Elements of arithmetic

47. I have said that all questions in arithmetic require nothing but addition and subtraction. I do not mean by this that no rule should ever be used except those given in the last section, but that all other rules only shew shorter ways of finding what might be found, if we pleased, by the methods there deduced. Even the last two rules themselves are only short and convenient ways of doing what may be done with a number of pebbles or counters.

48. I want to know the sum of five seventeens, or I ask the following question: There are five heaps of pebbles, and seventeen pebbles in each heap; how many are there in all? Write five seventeens in a column, and make the addition, which gives 85. In this case 85 is called the product of 5 and 17, and the process of finding the product is called multiplication, which gives nothing more than the addition of a number of the same quantities. Here 17 is called the multiplicand, and 5 is called the multiplier.

• 17
• 17
• 17
• 17
• 17
• 85

49. If no question harder than this were ever proposed, there would be no occasion for a shorter way than the one here followed. But if there were 1367 heaps of pebbles, and 429 in each heap, the whole number is then 1367 times 429, or 429 multiplied by 1367. I should have to write 429 1367 times, and then to make an addition of enormous length. To avoid this, a shorter rule is necessary, which I now proceed to explain.

50. The student must first make himself acquainted with the products of all numbers as far as 10 times 10 by means of the following table,[8] which must be committed to memory.

If from this table you wish to know what is 7 times 6, look in the first upright column on the left for either of them; 6 for example. Proceed to the right until you come into the column marked 7 at the top. You there find 42, which is the product of 6 and 7.

51. You may find, in this way, either 6 times 7, or 7 times 6, and for both you find 42. That is, six sevens is the same number as seven sixes. This may be shewn as follows: Place seven counters in a line, and repeat that line in all six times. The number of counters in the whole is 6 times 7, or six sevens, if I reckon the rows from the top to the bottom; but if I count the rows that stand side by side, I find seven of them, and six in each row, the whole number of which is 7 times 6, or seven sixes. And the whole number is 42, whichever way I count. The same method may be applied to any other two numbers. If the signs of (23) were used, it would be said that 7 × 6 = 6 × 7.

52. To take any quantity a number of times, it will be enough to take every one of its parts the same number of times. Thus, a sack of corn will be increased fifty-fold, if each bushel which it contains be replaced by 50 bushels. A country will be doubled by doubling every acre of land, or every county, which it contains. Simple as this may appear, it is necessary to state it, because it is one of the principles on which the rule of multiplication depends.

53. In order to multiply by any number, you may multiply separately by any parts into which you choose to divide that number, and add the results. For example, 4 and 2 make 6. To multiply 7 by 6 first multiply 7 by 4, and then by 2, and add the products. This will give 42, which is the product of 7 and 6. Again, since 57 is made up of 32 and 25, 57 times 50 is made up of 32 times 50 and 25 times 50, and so on. If the signs were used, these would be written thus:

7 × 6 = 7 × 4 + 7 × 2.
  50 × 57 = 50 × 32 + 50 × 25.

54. The principles in the last two articles may be expressed thus: If a be made up of the parts x, y, and x, ma is made up of mx, my, and mz; or,

ifa = x + y + z.

ma = mx + my + mz,

or,m(x + y + z) = mx + my + mz.

A similar result may be obtained if a, instead of being made up of x, y, and z, is made by combined additions and subtractions, such as x + y-z, x- y + z, x-y-z, &c. To take the first as an instance:

Leta = x + y - z,

thenma = mx + my - mz.

For, if a had been x + y, ma would have been mx + my. But since a is less than x + y by z, too much by z has been repeated every time that x + y has been repeated;—that is, mz too much has been taken; consequently, ma is not mx + my, but mx + my-mz. Similar reasoning may be applied to other cases, and the following results may be obtained:

m(a + b + c - d) = ma + mb + mc - md.

55. There is another way in which two numbers may be multiplied together. Since 8 is 4 times 2, 7 times 8 may be made by multiplying 7 and 4, and then multiplying that product by 2. To shew this, place 7 counters in a line, and repeat that line in all 8 times, as in figures I. and II.

The number of counters in all is 8 times 7, or 56. But (as in fig. I.) enclose each four rows in oblong figures, such as a and b. The number in each oblong is 4 times 7, or 28, and there are two of those oblongs; so that in the whole the number of counters is twice 28, or 28 x 2, or 7 first multiplied by 4, and that product multiplied by 2. In figure II. it is shewn that 7 multiplied by 8 is also 7 first multiplied by 2, and that product multiplied by 4. The same method may be applied to other numbers. Thus, since 80 is 8 times 10, 256 times 80 is 256 multiplied by 8, and that product multiplied by 10. If we use the signs, the foregoing assertions are made thus:

7 × 8 =   7 × 4 × 2  = 7 × 2 × 4.
256 × 80 = 256 × 8 × 10 = 256 × 10 × 8.

EXERCISES.

Shew that 2 × 3 × 4 × 5 = 2 × 4 × 3 × 5 = 5 × 4 × 2 × 3, &c.

Shew that 18 × 100 = 18 × 57 + 18 × 43.

56. Articles (51) and (55) may be expressed in the following way, where by ab we mean a taken b times; by abc, a taken b times, and the result taken c times.

ab = ba.  
abc = acb = bca = bac, &c.
abc = a × (bc) = b × (ca) = c × (ab).

If we would say that the same results are produced by multiplying by b, c, and d, one after the other, and by the product bcd at once, we write the following:

a × b × c × d = a × bcd.

The fact is, that if any numbers are to be multiplied together, the product of any two or more may be formed, and substituted instead of those two or more; thus, the product abcdef may be formed by multiplying

57. In order to multiply by 10, annex a cipher to the right hand of the multiplicand. Thus, 10 times 2356 is 23560. To shew this, write 2356 at length which is

2 thousands, 3 hundreds, 5 tens, and 6 units.

Take each of these parts ten times, which, by (52), is the same as multiplying the whole number by 10, and it will then become

2 tens of thou. 3 tens of hun. 5 tens of tens, and 6 tens,

which is

2 ten-thou. 3 thous. 5 hun. and 6 tens.

This must be written 23560, because 6 is not to be 6 units, but 6 tens. Therefore 2356 × 10 = 23560.

In the same way you may shew, that in order to multiply by 100 you must affix two ciphers to the right; to multiply by 1000 you must affix three ciphers, and so on. The rule will be best caught from the following table:

58. I now shew how to multiply by one of the numbers, 2, 3, 4, 5, 6, 7, 8, or 9. I do not include 1, because multiplying by 1, or taking the number once, is what is meant by simply writing down the number. I want to multiply 1368 by 8. Write the first number at full length, which is

1 thousand, 3 hundreds, 6 tens, and 8 units.

To multiply this by 8, multiply each of these parts by 8 (50) and (52), which will give

8 thousands, 24 hundreds, 48 tens, and 64 units.

Add these together, which gives 10944 as the product of 1368 and 8, or 1368 × 8 = 10944. By working a few examples in this way you will see for following rule.

59. I. Multiply the first figure of the multiplicand by the multiplier, write down the units’ figure, and reserve the tens.

II. Do the same with the second figure of the multiplicand, and add to the product the number of tens from the first; put down the units’ figure of this, and reserve the tens.

III. Proceed in this way till you come to the last figure, and then write down the whole number obtained from that figure.

IV. If there be a cipher in the multiplicand, treat it as if it were a number, observing that 0 × 1 = 0, 0 × 2 = 0, &c.

60. In a similar way a number can be multiplied by a figure which is accompanied by ciphers, as, for example, 8000. For 8000 is 8 × 1000, and therefore (55) you must first multiply by 8 and then by 1000, which last operation (57) is done by placing 3 ciphers on the right. Hence the rule in this case is, multiply by the simple number, and place the number of ciphers which follow it at the right of the product.

EXAMPLE.

61. EXERCISES.

What is 1007360 × 7?   Answer, 7051520.

123456789 × 9 + 10 and 123 × 9 + 4?—Ans. 1111111111 and 1111.

What is 136 × 3 + 129 × 4 + 147 × 8 + 27 × 3000?—Ans. 83100.

An army is made up of 33 regiments of infantry, each containing 800 men; 14 of cavalry, each containing 600 men; and 2 of artillery, each containing 300 men. The enemy has 6 more regiments of infantry, each containing 100 more men; 3 more regiments of cavalry, each containing 100 men less; and 4 corps of artillery of the same magnitude as those of the first: two regiments of cavalry and one of infantry desert from the former to the latter. How many men has the second army more than the first?—Answer, 13400.

62. Suppose it is required to multiply 23707 by 4567. Since 4567 is made up of 4000, 500, 60, and 7, by (53) we must multiply 23707 by each of these, and add the products.

which is the product required.

It will do as well if, instead of writing the ciphers at the end of each line, we keep the other figures in their places without them. If we take away the ciphers, the second line is one place to the left of the first, the third one place to the left of the second, and so on. Write the multiplier and the multiplicand over these lines, and the process will stand thus:

• 23707
•  4567
• 165949
• 142242
• 118535
• 94828
• 108269869

63. There is one more case to be noticed; that is, where there is a cipher in the middle of the multiplier. The following example will shew that in this case nothing more is necessary than to keep the first figure of each line in the column under the figure of the multiplier from which that line arises. Suppose it required to multiply 365 by 101001. The multiplier is made up of 100000, 1000 and 1. Proceed as before, and

and the whole process with the ciphers struck off is:

•  365
• 101001
•  365
• 365
•  365
• 36865365

64. The following is the rule in all cases:

I. Place the multiplier under the multiplicand, so that the units of one may be under those of the other.

II. Multiply the whole multiplicand by each figure of the multiplier (59), and place the unit of each line in the column under the figure of the multiplier from which it came.

III. Add together the lines obtained by II. column by column.

65. When the multiplier or multiplicand, or both, have ciphers on the right hand, multiply the two together without the ciphers, and then place on the right of the product all the ciphers that are on the right both of the multiplier and multiplicand. For example, what is 3200 × 3000? First, 3200 is 32 × 100, or one hundred times as great as 32. Again, 32 × 13000 is 32 × 13, with three ciphers affixed, that is 416, with three ciphers affixed, or 416000. But the product required must be 100 times as great as this, or must have two ciphers affixed. It is therefore 41600000, having as many ciphers as are in both multiplier and multiplicand.

66. When any number is multiplied by itself any number of times, the result is called a power of that number. Thus:

The second and third powers are usually called the square and cube, which are incorrect names, derived from certain connexions of the second and third power with the square and cube in geometry. As exercises in multiplication, the following powers are to be found.

67. It is required to multiply a + b by c + d, that is, to take a + b as many times as there are units in c + d. By (53) a + b must be taken c times, and d times, or the product required is (a + b)c + (a + b)d. But (52) (a + b)c is ac + bc, and (a + b)d is ad + bd; whence the product required is ac + bc + ad + bd; or,

(a + b)(c + d) = ac + bc + ad + bd.

By similar reasoning

(a - b)(c + d) is (a - b)c + (a - b)d; or,
(a - b)(c + d) = ac - bc + ad - bd.

To multiply a-b by c-d, first take a-b c times, which gives ac-bc. This is not correct; for in taking it c times instead of c-d times, we have taken it d times too many; or have made a result which is (a-b)d too great. The real result is therefore ac-bc-(a -b)d. But (a-b)d is ad- bd, and therefore

(a - b)(c - d) = ac - bc - ad - bd
 = ac - bc - ad + bd  (41)

From these three examples may be collected the following rule for the multiplication of algebraic quantities: Multiply each term of the multiplicand by each term of the multiplier; when the two terms have both + or both-before them, put + before their product; when one has + and the other-, put-before their product. In using the first terms, which have no sign, apply the rule as if they had the sign +.

68. For example, (a + b)(a + b) gives aa + ab + ab + bb. But ab + ab is 2ab; hence the square of a + b is aa + 2ab + bb. Again (a- b)(a-b) gives aa-ab-ab + bb. But two subtractions of ab are equivalent to subtracting 2ab; hence the square of a- b is aa-2ab + bb. Again, (a + b)(a-b) gives aa + ab-ab -bb. But the addition and subtraction of ab makes no change; hence the product of a + b and a- b is aa-bb.

Again, the square of a + b + c + d or (a + b + c + d)(a + b + c + d) will be found to be aa + 2ab + 2ac + 2ad + bb + 2bc + 2bd + cc + 2cd + dd; or the rule for squaring such a quantity is: Square the first term, and multiply all that come after by twice that term; do the same with the second, and so on to the end.

SECTION IV.
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.

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

therefore 91 + 39 + 26 contains 13 7 + 3 + 2 times, or 12 times.

Again, 156 is made up of 100, 50, and 6.

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

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
•    36
•    36
•     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.

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.

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).

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

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:

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

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.

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.

84. EXERCISES.

Shew that

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:

If n divide a and b without remainder,

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