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.

330

Reduce the fraction 239 to its lowest terms. ·

462

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 .

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 din visor and so on

Reduce the fraction to its lowest terms. 17 (9

1

1 is the greatest common divisor in this example. Therefore the fraction cannot be reduced.

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 5 = 45 and

45 and

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. = 15, and = 18.

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 × 3, and 18 = 2 x 3 x 3. 18 contains the factors 2 and 3 of 6 and 3 and 3 of 9 54 = 2 × 3 × 3x3

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 divid ed 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 3, to the least common denominator.

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

Divide 4 and 6 by 2, the quotients are 2 and 3. Then divide 3 and 9 by 3, the quotients are 1 and 3. Then multiply ing these quotients, and the undivided number 5, we have 2 × 1 × 3 × 530. Then multiplying 30 by the two common factors 2 and 3, we have 30 × 2 × 3 = 180, which is to be the common denominator.

5

5

41 91

Now to find how many 180ths each fraction is, take, %, and of 180. Or observe the factors of which 180 was made up in the multiplication above. Thus 2 × 1 × 3 × 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 denominator 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

144
180°

150
180 5

of a Lushel 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. In 12 there are 36, therefore 12 bushels wil! last 36 days.

At & of 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 fol lowing.

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.

23138 and are contained 27 times in 138. If 7 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 evi dent both numbers must be reduced to quarters. In this instance then, they must be reduced to 8ths.

7359 and 48 = 384

384 (59

In the three last examples, the purpose is to find how many times a fraction is contained in a whole number. This is dividing a whole number by a fraction, for which we find the following rule: Reduce the dividend to the same denomination as the divisor, and then divide by the numerator of the fraction.

Note. If the divisor is a mixed number, it must be re duced to an improper fraction.

N. B. The above rule amounts to this; multiply the dividend by the denominator of the divisor, and then divide it by the numerator.

At of a dollar a bushel, how many bushels of potatoes may be bought for 3 of a dollar.

is contained in as many times as 1 is contained in 3. Ans. 3 bushels.

If

of a ton of hay will keep a horse 1 month, how many horses will of a ton keep the same time?

3

9 10

are contained in as many times as 3 are contained in 9. Ans. 3 horses.

At of a dollar a pound, how many pounds of figs may be bought for 2 of a dollar?

5ths and 4ths are different denominations; before one can be divided by the other, they must be reduced to the same denomination; that is, reduced to a common denominator. } = and = are contained in as many times as 4 are contained in 15. Ans. 32 lb. At 7 dolls. a yard, how

20

bought for 57% dollars?
73=338 and 57 = 461.

many yards of cloth may be

5ths and 8ths are different de

nominations; they must, therefore, be reduced to a common denominator.

From the above examples we deduce the following rule, for dividing one fraction by another:

If the fractions are of the same denomination, divide the numerator of the dividend by the numerator of the divisor. If the fractions are of different denominations, they must first be reduced to a common denominator.