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2

2

CANCELLED.

2 3 2

2

20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11

10 × 9 × 8 × × 6 × 5 × 4 × 3 × 2 × 1 1 1 1 184756 Ans.

1

1

1

NOTE. In this question the product of the quotients of 2, 3, 2, and 2 is cancelled by the product of 4, 3, and 2 in the lower line. Any numbers may be cancelled, when their product is equal to the product of certain other numbers, as in the following question.

18. Divide the continued product of 4, 9, 3, 8, and 225 by the continued product 6, 6, 4, 6, and 11.

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As the product of 4 times 9 in the upper line is equal to the product of 6 times 6 in the under line, they cancel each other; and as the product of 3 times 8 in the upper line is equal to 4 times 6 in the under line, they cancel each other.

VI. To find the least common multiple of two or more numbers, that is, to find the least number, that may be divided by them without a remainder.

RULE.

Divide by such a number, as will divide most of the given numbers without a remainder, and set the several quotients with the several undivided numbers in a line beneath, and so continue to divide, until no number, greater than unity, will divide two or more of them. Then multiply all the divisors, quotients, and undivided numbers together, and the product is the least common multiple.

1. What is the least common multiple of 8, 4, 3, 6 ?

2)8 4 3 6 2)4 2 3 3 3) 2

1 3 3

2 1

1

It is evident, that 24 is a composite number, and that it is composed of the factors 2, 2, 3, and 2; and, therefore, it may be divided by any number, which is the

2 X2 X3 X 2=24 Ans.

product of any two of them; and, as the given numbers are either some one of these, or such a number as may be produced by the product of two or more of them, it is evident, therefore, that 24 may be divided by either of them without a remainder. Q. e. d.

2. What is the least common multiple of 7, 14, 21, and 15? Ans. 210. 3. What is the least common multiple of 3, 4, 5, 6, 7, and 8 ? Ans. 840. 4. What is the least number, that 10, 12, 16, 20, and 24 will divide without a remainder? Ans. 240. 5. Five men start from the same place to go round a certain island. The first can go round it in 10 days; the second in 12 days; the third in 16 days; the fourth in 18 days; the fifth in 20 days. In what time will they all meet at the place from which they started? Ans. 720 days.

VII. To reduce fractions to a common denominator; that is, to change fractions to other fractions, all having their denominators alike, yet retaining the same value.

1. Reduce 2, §, and 7 to a common denominator.

OPERATION.

First Method.

4)468 4×2×3=| 24 common denominator.

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Having first obtained a common multiple of all the denominators of the given fractions by the last rule, we assume this, as the common denominator required. This number (24) we divide by the denominators of the given fractions, 4, 6, and 8, and find their quotients to be 6, 4, and 3, which we place under the 24; these numbers we multiply by the numerators, 3, 5, and 7, and find their products to be 18, 20, and 21, and these numbers are the numerators of the fractions required.

H

OPERATION.

Second Method.

3 × 6 × 8=144 numerator for

5 × 4 × 8 = 160 numerator for

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7 × 4 × 6 = 168 numerator for = 168

192.

4 x 6 x8 = 192 common denominator.

NOTE. It will be perceived, that this method does not express the fractions in so low terms as the other.

From the above illustration we deduce the following

RULE.

Let compound fractions be reduced to simple fractions, mixed numbers to improper fractions, and whole numbers to improper fractions, by writing a unit under them; then find the least common multiple of all the denominators by the last rule, and it will be the denominator required. Divide the common multiple by each of the denominators, and multiply the quotients by the respective numerators of the fractions, and their products will be the numerators required.

Or, multiply each numerator into all the denominators except its own for a new numerator; and all the denominators into each other for a common denominator.

2. Reduce and § to a common denominator.

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5

980, 1980, 1980, 1980.

36

Ans. 25, 26, 10.

40 40 40.

18

Ans. 1, 8, 1, 482.

20 15

11. Reduce, 3, 8, 8, 7, and . Ans. 1,18,2,1,1,17.

12. Change,,,, t, and T.
13. Reduce, †, and 17.
14. Change 72, 561, 7, and 8.
15. Change, 4, 5, 7, and 9.

6 24

9

Ans. 18, 3, 1, 36, 36, 36.

Ans. 38, 18, t. Ans. 341, 244, 308, 352. Ans. 4, 16, 20, 28, 36.

VIII. To reduce fractions of a lower denomination to a

higher.

1. Reduce of a farthing to the fraction of a pound.

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farthings; therefore

of

ZIGO Ans. This question may be analyzed thus; since 4 farthings make a penny, there will be as many pence as of a farthing is

of a

penny. Again, as 12 pence make a shilling, there will be as many shillings as pence, therefore of of a penny is of a shilling. As 20 shillings make a pound, there will be as many pounds as shillings, therefore of Tog of a shilling is of a pound. Q. e. d.

The operation of this question may be abridged thus:

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Let the given fraction be reduced to a compound one by comparing it with all the denominations between the given one and the one to which it is required to reduce it; then reduce this compound fraction to a simple one.

of a grain Troy to the fraction of a pound.

2. Reduce

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3. What part of an ounce is of a scruple?

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4. What part of a ton is of an ounce ?

4 X 1 X 1 X 1 X 1

1

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Ans.

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=

1

9 × 272 × 40 × 4 × 3 294030

7. What part of 3hhds. is of a quart?

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Ans.

8. What part of 3 yards square, are 3 square yards?

Ans.

9. What part of of a solid foot is of a foot solid? Ans. .

IX. To reduce fractions of a higher denomination to a lower.

1. Reduce of a pound to the fraction of a farthing. Ans..

We explain this question in the following manner.

OPERATION.

d.

1700 X 20=1980 = 7bs.
70 × 12=43=
3X=3tqr. Ans.

As shillings are twentieths of a pound, there will be 20 times as many parts of a shilling in b of a pound, as there are parts of a pound; therefore of a pound is equal to 1 of 20 = 1200 = of a shilling. And as pence are twelfths of shillings, there will be twelve times as many parts of a penny in of a shilling, as there are parts of a shilling; therefore of a shilling is equal to 7% of 12 = +8= of a penny. Again, as farthings are fourths of a penny, there will be 4 times as many parts of a farthing in of a penny, as there are parts of a penny; therefore of a penny are equal to

of a farthing. Q. e. d.

of

=

The operation of this question may be facilitated by the

following manner.

OPERATION.

960 1400

1300 X 20 × 1 × 1 = {%=qr. Ans.

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