralph, the fourth from the top, is
the name he fixed upon.
DIVIDING THE BEER.
During the siege of Sebastopol, when the troops were
on "short allowance," a can of eight pints of porter was
ordered to be equally divided between two messes; but
having only a five pint can, and one which held three pints,
it was found impossible to make this division, till one of the
clever sappers suggested the following method; and, to
understand it, we will put down the contents of each of the
three cans at each stage of the process; commencing with
| |
|
8-pt. |
5-pt. |
3-pt. |
| |
The 8-pint can full, and the others empty, |
8 |
0 |
0 |
| 1. |
Filled the 5-pint can |
3 |
5 |
0 |
| 2. |
Filled the 3-pint can from the 5-pint |
3 |
2 |
3 |
| 3. |
Pour the contents of the 3-pint into the 8-pint |
6 |
2 |
0 |
| 4. |
Transfer the 2 pints from the 5-pint to the 3-pint |
6 |
0 |
2 |
| 5. |
Filled the 5-pint from the 8-pint |
1 |
5 |
2 |
| 6. |
Fill up the 3-pint from the 5-pint |
1 |
4 |
3 |
| 7. |
Poured the 3 pints into the 8-pint; completing the feat |
4 |
4 |
0 |
This was a dexterous expedient of the worthy sapper, the
only objections to it being the time the thirty men had to
wait, and the resulting flat condition of the beer.
THE DIFFICULT CASE OF WINE.
A gentleman had a bottle containing 12 pints of wine, 6
of which he was desirous of giving to a friend; but he had
nothing to measure it, except two other bottles, one of 7
pints, and the other of 5. How did he contrive to put 6
pints into the 7-pint bottle?
| |
|
12-pt. |
7-pt. |
6-pt. |
| |
Before he commenced, the contents of the bottles were |
12 |
0 |
0 |
| 1. |
He filled the 5-pint |
7 |
0 |
5 |
| 2. |
Emptied the 5-pint into the 7-pint |
7 |
5 |
0 |
| 3. |
Filled again the 5-pint from the 12-pint |
2 |
5 |
5 |
| 4. |
Filled up the 7-pint from the 5 |
2 |
7 |
3 |
| 5. |
Emptied the 7-pint into the 12-pint |
9 |
0 |
3 |
| 6. |
Poured the 3 pints from the 5 into the 7 |
9 |
3 |
0 |
| 7. |
Filled the 5-pint from the 12-pint |
4 |
3 |
5 |
| 8. |
Filled up the 7-pint from the 5-pint |
4 |
7 |
1 |
| 9. |
Emptied the 7-pint into the 12-pint |
11 |
0 |
1 |
| 10. |
Poured 1 pint from the 5-pint into the 7-pint |
11 |
1 |
0 |
| 11. |
Filled the 5-pint from the 12-pint |
6 |
1 |
5 |
| 12. |
Poured the contents of the 5-pint into the 7-pint |
6 |
6 |
0 |
ANOTHER DECIMATION OF FRUIT.
On the next visit of the youth to his uncle, the latter produced
thirty apples and ten oranges, and offered him the
favorite oranges, if his nephew
could arrange them in an oval,
so that by taking every twelfth
the apples should remain. But
this he could not accomplish,
and the old gentleman, being
well versed in the "Recreations
in Science," proceeded to arrange them thus:
The places which the oranges here occupy can be easily
remembered, being Nos. 7, 8, 11, 12, 21, 22, 24, 34, 36, 37.
THE WINE AND THE TABLES.
A certain hotel-keeper was dexterous in contrivances to
produce a large appearance with small means. In the dining-room
were three tables, between which he could divide
21 bottles, of which 7 only were full, 7 half full, and 7 apparently
just emptied, and in such a manner that each table
had the same number of bottles, and the same quantity of
wine. He did this in two ways:
| Table. |
Full Hf. |
full. |
Empty. |
|
Table. |
Full Hf. |
full. |
Empty. |
| 1 |
2 |
3 |
2 |
| |
1 |
3 |
1 |
3 |
| 2 |
2 |
3 |
2 |
| |
2 |
3 |
1 |
3 |
| 3 |
3 |
1 |
3 |
| |
3 |
1 |
5 |
1 |
He also performed a similar exploit with 24 bottles, 8 full,
8 half-full, and 8 empty:
| Table. |
Full Hf. |
full. |
Empty. |
|
Table. |
Full Hf. |
full. |
Empty. |
| 1 |
3 |
2 |
3 |
| |
1 |
2 |
4 |
2 |
| 2 |
3 |
2 |
2 |
| |
2 |
2 |
4 |
2 |
| 3 |
2 |
4 |
2 |
| |
3 |
4 |
0 |
4 |
Also with 27 bottles, 9 full, 9 half-full, and 9 empty:
| Table. |
Full Hf. |
full. |
Empty. |
|
Table. |
Full Hf. |
full. |
Empty. |
| 1 |
2 |
5 |
2 |
| |
1 |
1 |
7 |
1 |
 |
| 2 |
3 |
3 |
3 |
| |
2 |
4 |
1 |
4 |
|
| 3 |
4 |
1 |
4 |
| |
3 |
4 |
1 |
4 |
|
THE THREE TRAVELERS.
Three men met at a caravansary or inn, in Persia; and
two of them brought their provision along with them, according
to the custom of the country; but the third not
having provided any, proposed to the others that they should
eat together, and he would pay the value of his proportion.
This being agreed to, A produced 5 loaves, and B 3 loaves,
all of which the travelers ate together, and C paid 8 pieces
of money as the value of his share, with which the others
were satisfied, but quarreled about the division of it. Upon
this the matter was referred to the judge, who decided impartially.
What was his decision?
At first sight it would seem that the money should be divided
according to the bread furnished; but we must consider
that, as the 3 ate 8 loaves, each one ate 2-2/3 loaves of
the bread he furnished. This from 5 would leave 2-1/3 loaves
furnished the stranger by A; and 3 - 2-2/3 = 1/3 furnished by B,
hence 2-1/3 to 1/3 = 7 to 1, is the ratio in which the money is to
be divided. If you imagine A and B to furnish, and C to
consume all, then the division will be according to amounts
furnished.
WHICH COUNTER HAS BEEN THOUGHT OF OUT OF SIXTEEN?
Take sixteen pieces of card, and number them 1 to 16.
Arrange them in two rows, as at A B.
| A |
B |
C |
B |
D |
M |
E |
B |
F |
N |
G |
B |
H |
| 1 |
9 |
1 |
9 |
2 |
2 |
2 |
9 |
4 |
2 |
2 |
9 |
6 |
| 2 |
10 |
3 |
10 |
4 |
4 |
6 |
10 |
8 |
6 |
1 |
10 |
5 |
| 3 |
11 |
5 |
11 |
6 |
6 |
1 |
11 |
3 |
1 |
4 |
11 |
8 |
| 4 |
12 |
7 |
12 |
8 |
8 |
5 |
12 |
7 |
5 |
3 |
12 |
7 |
| 5 |
13 |
|
13 |
|
1 |
|
13 |
|
4 |
|
13 |
|
| 6 |
14 |
|
14 |
|
3 |
|
14 |
|
8 |
|
14 |
|
| 7 |
15 |
|
15 |
|
5 |
|
15 |
|
3 |
|
15 |
|
| 8 |
16 |
|
16 |
|
7 |
|
16 |
|
7 |
|
16 |
|
Desire a person to think of one of the numbers, and to
tell you in which row it is. Suppose he fixes on 6; he will
tell you that the row A contains the number he thought of.
Take up the row A, and arrange the numbers on each side
of the row B, as shown at C D, so that the first number of
the row A may be the first of the row C, the second of A be
the first of D, the third of A be the second of C, and so on.
Ask in which of the rows, C or D, is the number thought
of: in the case supposed it is in D.
Take up the rows C D, and put one underneath the other,
as at M, taking care that the half-row in which is the number
thought of, shall be above the other.
Divide it again into two rows, as at E F, on each side of
B, in the same way as before. Ask again in which row it
is: it is now in E.
Place one row under the other, as at N, and divide again
into two rows, which will now be as G H.
You will be informed that the number is in row H, and you
may then announce it to be the top number of that row.
The number thought of will always be at the top of one of
the rows after three transpositions. If there were 32 counters
it would be at the top after four transpositions.
MAGIC SQUARES.
The name "Magic Square" is given to a square divided
into several smaller squares, in which numbers are placed
in such a manner that every column of numbers, whether
vertical, horizontal, or from corner to corner, shall amount
to the same sum.
They are divided into three principal classes: 1st, Those
which have an odd number of squares in each band; 2d,
Those which have an even number of squares in each band,
this even number being divisible exactly by 4; 3d, Where
the even number of squares in each band cannot be divided
by 4 without a fraction.
ODD MAGIC SQUARES.
Squares of this kind are formed thus. Imagine an exterior
line of squares above the magic square you wish to
form, and another exterior line on the right hand of it
These two imaginary lines are shown in the figure.
Then attend to the two following rules:
1st. In placing the numbers in the squares we must go
in an ascending oblique direction from left to right; any
number which, by pursuing this direction, would fall into
the exterior line, must be carried along that line of squares,
whether vertical or horizontal, to the last square. Thus, 1
having been placed in the center of the top line, (see the
first table on p. 228,) 2 would fall into the exterior square
above the fourth vertical line; it must be therefore carried
down to the lowest square of that line; then, ascending
obliquely 3 falls into the square, but four falls out of it, to
the end of a horizontal line, and it must be carried along that
line to the extreme left, and there placed. Resuming our
oblique ascension to the right, we place 5, where the reader
sees it, and would place 6 in the middle of the top band,
but finding it occupied by 1, we look for direction to the
2d Rule, which prescribes that, when in ascending
obliquely, we come to a square already occupied, we must
place the number, which according to the first rule should
go into that occupied square, directly under the last number
placed. Thus, in ascending with 4, 5, 6, the 6 must be
placed directly under the 5, because the square next to 5 in
an oblique direction is "engaged."
Magic squares of this class, however large in the number
of compartments, can be easily filled up by attending to
these two rules.
We give opposite, a seven-placed square.
There are various other kinds of magic squares; but explanations
of them would be too lengthy for our work.
The invention of these contrivances has been traced
back to the early ages of science, and talismanic properties
were attributed to them. Modern philosophers have amused
themselves in bringing them to perfection, and none has
contributed so much as "the model of practical wisdom,"
Dr. Franklin.
THE SQUARE OF GOTHAM.
The wise men of Gotham, famous for their eccentric blunders,
once undertook the management of a school; they
arranged their establishment in the form of a square divided
into 9 rooms. The playground occupied the center, and
24 scholars the rooms around it, 3 being in each. In spite
of the strictness of discipline, it was suspected that the boys
were in the habit of playing truant, and it was determined
to set a strict watch. To assure themselves that all the
boys were on the premises, they visited the rooms, and found
three in each, or 9 in each row. Four boys then went out,
and the wise men soon after visited the rooms, and finding
9 in each row, thought all was right. The four boys then
came back, accompanied by four strangers; and the Gothamites,
on their third round, finding still 9 in each row,
entertained no suspicion of what had taken place. Then 4
more "chums" were admitted; but the clever men, on examining
the establishment a fourth time, still found 9 in
each row, and so came to an opinion that their previous
suspicions had been unfounded. How was all this possible?
The following figures represent the contents of each room
at the four different visits; the first, at the commencement
of the watch; the second, when four had gone out; the
third, when these 4, accompanied by another 4, had returned;
and the fourth, when 4 more had joined them.
On each change the boys arranged themselves in the
rooms in such a manner that, when the corner rooms were
counted as a part of two rows, each entire row of three
rooms contained the same number of boys. The illusion of
the wise men was due to their mistake in counting each
corner room twice.
THE MATHEMATICAL BLACKSMITH.
A blacksmith had a stone weighing 40 lbs. A mason coming
into the shop, hammer in hand, struck it and broke it
into four pieces. "There," says the smith, "you have ruined
my weight." "No," says the mason, "I have made it better,
for whereas you could before weigh but 40 lbs. with it,
now you can weigh every pound from 1 to 40." Required
size of the pieces?
Ans. 1, 3, 9, 27; for in any geometrical series proceeding
in a triple ratio, each term is 1 more than twice the sum of
all the preceding, and the above series might proceed to
any extent. In using the weights, they must be put in one
or both scales as may be necessary: as to weigh 2, put 1
in one scale, and 3 in the other.
CURIOUS PROPERTIES OF SOME FIGURES.
Select any two numbers you please, and you will find
that one of the two, their amount when added together, or
their difference, is always 3, or a number divisible by 3.
Thus, if the numbers are 3 and 8, the first number is 3;
let the numbers be 1 and 2, their sum is 3; let them be 4
and 7, the difference is 3. Again, 15 and 22, the first number
is divisible by 3: 17 and 26, their difference is divisible
by 3, &c.
All the odd numbers above 3, that can only be divided
by 1, can be divided by 6, by the addition or subtraction 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 |
| 365472 |
} |
9 |
| 6 |
} |
2 |
| ——— |
} |
3 |
| 219232 |
} |
2 |
| |
} |
— |
| |
} |
19 |
The amount of these numbers is 19; but 19, divided by
9, leaves a remainder of 1; you, therefore, want 8 to complete
another 9: 8, then, is the number erased.
The component figures of the product made by the multiplication
of every digit into the number 9, when added
together, make NINE.
The order of these component figures is reversed after
the said number has been multiplied by 5.
The component figures of the amount of the multipliers
(viz. 45,) when added together, make NINE.
The amount by the several products, or multiples of 9
(viz. 405,) when divided by 9, gives for a quotient, 45; that
is, 4 + 5 = NINE.
The amount of the first product (viz. 9,) when added to
the other product, whose respective component figures make
9, is 81; which is the square of NINE.
The said number 81, when added to the above mentioned
amount of the several products, or multiples of 9 (viz. 405)
makes 486, which, if divided by 9, gives for a quotient 54:
that is, 5 + 4 = NINE.
It is also observable, that the number of changes that
may be rung on nine bells is 362,880; which figures, added
together, make 27; that is, 2 + 7 = NINE.
And the quotient of 362,880, divided by 9, will be 40,320;
that is 4 + 0 + 3 + 2 + 0 = NINE.
If number 37 be multiplied by any of the progressive
numbers arising from the multiplication of 3 with any of
the units, the figures in the quotient will be similar, and the
result may be known beforehand by merely inspecting the
progressive numbers, thus, 3, 6, 9, 12, 15, 18, 21, 24, 27, &c.,
are the progressive numbers formed by 3 multiplied by the
units 1 to 9; and the result of the multiplication of any of
these numbers with 37 may be seen in the following examples:—37 × 3 = 111;
37 × 6 = 222; 37 × 12 = 444; 37 × 24 =
888; by which it appears that the numbers of which the
quotient is formed are the same as the units by which number
3 was multiplied to obtain the respective progressive
numbers. Thus—3 multiplied by 2 is equal to 6, and 37
multiplied by 9 is equal to 222; so, again, 4 multiplied by
3 produces 12, and 37 multiplied by 12 is equal to 444, and
so on.
THE INDUSTRIOUS FROG.
There was a well 30 feet deep, and at the bottom a frog
anxious to get out. He got up 3 feet per day, but regularly
fell back 2 feet at night. Required the number of days
necessary to enable him to get out?
The frog appears to have cleared one foot per day, and
at the end of 27 days, he would be 27 feet up, or within 3
feet of the top, and the next day he would get out. He
would therefore be 28 days getting out.
THE COUNCIL OF TEN.
Ten cards or counters, numbered from one to ten, or the
first ten playing cards of any suit disposed in a circular
form may be employed with great convenience for performing
this feat. The accompanying figure shows the cards
thus arranged, number one, or the ace, designated by A,
and the ten by K.
Having placed the cards in the above order, desire a
bystander to think of a card or number, and when he has
done so, to touch any other card or number. Request him
then to add to the number of the card touched the number
of the cards employed, which in this case is ten. Then
desire him to count the sum in an order contrary to that of
the natural numbers, beginning at the card he touched, and
assigning it the number of the card he thought of. By
counting in this manner, he will end at the number or card
he thought of, and consequently you will immediately
know it.
Thus, for example, suppose the person had thought of 3 C,
and touched 6 F; then, if 10 be added to 6, the sum will
be 16; and if that number be counted from F, the number
touched, towards E D B C A, and so on, in the retrograde
order, counting F three, the number thought of, E five, D
six, and so round to sixteen, that number will terminate at
C, showing that the person thought of 3, the number which
corresponds to C.
A greater or less number of cards or counters may be
employed at pleasure; but in every instance the whole
number of cards must be added to the number of the card
touched.
THE TWO TRAVELERS.
Two travelers trudged along the road together,
Talking, as Yankees do, about the weather;
When, lo! beside their path the foremost 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:
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 |
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 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,
3 1⁄2; and 2 1⁄2; 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 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.
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 51⁄2 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 1/2
a foot = 6 inches, from which we take 5 inches, and 1 remains;
we then carry 1/2 to 1, and borrowing 1/2 a yard = 1 1/2 feet,
we have 1 1/2 from 1 1/2 = 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 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
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.
67,856
47,218
52,781
———
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 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
2 1/3 times. In this case, we must, as before, multiply 4 by
2 1/3, which will give 9 1/3; to which, if 4 be added, we shall
have 13 1/3, or 40/3; if 1, then, be taken from 2 1/3, the remainder
will be 1 1/3, 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?
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.—1 1/4, 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.—6 6/6.
Add one to nine and make it twenty.
Ans. IX—cross the 1, it makes XX.
Place four fives so as to make six and a half. Ans.—5 5/5·5.
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; but suppose 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 99959/960 is £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 11 143/240, 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 of times, and it is obviously
impossible to bring value into 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
landlord 2/5 of the produce; but prior 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 71/5 bushels by such an arrangement,
as the rent would entitle him to 2/5 of the 18. The
tenant should give him 18 bushels from his own share after
the division is completed, otherwise the landlord would receive
but 2/7 of 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!