Hence, to divide by a number consisting of 9, or of several 98: I. DIVIDE REPEATEDLY BY A UNIT OF THE NEXT HIGHER ORDER, UNTIL YOU OBTAIN 0 FOR A QUOTIENT FIGURE.

II. ADD THE SEVERAL QUOTIENTS, AND LIKEWISE THE SEVERAL REMAINDERS THUS OBTAINED, CARRYING FROM THE REMAINDERS TO THE QUOTIENTS, AS IN COMMON ADDITION, AND PLACING THE FIGURE SO CARRIED, UNDER THE UNITS' COLUMN OF THE REMAINDERS.

III. TO THE TOTAL REMAINDER OBTAINED BY THIS ADDITION, ADD THIS FIGURE, AND IF THE SUM EQUAL OR EXCEED THE GIVEN DIVISOR, SUBTRACT THAT DIVISOR FROM IT, AND ADD 1 TO THE SUM OF THE QUOTIENTS. THE RESULTING QUOTIENT AND REMAINDER WILL BE THE TRUE ONES SOUGHT.

ART. IV. By a similar method we may abbreviate Division, when the units fig. ure of the divisor is not a 9. In order to understand this, we will premise, that the difference between any number and a unit of the next higher order, is called the complement of the number. Thus, 2 is the complement of 98, because 100-98

2. And 3 is the complement of 97, because 100-97-3, and so on. Now since 98 is 2 less than 100, 98 will be contained in any number just as often as 100 is contained in the same number, with just as many 2s over. Then if we find how often 98 is contained in these tiros over, and add this number to the number of 100s, we shall obtain the whole number of 988 in the given number. Of course, after dividing by 100, we must multiply the quotient by 2, and divide it by 98. And as we may use the same artifice as before, in this second division, and then in the third, and so on, we find that this case only differs from that in Art. III, in requiring us to multiply by 2 our successive quotients, obtained from repeated divisions by 100.

This will be rendered clear by an illustration.

Divide 75,432 by 98.

Divide by 100, thus,

754X2 1,508.
15X2=30.

1508100=

30+100

Quo. 769 70 Rem.

We will now give a case, in which it will be necessary to carry from the remainders to the quotients.

Divide 49,999 by 98.

Divide by 100, thus,

499X2 998. 998-100-
9X2=18.

18+100

Here, we had 2 to carry to the quotients. As this 2 was 2 hundreds, and as cach hundred is 2 more than the divisor, 98, we must increase the remainder by 2X2=4, which is accordingly done above. Instead, then, of adding to the remainder, as before, the figure which we carry, we add that figure, multiplied by the complement 2.

It is plain, that if we were dividing by 97, we should be obliged to multiply by 3, wherever, in the above instances, we multiplied by 2. If by 96, we should be obliged to multiply by 4; if by 95, by 5; and so on: that is, we should always multiply by the complement of the divisor. Hence, to divide by a number which consists of 98, except the unit figure,

PROCEED BY THE LAST RULE, EXCEPT THAT YOU MULTIPLY EACH SUC CESSIVE QUOTIENT BY THE COMPLEMENT OF THE DIVISOR, BEFORE YOU DIVIDE AGAIN; AND THAT YOU ADD TO THE REMAINDER, THE PRODUCT

OF THE COMPLEMENT AND THE NUMBER CARRIED TO THE QUOTIENT, INSTEAD OF THE NUMBER CARRIED, ITSELF.

764)98532(128 2213

ART. V. When we employ long division, we multiply the divisor by each quotient figure, and set down the product before subtracting it from the partial dividend. We may somewhat shorten the process, by subtracting each figure of this product, as we multiply, and noting down only the remainder, thus: Here our first quotient figure is 1. We multiply the divisor by it, and subtract each figure of the product, as we multiply, setting down only the remainder, 221. To this we annex the next figure of the dividend, 3. We then multiply and subtract, as before, and so on. Our final remainder is 740; our quotient 128. In this operation, there is so much to be retained in the mind, that it requires considerable acquaintance with numerical calculations, to render it very useful.

6852

740

ART. VI. When the divisor and dividend are large, it will often be found useful to make a table of products of the divisor, by the nine digits. For each quotient figure must be one of these digits or a cypher; and, by means of the table, we can tell, at a glance, how many times the divisor is contained in each partial dividend. We have, then, the quotient figure and its corresponding product, and, of course, have nothing to do but subtract.

These are the most important contractions in Division. We have now been attending to Notation or Numeration, Addition, Multiplication, Subtraction, and Division. These are called the fundamental or ground rules of Arithmetic, because, by their various combinations, all arithmetical operations are performed. EvoLUTION might, perhaps, seem an exception to this remark; but a little reflection will convince any one, that this process is only a case of division, in which the divisor and quotient are both unknown, but are required to be equal. For division consists in resolving a number into factors, and evolution, in resolving a number into equal factors.

COMPOUND NUMBERS.

ADDITION.

MENTAL EXERCISES.

XXXV. 1. Four boys gathered chesnuts. The first gathered 2 qts. ; the second, 3 qts.; the third, 7 qts., and the fourth, 5 qts. How many pecks did they all gather? Then, 2 qts. and 3 qts. and 7 qts. and 5 qts. are how many pecks?

2. A silversmith melted together several pieces of silver. as follows; one weighing 10 oz.; another 11 oz.; another, 5 oz.; and another 6 oz. How many lbs. did he melt. Then, 10 oz. and 11 oz. and 5 oz. and 6 oz. are how many lbs. ?

3. In 3s. 4d. and 5s. 6d. and 2s. 1d. how many shil lings and pence?

4. In 4s. 1d. and 8s. 3d. and 2s. 11d. how many shillings and pence?

5. In 16s. 9d. and 7s 10d. how many pounds, shillings and pence?

6. In 3 qts. 1 pt. and 5 qts 1 pt. how many gal. and qts. ?

7. In 3 qts. 1 pt. and 2 qts. I pt. and 1 qt. 1 pt. how many gal. qts. and pts. ?

8. In 2 yds. 2 ft. and 7 yds. 2 ft. how many yds. and ft.? 9. In 6 fur. 35 rds. and 7 fur. 15 rds. how many mls. fur. and rds.?

In the above examples, the numbers have consisted of several denominations. Such numbers are called COMPOUND NUMBERS. Numbers consisting of one denomination only, are called SIMPLE NUMBERS. Write the following examples.

d.

qrs.

5

4

2

2

3

3

5

1

1

2

10. A boy paid for a book 5s. 4d. 2 qrs.; for a bunch of quills, 2s. 3d. 3 qrs.; and for a penknife 3s. 5d. 1 qr. What cost the whole ? Place like denominations under each other; then add the qrs. 1+3+2=6qrs. But 6 qrs.=ld. 2 qrs., because 4 qrs. 1d. Set down the 2 qrs. and carry the 1d. to the pence. There are then 13d. 1s. 1d. Set down the 1d. and carry the 1s. to the shillings. The shillings, added, are 11s., which, as they are not enough to maxe a £. set down under shillings. In like manner, perform the following.

3

11

11. Bought a watch for 5£. 6s. and a chain for 1£. 19s. How much did I give for both? Ans, 7£, 5s.

12. A man has four farms. The first contains 100 acres, 1 rood, 20 rods; the second 75 acres, 2 roods, 10 rods; the third 150 acres, 15 rods; the fourth 125 acres. 1 rood, 5 rods. How many acres in all? Ans. 451 acres, 1 rood, 10 reds.

13. A merchant bought 4 pieces of cloth, the first containing 10 yds. 2 qrs. 1 nl.; the second 25 yds. 2 qrs. 1 nl.; the third 20 yds. 2 qrs. 2 nls. and the fourth 10 yds. 1 qr. 1 nl. How much in all? Ans. 67 yds. 1 nl.

14. Add together 38 gals. 2 qts. 1 pt. 2 gi.; 16 gals. 1 qt. 3 gi,; 20 gals. 2 qts. 1 pt. 1 gi.; 18 gals. 1 qt. 1 pt.;. 7 gals. 1 qt. 2 gi.; 30 gals. 2 qts. 1 pt. Ans. 132 gals.

From these examples we derive the rule to perform Addition of Compound numbers.

I. PLACE THE SAME DENOMINATIONS UNDER EACH OTHER.

II. ADD, FIRST, THE LOWEST DENOMINATION. FIND HOW MANY OF

THE NEXT HIGHER ARE CONTAINED IN THE SUM, WHICH NUMBER CAR, RY TO THE NEXT HIGHER, AND PLACE THE REMAINDER UNDER The de, NOMINATION ADDED.

III. PROCEED THUS WITH ALL THE DENOMINATIONS,

EXAMPLES FOR PRACTICE.

15. A man bought land to the amount of 69£ 13s. 5d.; farming implements to the amount of 11£ 10s. ; a yoke of oxen for 15£ 6s.; a horse for 13£. 0s. 4d.; a cart for 4£ 17s. 8d. and a saddle for 198. 4d. 2 qrs. What did the whole cost him; Ans. £115; 6; 91.

16. A silversmith purchased seven ingots of silver. The first weighed 11 lb. 6 oz. 18 dwt.; the second 9 lb. 3 oz. 7 dwt.; the third 4 lb. 7 oz. 9 dwt. 5 gr.; the fourth, 8 lb. 4 oz. 6 dwt.; the fifth, 10 lb. 3 oz. 5 dwt. 19 gr.; the sixth, 7 lb. 9 oz. 0 dwt. 18 gr.; and the seventh 8 lb. 11 oz. 10 dwt. What did the whole weigh; Ans. 60 lb. 9 oz. 16 dwt. 18 gr.

17. George lived in Hartford until he was 14 yrs. 3 mo. 4 d. old; then he went to New-Haven, where he remained 8 yrs. 5 mo.; then he went to New York, where he remained 3 yrs.; then to Philadelphia, where he staid 3 yrs. 2 mo. 1 d. His journeys occupied 4 days? How old was he then ? Ans. 28 yrs. 10 mo. I w. 2 d.

18. A man has, in real estate £304; 5, in one place, and £247; 0; 11, in another; and in personal property, the several sums £34; 19; 7, £7; 18; 5, £45; 0; 6, and 19s. Od. 3 qrs. How much in all? Ans. £640; 3; 5; 3.

19. A man brings to market 4 loads of wood, containing, the first 1 cord 60 ft. 860 in.; the second 1 cord 67 ft. 68 in.; the third 1 cord 30 ft. 300 in.; the fourth 1 cord 30 ft. 631 in. How much in all? Ans. 5 cords, 60 ft. 131 in

20. Bought a quantity of goods for £125; 10: paid for freight £3; 19; 6 for transportation, £2; 5: for duties £1; 15; 10: my expenses were £2; 13; 9. What was the whole expense of the goods to me?

MULTIPLICATION.

MENTAL EXERCISES.

§ XXXVI. 1. A man gave 6d. apiece to four of his children. How many shillings did he give them?

2. 5 boys gathered 3 quarts of walnuts apiece. How many pecks did they all have?

3. Three baskets hold 1 pk. 4 qts. each. How many pks. and qts. will all hold?

4. If one bushel of grain cost 2s. 6d. how much will 2 cost? How much will 3? will 4? 5? 6? 7? 8?

5. Multiply 3 qrs. 4 nls. by 2; by 3; by 4; by 5; by 6; by 7; by 8;

6. Multiply 15 min. by 15 7. Multiply 1 pt. 3 gi. by 2; 8. Multiply 0 oz. 12 dwt. 6; 7; 8; 9;

sec. by 4; by 8; by 12; by 3; 4; 5; 6; 7; 8; 9; 13 gr. by 2; by 3; 4; 5;

9. Multiply 2 in. 2 b. c. by 2; by 3; by 4; 5; 6; 7; 8; 9;

10. Multiply 3 yds. 2 ft. 8 in. by 2; 3; 4; 5; 6; 7 8; 9;

Perform the following on your slate.

11. A man bought 3 sheep at 1£. 10s. apiece. whole? Set the multiplier under the multiplicand, and multiply as in simple numbers, observing only to carry from one denomination to another, as in Addition of Compound Numbers.

What cost the

£.

S.

1

10

3

10

4

12. How much wool in 3 packs, each pack weighing 2 cwt. 2 qrs, 13 lb. A. 7 cwt. 3 qrs. 11 lb. 13. What is the value of 5 cwt. of raisins, at 2£. 1s. 8d. pr. cwt. A. 10£. 8s. 4d.

14. What is the weight of 3 doz. silver spoons, each doz. weighing 2 lb. 6 oz. 12 dwt. 3 gr. A. 7 lb. 7 oz. 18 dwt. 9 grs. 15. In 8 bales of cloth, each bale containing 12 pieces, and each piece 27 yds. 1 qr. 2 nls., how many yds. ? From these examples, we derive the following rule.

A. 2,628.

PLACE THE MULTIPLIER UNDER THE MULTIPLICAND, MULTIPLY EACH DENOMINATION SEPARATELY, BEGINNING WITH THE LOWEST, AND CARRY AS IN ADDITION.

EXAMPLES FOR PRACTICE.

16. What cost 12 bu. of apples, at 1s. 9d. per bu.? A. £1. Is.. 17. What cost 9 lbs. of cinnamon, at 11s. 4 d. per lb. ?