When two or more numbers can be divided by the same number without a remainder, that number is called their common divisor, and the greatest number which will divide them so, is called their greatest common divisor. When two or more numbers have several common divisors, it is evident that the greatest common divisor will be the product of them all.

In order to reduce a fraction to the lowest terms possible, it is necessary to divide the numerator and denominator by all their common divisors, or by their greatest common divisor at first.

126

Reduce 1 to its lowest terms.

I observe in the first place that both numerator and denominator are divisible by 9, because the sum of the figures in each is 9. I observe also, that both are divisible by 2, because the right hand figure of each is so; therefore they are both divisible by 18. But it is most convenient to divide by them separately.

126 (9 (2=77.
= 14
342

7 and 19 have no common divisor, therefore cannot be reduced to lower terms.

The greatest common divisor cannot always be found by the above method. It will therefore be useful to find a rule by which it may always be discovered.

Let us take the same numbers 126 and 342.

126 is a number of even 18s, and 342 is a number of even 188; therefore if 126 be subtracted from 342, the remainder 216 must be a number of even 18s. And if 126 be subtracted from 216, the remainder 90 must be a number of even 18s. Now I cannot subtract 126 from 90, but since 90 is a number of even 18s, if I subtract it from 126, the remainder 36 must be a number of even 18s. Now if 36 be subtracted from 90, the remainder 54 must be a number of even 18s. Subtracting 36 from 54, the remainder is 18. Thus by subtracting one number from the other, a smaller number was obtained every time, which was always a number of even 18s, until at last I came to 18 itself. If 18 be subtracted twice from 36 there will be no remainder. It is easy to see, that whatever be the common divisor, since each number is a certain number of times the common divisor, if one be subtracted from the other, the remainder will be a certain number of times the common divisor, that is, it will have the same divisor as the numbers themselves. And every

time the subtraction is made, a new number, smaller than the last, is obtained, which has the same divisor; and at length the remainder must be the common divisor itself; and if this be subtracted from the last smaller number as many times as it can be, there will be no remainder. By this it may be known when the common divisor is found. It is the number which being subtracted leaves no remainder. When one number is considerably larger than the other, division may be substituted for subtraction. The remainders only are to be noticed, no regard is to be paid to the quotient.

Reduce the fraction 239 to its lowest terms.

Subtracting 330 from 462, there remains 132. 132 may be subtracted twice, or which is the same thing, is contained twice in 330, and there is 66 remainder. 66 may be subtracted twice from 132, or it is contained twice in 132, and leaves no remainder; 66 therefore is the greatest common divisor. Dividing both numerator and denominator by 66, the fraction is reduced to 4.

From the above examples is derived the following general rule, to find the greatest common divisor of two numbers: Divide the greater by the less, and if there is no remainder, that number is itself the divisor required; but if there is a remainder, divide the divisor by the remainder, and then divide the last divisor by that remainder, and so on, until there is no remainder, and the last divisor is the divisor required.

If there be more than two numbers of which the greatest common divisor is to be found; find the greatest common divisor of two of them, and then take that common divisor and one of the other numbers, and find their greatest common divisor, and so on.

Reduce the fraction to its lowest terms.

17 (9

9

1

9 (8

8

1

1 is the greatest common divisor in this example.

Therefore the fraction

cannot be reduced.

1 (1

1

XXII. The method for finding the common denominator, given in Art. XIX. though always certain, is not always the best; for it frequently happens that they may be reduced to a common denominator, much smaller than the one obtained by that rule.

Reduce and to a common denominator.

According to the rule in Art. XIX., the common denominator will be 54, and § and 2.

It was observed Art. XIX., that the common denominator may be any number, of which all the denominators are factors. 6 and 9 are both factors of 18, therefore they may be both reduced to 18ths.ğ 13, and

4.

When the fractions consist of small numbers, the least denominator to which the fractions can be reduced, may be easily discovered by trial; but when they are large it is more difficult. It will, therefore, be useful to find a rule for it.

Any number, which is composed of two or more factors, is called a multiple of any one of those factors. Thus 18 is a multiple of 2, or of 3, or of 6, or of 9. It is also a common multiple of these numbers, that is, it may be produced by multiplying either of them by some number. The least common multiple of two or more numbers, is the least number of which they are all factors. 54 is a common multiple of 6 and 9, but their least common multiple is 18.

The least common denominator of two or more fractions will be the least common multiple of all the denominators; the fractions being previously reduced to their lowest terms.

One number will always be a multiple of another, when the former contains all the factors of the latter. 62 × 3, and 9 3 x 3, and 18 2 × 3 × 3.18 contains the factors 2 and 3 of 6, and 3 and 3 of 9. 54 = 2 × 3 × 3 × 3.

54, which is produced by multiplying 6 and 9, contains all these factors, and one of them, viz. 3, repeated. The reason why 3 is repeated is because it is a factor of both 6 and 9. By reason of this repetition, a number is produced 3 times as large as is necessary for the common multiple.

When the least common multiple of two or more numbers is to be found, if two or more of them have a common factor, it may be left out of all but one, because it will be sufficient that it enters once into the product.

These factors will enter once into the product, and only once, if all the numbers which have common factors be divided by those factors; and then the undivided numbers, and these quotients be multiplied together, and the product multiplied by the common factors.

If any of the quotients be found to have a common factor with either of the numbers, or with each other, they may be divided by that also.

Reduce 3, 3, 5, and, to the least common denominator.

The least common denominator will be the least common multiple of 4, 9, 6, and 5.

Then

Divide 4 and 6 by 2, the quotients are 2 and 3. divide 3 and 9 by 3, the quotients are 1 and 3. Then multiplying these quotients, and the undivided number 5, we have 2 × 1 × 3 × 5 = 30. Then multiplying 30 by the two common factors 2 and 3, we have 30 × 2 × 3 = 180, which is to be the common denominator.

Now to find how many 180ths each fraction is, take 3, §, %, and of 180. Or observe the factors of which 180 was made up in the multiplication above. Thus 2 x 1 x 3 x 5 X 2 X 3 180. Then multiply the numerator of each fraction by the numbers by which the factors of its denominator were multiplied.

The factors 2 and 2 of the den.ominator of the first fraction, were multiplied by 1, 3, 3, and 5. The factors 3 and 3, of the second, were multiplied by 2, 1, 5, and 2. The factors 2 and 3, of the third, were multiplied by 2, 1, 3, 5; and 5, the denominator of the fourth, was multiplied by 2, 2, 1, 3, and 3.

XXIII. If a horse will eat of a bushel of oats in a day, how long will 12 bushels last him?

In this question it is required to find how many times

of a bushel is contained in 12 bushels. therefore 12 bushels wil! last 36 days.

In 12 there are 36,

At of a dollar a bushel, how many bushels of corn may be bought for 15 dollars?

First find how many bushels might be bought at of a dollar a bushel. It is evident, that each dollar would buy 5 bushels; therefore 15 dollars would buy 15 times 5, that is, 75 bushels. But since it is instead of of a dollar a bushel, it will buy only as much, that is, 25 bushels.

This question is to find how many times of a dollar, are contained in 15 dollars. It is evident, that 15 must be reduced to 5ths, and then divided by 3.

15

5

75 (3

25 bushels.

The above question is on the same principle as the following.

How much corn, at 5 shillings a bushel, may be bought for 23 dollars?

The dollars in this example must be reduced to shillings, before we can find how many times 5 shillings are contained in them; that is, they must be reduced to 6ths, before we can find how many times are contained in them.

23

6

138 (5

23=

138 and

Ans. 273 bushels.

are contained 273 times in 138.

If 73 yds. of cloth will make 1 suit of clothes, how many suits will 48 yards make?

If the question was given in yards and quarters, it is evident both numbers must be reduced to quarters. In this instance then, they must be reduced to 8ths.

384

7359 and 48 = 384

384 (59