DIVISION.
69. Suppose I ask whether 156 can be divided into a number of parts
each of which is 13, or how many thirteens 156 contains; I propose a
question, the solution of which is called DIVISION. In this case, 156
is called the _dividend_, 13 the _divisor_, and the number of parts
required is the _quotient_; and when I find the quotient, I am said to
divide 156 by 13.
70. The simplest method of doing this is to subtract 13 from 156,
and then to subtract 13 from the remainder, and so on; or, in common
language, to _tell off_ 156 by thirteens. A similar process has already
occurred in the exercises on subtraction, Art. (46). Do this, and
mark one for every subtraction that is made, to remind you that each
subtraction takes 13 once from 156, which operations will stand as
follows:
156
13 1
------
143
13 1
------
130
13 1
------
117
13 1
------
104
13 1
------
91
13 1
------
78
13 1
------
65
13 1
------
52
13 1
------
39
13 1
------
26
13 1
------
13
13 1
------
0
Begin by subtracting 13 from 156, which leaves 143. Subtract 13 from
143, which leaves 130; and so on. At last 13 only remains, from which
when 13 is subtracted, there remains nothing. Upon counting the number
of times which you have subtracted 13, you find that this number is 12;
or 156 contains twelve thirteens, or contains 13 twelve times.
This method is the most simple possible, and might be done with
pebbles. Of these you would first count 156. You would then take 13
from the heap, and put them into one heap by themselves. You would
then take another 13 from the heap, and place them in another heap by
themselves; and so on until there were none left. You would then count
the number of heaps, which you would find to be 12.
71. Division is the opposite of multiplication. In multiplication you
have a number of heaps, with the same number of pebbles in each, and
you want to know how many _pebbles_ there are in all. In division you
know how many there are in all, and how many there are to be in each
heap, and you want to know how many _heaps_ there are.
72. In the last example a number was taken which contains an exact
number of thirteens. But this does not happen with every number. Take,
for example, 159. Follow the process of (70), and it will appear that
after having subtracted 13 twelve times, there remains 3, from which
13 cannot be subtracted. We may say then that 159 contains twelve
thirteens and 3 _over_; or that 159, when divided by 13, gives a
_quotient_ 12, and a _remainder_ 3. If we use signs,
159 = 13 × 12 + 3.
EXERCISES.
146 = 24 × 6 + 2, or 146 contains six twenty-fours and 2 over.
146 = 6 × 24 + 2, or 146 contains twenty-four sixes and 2 over.
300 = 42 × 7 + 6, or 300 contains seven forty-twos and 6 over.
39624 = 7277 × 5 + 3239.
73. If _a_ contain _b_ _q_ times with a remainder _r_, _a_ must be
greater than _bq_ by _r_; that is,
_a_ = _bq_ + _r_.
If there be no remainder, _a_ = _bq_. Here _a_ is the dividend, _b_ the
divisor, _q_ the quotient, and _r_ the remainder. In order to say that
_a_ contains _b_ _q_ times, we write,
_a_/_b_ = _q_, or _a_ : _b_ = _q_,
which in old books is often found written thus:
_a_ ÷ _b_ = _q_.
74. If I divide 156 into several parts, and find how often 13 is
contained in each of them, it is plain that 156 contains 13 as often as
all its parts together. For example, 156 is made up of 91, 39, and 26.
Of these
91 contains 13 7 times,
39 contains 13 3 times,
26 contains 13 2 times;
therefore 91 + 39 + 26 contains 13 7 + 3 + 2 times, or 12 times.
Again, 156 is made up of 100, 50, and 6.
Now 100 contains 13 7 times and 9 over,
50 contains 13 3 times and 11 over,
6 contains 13 0 times[9] and 6 over.
[9] To speak always in the same way, instead of saying that 6 does not
contain 13, I say that it contains it 0 times and 6 over, which is
merely saying that 6 is 6 more than nothing.
Therefore 100 + 50 + 6 contains 13 7 + 3 + 0 times and 9 + 11 + 6 over;
or 156 contains 13 10 times and 26 over. But 26 is itself 2 thirteens;
therefore 156 contains 10 thirteens and 2 thirteens, or 12 thirteens.
75. The result of the last article is expressed by saying, that if
_a_ = _b_ + _c_ + _d_, then _a_/_m_ = _b_/_m_ + _c_/_m_ + _d_/_m_
76. In the first example I did not take away 13 more than once at a
time, in order that the method might be as simple as possible. But
if I know what is twice 13, 3 times 13, &c., I can take away as many
thirteens at a time as I please, if I take care to mark at each step
how many I take away. For example, take away 13 ten times at once from
156, that is, take away 130, and afterwards take away 13 twice, or take
away 26, and the process is as follows:
156
130 10 times 13.
---
26
26 2 times 13.
---
0
Therefore 156 contains 13 10 + 2, or 12 times.
Again, to divide 3096 by 18.
3096
1800 100 times 18.
----
1296
900 50 times 18.
----
396
360 20 times 18.
----
36
36 2 times 18.
----
0
Therefore 3096 contains 18 100 + 50 + 20 + 2, or 172 times.
77. You will now understand the following sentences, and be able to
make similar assertions of other numbers.
450 is 75 × 6; it therefore contains any number, as 5, 6 times as often
as 75 contains it.
135 contains 3 more than 26 times; therefore,
Twice 135 ” 3 ” 52 or twice 26 times.
10 times 135 ” 3 ” 260 or 10 times 26
50 times 135 ” 3 ” 1300 or 50 times 26
472 contains 18 more than 21 times; therefore,
4720 contains 18 more than 210 times,
47200 contains 18 more than 2100 times,
472000 contains 18 more than 21000 times,
32 contains 12 more than 2 times, and less than 3 times.
320 ” 12 ” 20 30
3200 ” 12 ” 200 300
32000 ” 12 ” 2000 3000
&c. &c. &c.
78. The foregoing articles contain the principles of division. The
question now is, to apply them in the shortest and most convenient way.
Suppose it required to divide 4068 by 18, or to find 4068/18 (23).
If we divide 4068 into any number of parts, we may, by the process
followed in (74), find how many times 18 is contained in each of these
parts, and from thence how many times it is contained in the whole.
Now, what separation of 4068 into parts will be most convenient?
Observe that 4, the first figure of 4068, does not contain 18; but that
40, the first and second figures together, _does contain 18 more than
twice, but less than three times_.[10] But 4068 (20) is made up of 40
hundreds, and 68; of which, 40 hundreds (77) contains 18 more than 200
times, and less than 300 times. Therefore, 4068 also contains more than
200 times 18, since it must contain 18 more times than 4000 does. It
also contains 18 less than 300 times, because 300 times 18 is 5400, a
greater number than 4068. Subtract 18 200 times from 4068; that is,
subtract 3600, and there remains 468. Therefore, 4068 contains 18 200
times, and as many more times as 468 contains 18.
[10] If you have any doubt as to this expression, recollect that it
means “contains more than two eighteens, but not so much as three.”
It remains, then, to find how many times 468 contains 18. Proceed
exactly as before. Observe that 46 contains 18 more than twice, and
less than 3 times; therefore, 460 contains it more than 20, and less
than 30 times (77); as does also 468. Subtract 18 20 times from 468,
that is, subtract 360; the remainder is 108. Therefore, 468 contains
18 20 times, and as many more as 108 contains it. Now, 108 is found to
contain 18 6 times exactly; therefore, 468 contains it 20 + 6 times,
and 4068 contains it 200 + 20 + 6 times, or 226 times. If we write down
the process that has been followed, without any explanation, putting
the divisor, dividend, and quotient, in a line separated by parentheses
it will stand, as in example(A).
Let it be required to divide 36326599 by 1342 (B).
A. B.
18)4068(200 + 20 + 6 1342)36326599(20000 + 7000 + 60 + 9
3600 26840000
---- --------
468 9486599
360 9394000
--- -------
108 92599
108 80520
--- -----
0 12079
12078
-----
1
As in the previous example, 36326599 is separated into 36320000 and
6599; the first four figures 3632 being separated from the rest,
because it takes four figures from the left of the dividend to make
a number which is greater than the divisor. Again, 36320000 is found
to contain 1342 more than 20000, and less than 30000 times; and 1342
× 20000 is subtracted from the dividend, after which the remainder is
9486599. The same operation is repeated again and again, and the result
is found to be, that there is a quotient 20000 + 7000 + 60 + 9, or
27069, and a remainder 1.
Before you proceed, you should now repeat the foregoing article at
length in the solution of the following questions. What are
10093874 66779922 2718218
-------- , -------- , ------- ?
3207 114433 13352
the quotients of which are 3147, 583, 203; and the remainders 1445,
65483, 7762.
79. In the examples of the last article, observe, 1st, that it is
useless to write down the ciphers which are on the right of each
subtrahend, provided that without them you keep each of the other
figures in its proper place: 2d, that it is useless to put down the
right hand figures of the dividend so long as they fall over ciphers,
because they do not begin to have any share in the making of the
quotient until, by continuing the process, they cease to have ciphers
under them: 3d, that the quotient is only a number written at length,
instead of the usual way. For example, the first quotient is 200 + 20
+ 6, or 226; the second is 20000 + 7000 + 60 + 9, or 27069. Strike
out, therefore, all the ciphers and the numbers which come above them,
except those in the first line, and put the quotient in one line; and
the two examples of the last article will stand thus:
18)4068(226 1342)36326599(27069
36 2684
--- -----
46 9486
36 9394
--- -----
108 9259
108 8052
--- -----
0 12079
12078
-----
1
80. Hence the following rule is deduced:
I. Write the divisor and dividend in one line, and place parentheses on
each side of the dividend.
II. Take off from the left-hand of the dividend the least number of
figures which make a number greater than the divisor; find what number
of times the divisor is contained in these, and write this number as
the first figure of the quotient.
III. Multiply the divisor by the last-mentioned figure, and subtract
the product from the number which was taken off at the left of the
dividend.
IV. On the right of the remainder place the figure of the dividend
which comes next after those already separated in II.: if the remainder
thus increased be greater than the divisor, find how many times the
divisor is contained in it; put this number at the right of the first
figure of the quotient, and repeat the process: if not, on the right
place the next figure of the dividend, and the next, and so on until it
is greater; but remember to place a cipher in the quotient for every
figure of the dividend which you are obliged to take, except the first.
V. Proceed in this way until all the figures of the dividend are
exhausted.
In judging how often one large number is contained in another, a first
and rough guess may be made by striking off the same number of figures
from both, and using the results instead of the numbers themselves.
Thus, 4,732 is contained in 14,379 about the same number of times
that 4 is contained in 14, or about 3 times. The reason is, that 4
being contained in 14 as often as 4000 is in 14000, and these last
only differing from the proposed numbers by lower denominations, viz.
hundreds, &c. we may expect that there will not be much difference
between the number of times which 14000 contains 4000, and that which
14379 contains 4732: and it generally happens so. But if the second
figure of the divisor be 5, or greater than 5, it will be more accurate
to increase the first figure of the divisor by 1, before trying the
method just explained. Nothing but practice can give facility in this
sort of guess-work.
81. This process may be made more simple when the divisor is not
greater than 12, if you have sufficient knowledge of the multiplication
table (50). For example, I want to divide 132976 by 4. At full length
the process stands thus:
4)132976(33244
12
---
12
12
---
9
8
--
17
16
---
16
16
--
0
But you will recollect, without the necessity of writing it down,
that 13 contains 4 three times with a remainder 1; this 1 you will
place before 2, the next figure of the dividend, and you know that 12
contains 4 3 times exactly, and so on. It will be more convenient to
write down the quotient thus:
4)132976
-------
33244
While on this part of the subject, we may mention, that the shortest
way to multiply by 5 is to annex a cipher and divide by 2, which is
equivalent to taking the half of 10 times, or 5 times. To divide by
5, multiply by 2 and strike off the last figure, which leaves the
quotient; half the last figure is the remainder. To multiply by 25,
annex two ciphers and divide by 4. To divide by 25, multiply by 4 and
strike off the last two figures, which leaves the quotient; one fourth
of the last two figures, taken as one number, is the remainder. To
multiply a number by 9, annex a cipher, and subtract the number, which
is equivalent to taking the number ten times, and then subtracting it
once. To multiply by 99, annex two ciphers and subtract the number, &c.
In order that a number may be divisible by 2 without remainder, its
units’ figure must be an even number.[11] That it may be divisible by
4, its last two figures must be divisible by 4. Take the example 1236:
this is composed of 12 hundreds and 36, the first part of which, being
hundreds, is divisible by 4, and gives 12 twenty-fives; it depends then
upon 36, the last two figures, whether 1236 is divisible by 4 or not.
A number is divisible by 8 if the last three figures are divisible by
8; for every digit, except the last three, is a number of thousands,
and 1000 is divisible by 8; whether therefore the whole shall be
divisible by 8 or not depends on the last three figures: thus, 127946
is not divisible by 8, since 946 is not so. A number is divisible by 3
or 9 only when the sum of its digits is divisible by 3 or 9. Take for
example 1234; this is
[11] Among the even figures we include 0.
1 thousand, or 999 and 1
2 hundred, or twice 99 and 2
3 tens, or three times 9 and 3
and 4 or 4
Now 9, 99, 999, &c. are all obviously divisible by 9 and by 3, and so
will be any number made by the repetition of all or any of them any
number of times. It therefore depends on 1 + 2 + 3 + 4, or the sum of
the digits, whether 1234 shall be divisible by 9 or 3, or not. From
the above we gather, that a number is divisible by 6 when it is even,
and when the sum of its digits is divisible by 3. Lastly, a number is
divisible by 5 only when the last figure is 0 or 5.
82. Where the divisor is unity followed by ciphers, the rule becomes
extremely simple, as you will see by the following examples:
100)33429(334
300
----
342
300
----
429
400
---
29
This is, then, the rule: Cut off as many figures from the right hand of
the dividend as there are ciphers. These figures will be the remainder,
and the rest of the dividend will be the quotient.
10)2717316
--------
271731 and rem. 6.
Or we may prove these results thus: from (20), 2717316 is 271731 tens
and 6; of which the first contains 10 271731 times, and the second not
at all; the quotient is therefore 271731, and the remainder 6 (72).
Again (20), 33429 is 334 hundreds and 29; of which the first contains
100 334 times, and the second not at all; the quotient is therefore
334, and the remainder 29.
83. The following examples will shew how the rule may be shortened when
there are ciphers in the divisor. With each example is placed another
containing the same process, all unnecessary figures being removed; and
from the comparison of the two, the rule at the end of this article is
derived.
I. 1782000)6424700000(3605 1782)6424700(3605
5346000 5346
-------- ----
10787000 10787
10692000 10692
---------- -------
9500000 9500
8910000 8910
------- -------
590000 590000
II. 12300000)42176189300(3428 123)421761(3428
36900000 369
--------- ----
52761893 527
49200000 492
--------- ----
35618930 356
24600000 246
--------- ----
110189300 1101
98400000 984
-------- ----------
11789300 11789300
The rule, then, is: Strike out as many _figures_[12] from the right of
the dividend as there are _ciphers_ at the right of the divisor. Strike
out all the ciphers from the divisor, and divide in the usual way; but
at the end of the process place on the right of the remainder all those
figures which were struck out of the dividend.
[12] Including both ciphers and others.
84. EXERCISES.
Dividend. | Divisor. |Quotient.|Remainder.
9694 | 47 | 206 | 12
175618 | 3136 | 56 | 2
23796484 | 130000 | 183 | 6484
14002564 | 1871 | 7484 | 0
310314420 | 7878 | 39390 | 0
3939040647 | 6889 | 571787 | 4
22876792454961 | 43046721 | 531441 | 0
Shew that
100 × 100 × 100 - 43 × 43 × 43
I. ------------------------------ = 100 × 100 + 100 × 43 + 43 × 43.
100 - 43
100 × 100 × 100 + 43 × 43 × 43
II. ------------------------------ = 100 × 100 - 100 × 43 + 43 × 43.
100 + 43
76 × 76 + 2 × 76 × 52 + 52 × 52
III. -------------------------------- = 76 + 52.
76 + 52
12 × 12 × 12 × 12 - 1
IV. 1 + 12 + 12 × 12 + 12 × 12 × 12 = ----------------------.
12 - 1
What is the nearest number to 1376429 which can be divided by 36300
without remainder?--_Answer_, 1379400.
If 36 oxen can eat 216 acres of grass in one year, and if a sheep eat
half as much as an ox, how long will it take 49 oxen and 136 sheep
together to eat 17550 acres?--_Answer_, 25 years.
85. Take any two numbers, one of which divides the other without
remainder; for example, 32 and 4. Multiply both these numbers by any
other number; for example, 6. The products will be 192 and 24. Now,
192 contains 24 just as often as 32 contains 4. Suppose 6 baskets,
each containing 32 pebbles, the whole number of which will be 192.
Take 4 from one basket, time after time, until that basket is empty.
It is plain that if, instead of taking 4 from that basket, I take 4
from each, the whole 6 will be emptied together: that is, 6 times 32
contains 6 times 4 just as often as 32 contains 4. The same reasoning
applies to other numbers, and therefore _we do not alter the quotient
if we multiply the dividend and divisor by the same number_.
86. Again, suppose that 200 is to be divided by 50. Divide both the
dividend and divisor by the same number; for example, 5. Then, 200 is 5
times 40, and 50 is 5 times 10. But by (85), 40 divided by 10 gives the
same quotient as 5 times 40 divided by 5 times 10, and therefore _the
quotient of two numbers is not altered by dividing both the dividend
and divisor by the same number_.
87. From (55), if a number be multiplied successively by two others, it
is multiplied by their product. Thus, 27, first multiplied by 5, and
the product multiplied by 3, is the same as 27 multiplied by 5 times
3, or 15. Also, if a number be divided by any number, and the quotient
be divided by another, it is the same as if the first number had been
divided by the product of the other two. For example, divide 60 by 4,
which gives 15, and the quotient by 3, which gives 5. It is plain, that
if each of the four fifteens of which 60 is composed be divided into
three equal parts, there are twelve equal parts in all; or, a division
by 4, and then by 3, is equivalent to a division by 4 × 3, or 12.
88. The following rules will be better understood by stating them in
an example. If 32 be multiplied by 24 and divided by 6, the result is
the same as if 32 had been multiplied by the quotient of 24 divided
by 6, that is, by 4; for the sixth part of 24 being 4, the sixth part
of any number repeated 24 times is that number repeated 4 times; or,
multiplying by 24 and dividing by 6 is equivalent to multiplying by 4.
89. Again, if 48 be multiplied by 4, and that product be divided by
24, it is the same thing as if 48 were divided at once by the quotient
of 24 divided by 4, that is, by 6. For, every unit which is repeated 6
times in 48 is repeated 4 times as often, or 24 times, in 4 times 48,
or the quotient of 48 and 6 is the same as the quotient of 48 × 4 and 6
× 4.
90. The results of the last five articles may be algebraically
expressed thus:
_ma_ _a_
---- = ---- (85)
_mb_ _b_
If _n_ divide _a_ and _b_ without remainder,
_a_
----
_n_ _a_
------ = ---- (86)
_b_ _b_
----
_n_
_a_
----
_b_ _a_
------ = ---- (87)
_c_ _bc_
_ab_ _b_
------ = _a_ × ---- (88)
_c_ _c_
_ac_ _a_
----- = ------ (89)
_b_ _b_
----
_c_
It must be recollected, however, that these have only been proved in
the case where all the divisions are without remainder.
91. When one number divides another without leaving any remainder,
or is contained an exact number of times in it, it is said to be a
_measure_ of that number, or to _measure_ it. Thus, 4 is a measure of
136, or measures 136; but it does not measure 137. The reason for
using the word measure is this: Suppose you have a rod 4 feet long,
with nothing marked upon it, with which you want to measure some
length; for example, the length of a street. If that street should
happen to be 136 feet in length, you will be able to _measure_ it with
the rod, because, since 136 contains 4 34 times, you will find that the
street is exactly 34 times the length of the rod. But if the street
should happen to be 137 feet long, you cannot measure it with the rod;
for when you have measured 34 of the rods, you will find a remainder,
whose length you cannot tell without some shorter measure. Hence 4 is
said to measure 136, but not to measure 137. A measure, then, is a
divisor which leaves no remainder.
92. When one number is a measure of two others, it is called a _common
measure_ of the two. Thus, 15 is a common measure of 180 and 75. Two
numbers may have several common measures. For example, 360 and 168
have the common measures 2, 3, 4, 6, 24, and several others. Now, this
question maybe asked: Of all the common measures of 360 and 168, which
is the greatest? The answer to this question is derived from a rule of
arithmetic, called the rule for finding the GREATEST COMMON MEASURE,
which we proceed to consider.
93. If one quantity measure two others, it measures their sum and
difference. Thus, 7 measures 21 and 56. It therefore measures 56 + 21
and 56-21, or 77 and 35. This is only another way of saying what was
said in (74).
94. If one number measure a second, it measures every number which the
second measures. Thus, 5 measures 15, and 15 measures 30, 45, 60, 75,
&c.; all which numbers are measured by 5. It is plain that if
15 contains 5 3 times,
30, or 15 + 15 contains 5 3 + 3 times, or 6 times,
45, or 15 + 15 + 15 contains 5 3 + 3 + 3 or 9 times;
and so on.
95. Every number which measures both the dividend and divisor measures
the remainder also. To shew this, divide 360 by 112. The quotient is
3, and the remainder 24, that is (72) 360 is three times 112 and 24,
or 360 = 112 × 3 + 24. From this it follows, that 24 is the difference
between 360 and 3 times 112, or 24 = 360-112 × 3. Take any number which
measures both 360 and 112; for example, 4. Then
4 measures 360,
4 measures 112, and therefore (94) measures 112 × 3,
or 112 + 112 + 112.
Therefore (93) it measures 360-112 × 3, which is the remainder 24. The
same reasoning may be applied to all other measures of 360 and 112; and
the result is, that every quantity which measures both the dividend and
divisor also measures the remainder. Hence, every _common measure_ of
a dividend and divisor is also a _common measure_ of the divisor and
remainder.
96. Every common measure of the divisor and remainder is also a
common measure of the dividend and divisor. Take the same example,
and recollect that 360 = 112 × 3 + 24. Take any common measure of the
remainder 24 and the divisor 112; for example, 8. Then
8 measures 24;
and 8 measures 112, and therefore (94) measures 112 × 3.
Therefore (93) 8 measures 112 × 3 + 24, or measures the dividend 360.
Then every common measure of the remainder and divisor is also a common
measure of the divisor and dividend, or there is no common measure of
the remainder and divisor which is not also a common measure of the
divisor and dividend.
97. I. It is proved in (95) that the remainder and divisor have all the
common measures which are in the dividend and divisor.
II. It is proved in (96) that they have no others.
It therefore follows, that the greatest of the common measures of the
first two is the greatest of those of the second two, which shews how
to find the greatest common measure of any two numbers,[13] as follows:
98. Take the preceding example, and let it be required to find the g.
c. m. of 360 and 112, and observe that
360 divided by 112 gives the remainder 24,
112 divided by 24 gives the remainder 16,
24 divided by 16 gives the remainder 8,
16 divided by 8 gives no remainder.
[13] For shortness, I abbreviate the words _greatest common measure_
into their initial letters, g. c. m.
Now, since 8 divides 16 without remainder, and since it also divides
itself without remainder, 8 is the g. c. m. of 8 and 16, because it is
impossible to divide 8 by any number greater than 8; so that, even if
16 had a greater measure than 8, it could not be _common_ to 16 and 8.
Therefore 8 is g. c. m. of 16 and 8,
(97) g. c. m. of 16 and 8 is g. c. m. of 24 and 16,
g. c. m. of 24 and 16 is g. c. m. of 112 and 24,
g. c. m. of 112 and 24 is g. c. m. of 360 and 112,
Therefore 8 is g. c. m. of 360 and 112.
The process carried on may be written down in either of the following
ways:
112)360(3
336
---
24)112(4 112 | 360 3
96 96 | 336 4
--- ----+-------
16)24(1 16 | 24 1
16 16 | 16 2
-- ----+-------
8)16(2 0 | 8
16
--
0
The rule for finding the greatest common measure of two numbers is,
I. Divide the greater of the two by the less.
II. Make the remainder a divisor, and the divisor a dividend, and find
another remainder.
III. Proceed in this way until there is no remainder, and the last
divisor is the greatest common measure required.
99. You may perhaps ask how the rule is to shew when the two numbers
have no common measure. The fact is, that there are, strictly speaking,
no such numbers, because all numbers are measured by 1; that is,
contain an exact number of units, and therefore 1 is a common measure
of every two numbers. If they have no other common measure, the last
divisor will be 1, as in the following example, where the greatest
common measure of 87 and 25 is found.
25)87(3
75
--
12)25(2
24
--
1)12(12
12
--
0
EXERCISES.
Numbers. g. c. m.
6197 9521 1
58363 2602 1
5547 147008443 1849
6281 326041 571
28915 31495 5
1509 300309 3
What are 36 × 36 + 2 × 36 × 72 + 72 × 72
and 36 × 36 × 36 + 72 × 72 × 72;
and what is their greatest common measure?--_Answer_, 11664.
100. If two numbers be divisible by a third, and if the quotients be
again divisible by a fourth, that third is not the greatest common
measure. For example, 360 and 504 are both divisible by 4. The
quotients are 90 and 126. Now 90 and 126 are both divisible by 9,
the quotients of which division are 10 and 14. By (87), dividing a
number by 4, and then dividing the quotient by 9, is the same thing
as dividing the number itself by 4 × 9, or by 36. Then, since 36 is
a common measure of 360 and 504, and is greater than 4, 4 is not the
greatest common measure. Again, since 10 and 14 are both divisible by
2, 36 is not the greatest common measure. It therefore follows, that
when two numbers are divided by their greatest common measure, the
quotients have no common measure except 1 (99). Otherwise, the number
which was called the greatest common measure in the last sentence is
not so in reality.
101. To find the greatest common measure of three numbers, find the g.
c. m. of the first and second, and of this and the third. For since
all common divisors of the first and second are contained in their g.
c. m., and no others, whatever is common to the first, second, and
third, is common also to the third and the g. c. m. of the first and
second, and no others. Similarly, to find the g. c. m. of four numbers,
find the g. c. m. of the first, second, and third, and of that and the
fourth.
102. When a first number contains a second, or is divisible by it
without remainder, the first is called a multiple of the second. The
words _multiple_ and _measure_ are thus connected: Since 4 is a
measure of 24, 24 is a multiple of 4. The number 96 is a multiple of
8, 12, 24, 48, and several others. It is therefore called a _common
multiple_ of 8, 12, 24. 48, &c. The product of any two numbers is
evidently a common multiple of both. Thus, 36 × 8, or 288, is a common
multiple of 36 and 8. But there are common multiples of 36 and 8 less
than 288; and because it is convenient, when a common multiple of two
quantities is wanted, to use the least of them, I now shew how to find
the least common multiple of two numbers.
103. Take, for example, 36 and 8. Find their greatest common measure,
which is 4, and observe that 36 is 9 × 4, and 8 is 2 × 4. The quotients
of 36 and 8, when divided by their greatest common measure, are
therefore 9 and 2. Multiply these quotients together, and multiply the
product by the greatest common measure, 4, which gives 9 × 2 × 4, or
72. This is a multiple of 8, or of 4 × 2 by (55); and also of 36 or of
4 × 9. It is also the least common multiple; but this cannot be proved
to you, because the demonstration cannot be thoroughly understood
without more practice in the use of letters to stand for numbers. But
you may satisfy yourself that it is the least in this case, and that
the same process will give the least common multiple in any other case
which you may take. It is not even necessary that you should know it is
the least. Whenever a common multiple is to be used, any one will do as
well as the least. It is only to avoid large numbers that the least is
used in preference to any other.
When the greatest common measure is 1, the least common multiple of the
two numbers is their product.
The rule then is: To find the least common multiple of two numbers,
find their greatest common measure, and multiply one of the numbers by
the quotient which the other gives when divided by the greatest common
measure. To find the least common multiple of three numbers, find the
least common multiple of the first two, and find the least common
multiple of that multiple and the third, and so on.
EXERCISES.
Numbers proposed. | Least common multiple.
14, 21 | 42
16, 5, 24 | 240
1, 2, 3, 4, 5, 6, 7, 8, 9, 10 | 2520
6, 8, 11, 16, 20 | 2640
876, 864 | 63072
868, 854 | 52948
A convenient mode of finding the least common multiple of several
numbers is as follows, when the common measures are easily visible:
Pick out a number of common measures of two or more, which have
themselves no divisors greater than unity. Write them as divisors,
and divide every number which will divide by one or more of them.
Bring down the quotients, and also the numbers which will not divide
by any of them. Repeat the process with the results, and so on until
the numbers brought down have no two of them any common measure except
unity. Then, for the least common multiple, multiply all the divisors
by all the numbers last brought down. For instance, let it be required
to find the least common multiple of all the numbers from 11 to 21.
2, 2, 3, 5, 7)11 12 13 14 15 16 17 18 19 20 21
---------------------------------
11 1 13 1 1 4 17 3 19 1 1
There are now no common measures left in the row, and the least common
multiple required is the product of 2, 2, 3, 5, 7, 11, 13, 4, 17, 3,
and 19; or 232792560.