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.
[Illustration]
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.
=E=mma. =D=orothea. =I=saac. =T=heodore. =H=erschel.
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.
=F=ather. =L=ily. =O=rder. =W=ashington. =N=apoleon.
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:
[Illustration]
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 2⅔ loaves of the bread he furnished. This from 5 would
leave 2⅓ loaves furnished the stranger by A; and 3 - 2⅔ = ⅓
furnished by B, hence 2⅓ to ⅓ = 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."
[Illustration:
...........................................
. . . . . . .
. . . . . . .
. . 18 . 25 . 2 . 9 . .
. . . . . . .
+----------------------------------+.......
| | | | | | .
| | | | | | .
| 17 | 24 | 1 | 8 | 15 | 17 .
| | | | | | .
|------+------+------+------+------+.......
| | | | | | .
| | | | | | .
| 23 | 5 | 7 | 14 | 16 | 23 .
| | | | | | .
|------+------+------+------+------+.......
| | | | | | .
| | | | | | .
| 4 | 6 | 13 | 20 | 22 | 4 .
| | | | | | .
|------+------+------+------+------+.......
| | | | | | .
| | | | | | .
| 10 | 12 | 19 | 21 | 3 | 10 .
| | | | | | .
|------+------+------+------+------+.......
| | | | | | .
| | | | | | .
| 11 | 18 | 25 | 2 | 9 | .
| | | | | | .
+------+------+------+------+------+.......
]
[Illustration:
+------+------+------+------+------+------+------+
| | | | | | | |
| | | | | | | |
| 30 | 39 | 48 | 1 | 10 | 19 | 28 |
| | | | | | | |
|------+------+------+------+------+------+------|
| | | | | | | |
| | | | | | | |
| 38 | 47 | 7 | 9 | 18 | 27 | 29 |
| | | | | | | |
|------+------+------+------+------+------+------|
| | | | | | | |
| | | | | | | |
| 46 | 6 | 8 | 17 | 26 | 35 | 37 |
| | | | | | | |
|------+------+------+------+------+------+------|
| | | | | | | |
| | | | | | | |
| 5 | 14 | 16 | 25 | 34 | 36 | 45 |
| | | | | | | |
|------+------+------+------+------+------+------|
| | | | | | | |
| | | | | | | |
| 13 | 15 | 24 | 33 | 42 | 44 | 4 |
| | | | | | | |
|------+------+------+------+------+------+------|
| | | | | | | |
| | | | | | | |
| 21 | 23 | 32 | 41 | 43 | 3 | 12 |
| | | | | | | |
|------+------+------+------+------+------+------|
| | | | | | | |
| | | | | | | |
| 22 | 31 | 40 | 49 | 2 | 11 | 20 |
| | | | | | | |
+------+------+------+------+------+------+------+
]
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.
I. II. III. IV.
___________ ___________ ___________ ___________
| | | | | | | | | | | | | | | |
| 3 | 3 | 3 | | 4 | 1 | 4 | | 2 | 5 | 2 | | 1 | 7 | 1 |
|___|___|___| | __|___|___| | __|___|___| | __|___|___|
| | | | | | | | | | | | | | | |
| 3 | | 3 | | 1 | | 1 | | 5 | | 5 | | 7 | | 7 |
|___|___|___| | __|___|___| | __|___|___| | __|___|___|
| | | | | | | | | | | | | | | |
| 3 | 2 | 3 | | 4 | 1 | 4 | | 2 | 5 | 2 | | 1 | 7 | 1 |
|___|___|___| | __|___|___| | __|___|___| | __|___|___|
On each change the boys arranged themselves in the rooms in such a manner
that, when the corner rooms were counted as a part of two rows, each
entire row of three rooms contained the same number of boys. The illusion
of the wise men was due to their mistake in counting each corner room
twice.
THE MATHEMATICAL BLACKSMITH.
A blacksmith had a stone weighing 40 lbs. A mason coming into the shop,
hammer in hand, struck it and broke it into four pieces. "There," says the
smith, "you have ruined my weight." "No," says the mason, "I have made it
better, for whereas you could before weigh but 40 lbs. with it, now you
can weigh every pound from 1 to 40." Required size of the pieces?
_Ans._ 1, 3, 9, 27; for in any geometrical series proceeding in a triple
ratio, each term is 1 more than twice the sum of all the preceding, and
the above series might proceed to any extent. In using the weights, they
must be put in one or both scales as may be necessary: as to weigh 2, put
1 in one scale, and 3 in the other.
CURIOUS PROPERTIES OF SOME FIGURES.
Select any two numbers you please, and you will find that one of the two,
their amount when added together, or their difference, is always 3, or a
number divisible by 3.
Thus, if the numbers are 3 and 8, the first number is 3; let the numbers
be 1 and 2, their sum is 3; let them be 4 and 7, the difference is 3.
Again, 15 and 22, the first number is divisible by 3: 17 and 26, their
difference is divisible by 3, &c.
All the odd numbers above 3, that can only be divided by 1, can be divided
by 6, by the addition or subtraction of a unit. For instance, 13 can only
be divided by 1; but after deducting 1, the remainder can be divided by 6;
for example, 5 + 1 = 6 ; 7 - 1 = 6; 17 + 1 = 18; 19 - 1 = 18; 25 - 1 = 24,
and so on.
If you multiply 5 by itself, and the quotient again by itself, and the
second quotient by itself, the last figure of each quotient will always
be 5. Thus 5 × 5 = 25; 25 × 25 = 125; 125 × 125 = 625, &c. Again, if
you proceed in the same manner with the figure 6, the last figure will
constantly be 6; thus, 6 × 6 = 36; 36 × 36 = 216; 216 × 216 = 1,296, and
so on.
To multiply by 2 is the same as to multiply by 10 and divide by 5.
Any number of figures you may wish to multiply by 5, will give the same
result if divided by 2--a much quicker operation than the former; but you
must remember to annex a cipher to the answer where there is no remainder,
and where there is a remainder, annex a 5 to the answer. Thus, multiply
464 by 5, the answer will be 2320; divide the same number by 2, and you
have 232, and as there is no remainder you add a cipher. Now, take 357,
and multiply by 5--the answer is 1785. On dividing 357 by 2, there is 178,
and a remainder; you therefore place 5 at the right of the line, and the
result is again 1785.