Again, multiply £12 8s. 34d. by 39. Since the multiplier in this case cannot be divided into two entire factors, we take the number nearest to 39, which can be divided into factors, viz. 36=6×6, and multiply by each of these factors; then multiply the given number by the difference between the assumed number and the given multiplier, and add this product to that obtained by the multiplication by the two factors 6 and 6. The reason of this is plain, for as in multiplication the multiplicand has to be repeated as many times as there are units in the multiplier, the multiplication by the factors used only repeats the multiplicand 36 times;

and then multiplying it by 3, it was repeated 3 times more, that is, 36+3=39 times.

From these illustrations we deduce the following

RULE.

I. Set the multiplier under the lowest unit of the multiplicand, and draw a line beneath.

II. Multiply the number on the right by the multiplierand divide this product by as many units as make one of the next higher units. Set the remainder, if any, under the number multiplied, and carry the quotient into the next product. Multiply the number of the next higher units by the multiplier, and to this product add the number carried. Divide this sum by as many units as make one of the next higher units, and proceed in this way until all the units are multiplied, setting down the entire product when the highest unit is multiplied.

III. When the multiplier exceeds 12, and can be divided into factors, multiply the given number by one of these fac tors, and this product by the second factor, and so on until all the factors are used: the last product will be the required product.

IV. When the multiplier exceeds 12, and cannot be divided into entire factors, take the nearest less number to the given multiplier which can be divided into factors, and multiply by them as before; then multiply the given multiplicand by the difference between the given multiplier and the assumed number, and add this product to the last product: the sum will be the required product.

The proof is the same as in multiplication of simple numbers.

Q. What is multiplication of compound numbers? How are the numbers set down? Where do you commence to multiply? Do you set down the whole product? What do you set down? What do you carry? When the multiplier exceeds 12, and can be divided into factors, how may you multiply? When the multiplier exceeds 12, and cannot be divided into factors? What is the reason of this? What is the rule for multiplication of compound numbers? What is the proof?

8. Multiply 14 A. 3 R. 27 P. by 47. 9. Multiply 10 cords 4 cu. ft. 12 cu. in. by 49. 10. Multiply 41 gals. 2 qt. 1 pt. by 63.

11. Multiply 10 mo. 2 wk. 6 dy. 14 h. by 108. 12. Multiply 70° 4′ 22′′ by 180.

DIVISION OF COMPOUND NUMBERS.

OPERATION.

160. LET it be required to divide £18 88. 4d. by 2. We place the divisor on the left of the dividend, as in division of simple numbers; draw a curve line to sepa. rate them, and then a straight line beneath: 2 in 18 pounds goes 9 pounds; set down 9 under pounds: 2 in 8 shil

£ S. d. 2)18 8 4

Ans. 9 4 2

lings goes 4 shillings; set down the 4 under shillings: 2 in 4 pence goes 2 pence; set down 2 under pence: the quotient is £9 4s. 2d.

Again, divide £17 5s. 4d. by 4.

OPERATION. £ 8. d.

4) 17 5 4

Ans. 4 6 4

In this example 4 in 17 pounds goes 4 pounds, and 1 pound over; set down 4 under pounds. But since the 1 pound equals 20 shillings, and has not been divided, we add 20 to 5 shillings, which make 25 shillings; 4 in 25 shillings goes 6 shillings, and 1 shilling over; 1 shilling is equal to 12 pence, and adding the 12 pence to the 4 pence makes 16 pence, which divided by 4 gives 4 pence. The answer is £4 68. 4d.

OPERATION.

£ 8. d.

16) 65 14 8(4 pounds. 64

Again, divide £65 14s. 8d. by 16. The divisor in this example being greater than 12, we arrange the numbers as in division of simple numbers. Dividing the 65 pounds by 16, we get 4 pounds for the quotient, and 1 pound for a remainder. Multiplying this remainder by 20 to bring it to shillings (Art. 118), and adding to them the 14 shillings which have not been divided, gives us 34 shillings. Taking this number as a new dividend, and dividing again by 16, we get 2 shillings for

mtient and 2 billió mo

1

20

16) 34 (2 shillings.

32

2

(

12

16) 32 (2 pence.

32

Ans. £4 28. 2d.

over. Multiplying this remainder by 12 to bring the shillings to pence, and adding in the 8 pence, gives us 32 pence; which, divided by 16, gives 2 pence and no remainder. The answer is £4 28. 2d.

From these illustrations we deduce the following

RULE.

I. Arrange the numbers as in division of simple numbers II. Divide the number corresponding to the highest unit by the divisor: the quotient will be the number corresponding to the highest unit in the quotient.

III. Multiply the remainder by as many of the next lower units as make one of the higher, and to this product add the number of the next lower units in the dividend; divide as before; the quotient will be the number corresponding to the next lower unit in the quotient.

IV. Proceed in this manner through all the different units, and set the last remainder, if there be any, as in division of simple numbers.

The proof is the same as for simple numbers.

Q. How are the numbers arranged for division of compound numbers? Where do you commence to divide? Suppose there is a remainder, how do you proceed? After multiplying this remainder, what do you add to it? What do you do with the last remainder? What is the proof for simple numbers?

161. When the divisor can be separated into factors, we may divide by each of these factors separately, as in division of simple numbers (Art. 56).

Thus, divide 14 cwt. 2 qrs. 20 lbs. by 24.

Q. How may you perform the division when the divisor can be separated into its factors?

9. Divide 12 mos. 2 wk. 2 dy. by 16. Ans. 0 m. 3 w. 1 d. 10. Divide 180° by 144.

11. Divide 116 m. 2 fur. by 28.

12. Divide 34 bbls. 21 galls. by 42.

REDUCTION OF VULGAR FRACTIONS

WHICH ARE EXPRESSED IN TERMS OF DIFFERENT UNITS.

162. THE Vulgar Fractions which have been heretofore considered, have expressed parts of the same kind of unit. But in arithmetical operations, fractions are frequently presented for addition, &c., in which parts of different units are considered.

of a

Thus, it may be required to add of a pound to shilling; and it is evident that before this addition can be effected, the of a pound must be reduced to the same unit as of a shilling, or the reverse, the fraction of the shilling must be brought to the same unit as that of the pound.

Thus, it appears that to prepare fractions of different units for the operations of addition, &c., they must first be reduced to the same unit, and this reduction may be effected in two ways:

1st. By reducing a fraction of a higher unit to a fraction of a lower.