the compound number by the simple number, and set their products separately, as follows.

Here we have 96lb. 64oz. 48pwt. 128grs; now if we begin with 128grs. and divide by 24, the number of grains it takes to make a penny-weight, we get the number of penny weights there are in 128 grains, which is 5, and 8 remainder; if we set down the remainder and add these 5pwt. to 48pwt we shall have 96 lb 64 oz. 53 pwt. and 8 grains. If we divide 53 pwt by 20, the number of penny-weights it takes to make an ounce, we shall get the number of ounces there are in 53 pwt. which is 2 and 13 remainder; if we add this quotient 2, to 64, we shall have the number of ounces there are in 64 oz 48 pwt. 128 grs. viz. 66 oz. If we now divide 66 oz. by 12, the number of ounces it takes to make a pound, the quotient will be the number of pounds contained in 66 oz. viz. 5 lb. and 6 oz over; add 5 lb to 96 lb and there will be 101 lb. 6 oz. 13 pwt. 8 grs. the same as by the first pro

cess.

Pup. The reason, then, of multiplying and dividing as we do, is the same as for carrying for ten in simple multiplication.

Tut. It is the same; the only difference being, that the numbers do not increase in a uniform ratio. and require that we carry, from one denomination to another, such a number as it takes of the former to make one of the latter.

Multiply 16 cwt. 2 qr. 24 lb. by 6.

Ans. 100 cwt. 1 qr. 4 lb.

Multiply 125 d. 10 h. 14' 42" by 9.

Ans. 1128 J. 20 h. 12' 18".

Multiply 12 lb. 9 oz 13 pwt. 23 grs. by 7.

Ans. 89 lb. 7 oz. 16 pwt. 17 grs.

Ans. 37 cwt.

What is the weight of 8 hhds. sugar, each weighing

4 cwt. 2 qr. 14 lb.?

When it is required to multiply a number, exceeding 12; resolve the number into two or more factors; then multiply first by one of them, and then their product by one of the others, and so on with all the factors, and the last product will be the answer.

Pup. Will this give the same answer as it would to multiply by 16?

Tut. The answer will be the same; for multiplying by 8 gives 8 times the multiplicand, and then multiplying by 2 doubles this product, which is the product by 16, as twice 8 is 16.

Pup. I now understand it, and see that it must be much easier than to multiply by 16.

Tut. You may multiply the following numbers together.

Multiply 14 T. 12 cwt. 2 qrs. 16 lb. by 36.

as. 526 T. 15 cwt. O qrs. 16 lb.

Multiply 126 d. 22 h. 48' 12" by 56.

Ans. 7110 d. 16 h. 59′ 12′′.

Frequently the multiplier cannot be resolved into factors, and express the exact number. When this is the case, take the number nearest to it which can be resolved into factors, and find the product of the multiplicand by this number. Then add or subtract from this product so many times the multiplicand as the number resolved exceeds or falls short of the given number..

Here we find that 17 cannot be resolved into factors, but 16 can, 4 times 4 being equal to 16. The product by 16 falls short of the product by 17 by once the multiplicand, consequently we must add the multiplicand to its product by 16, and it gives the product by 17. 18 might have been taken instead of 16, to resolve into factors, which would have given a product too large by once the multiplicand, and the multiplicand must have been subtracted from the product.

Multiply 12 lb. 8 oz. 6 pwt. 18 grs. by 44.

Multiply 2 d. 16 h. 24' by 37.

Ans. 558 lb. 6 oz. 17 pwt.

Ans. 99 d. 8 h. 48'.

When the multiplier exceeds 144, first multiply the multiplicand by 10, then that product by 10, which gives the product of the multiplicand by 100, and if the multiplier be even hundreds, mutiply this product by the number of hundreds, and the product will be the answer. If there be odd numbers, multiply the product of 10 by the number of tens, and the multiplicand or given number by the number of units; then the sum of these products will be the answer.

Multiply 12 cwt. 2 qr. 16 lb. by 160.

Here I first find the product of the multiplicand by 10; then I multiply this product by 10, which gives the product by 100; I then find the product of the multiplicand by 60, by multiplying the product by 10, by 6; and the product of the mu tiplicand by 60, added to the product of the same by 100, is the answer.

What will 16 cwt. 3 qr. 14 lb. of sugar cost at $10,50 per cwt. ? Ans. $117,18. What will 4 cwt. 2 qr. 18 lb. tobacco cost at $6,25 per cwt.? Ans. $29,12.

When the numbers are such that the lower denominations can be resolved into even parts of the highest denomination; to obtain the price of the lower denomination, take such parts of the price as the lower denominations are parts of the highest denomination, and these several parts, added to the product of the highest denomination by the price, will be the answer.

What will 12 cwt. 2 qr. 14 lb. sugar cost at $8,25 per cwt. ?

8-2) 8,25

12

99,00 price of 12 cwt.

4,12 price of half cwt. or 2 qr. 1,03 price of of a cwt. or 14 lb.

Ans. $104,15

What will the freight of 42 cwt. 3 qrs. 21 lb. be, at $0,42 per cwt. ? Ans $18,03. What will 8 yds. 3 qrs. 2 nls. cost, at $7,25 per yard? Ans. $64,34.

Compound numbers may be divided by considering each denomination as a distinct dividend; and as we multiplied each separately, we will reverse the order of the work,

and divide each separately. If we take the example given in multiplication, and reverse the order of the work, we shall obtain the former number.

8)96 8)64 8)48 8)128

12

8

6

16 12lb. 8oz. 6pwt. 16gr.

As in multiplication we increased each number 8 times, so in division we decrease each number 8 times.

But as we have, in multiplication, a shorter method of performing the work, so in division we can divide a compound number by a simple number without the trouble of the above process. The following is the general

rule for

COMPOUND DIVISION.

Place the numbers as in simple division; then divide the highest denomination by the divisor, and reduce the remainder to the next lower denomination, adding to the product the given number of that denomination to which it is reduced; then divide this number by the divisor, and reduce the remainder to the next lower denomination, adding the given number of this denomination as before, and proceed in this manner with all the denominations, when the remainder, if there is any, must be expressed by a fraction; then the sum of the several quotients will be the answer.

If we take the product from the example given in multiplication, and divide it by 8, we shall obtain the number there given, and the work will stand as follows,

Here we first find how often 8 is contained in 101, and find it is 12 times and 5 over; 5 here means 5 pounds, and must be reduced to ounces, and the given ounces added to them, which make 66 oz. ; 66 divided by 8, give a quotient of 8, and 2 remainder; this remainder must now be reduced to penny-weights, and the given pennyweights added, which make 53; 53 divided by 8, give a quotient of 6, and 5 over; 5, reduced to grains, make,