is plain that each parcel must contain 2 pints, or 1 quart; that is, every unit of the fi:st quotient contains 2 of the units of the given dividend; consequently, every unit of it that remains will contain the same; (Art. 113. Obs. 2;) therefore this remainder must be multiplied by 2, in order to find the units of the given dividend which it contains.

2. Dividing the quotient 253 parcels, by 8, will distribute them into 31 other equal parcels, each of which will evidently contain 8 times the quantity of the preceding, viz: 8 times 1 quart =8 quarts, or 1 peck; that is, every unit of th second quotient contains 8 of the units in the first quotient, or 8 times 2 of th units in the given dividend; therefore what remains of it, must be multiplied by 8×2, or 16, to find the units of the given dividend which it contains.

3. In like manner, it may be shown, that dividing by each successive factor reduces each quotient to a class of units of a higher value than the preceding; that every unit which remains of any quotient, is of the same value as that quotient, and must therefore be multiplied by all the preceding divisors, in order to find the units of the given dividend which it contains.

4. Finally, the several remainders being reduced to the same units as those of the given dividend according to the rule, their sum must evidently be the true remainder. (Ax.11.)

3. How many acres of land, at 35 dollars an acre, can you buy for 4650 dollars?

4. Divide 16128 by 24.

6. Divide 17220 by 84.

5. Divide 25760 by 56.

7. Divide 91080 by 72.

CASE II.-When the divisor is 1 with ciphers annexed to it.

130. It has been shown that annexing a cipher to a number increases its value ten times, or multiplies it by 10. (Art. 98.) Reversing this process; that is, removing a cipher from the right hand of a number, will evidently diminish its value ten times, or divide it by 10; for, each figure in the number is thus restored to its original place, and consequently to its original value. Thus, annexing a cipher to 15, it becomes 150, which is the same as 15X10. On the other hand, removing the cipher from 150, it becomes 15, which is the same as 150-10.

In the same manner it may be shown, that removing two ciphers from the right of a number, divides it by 100; removing three, divides it by 1000; removing four, divides it by 10000, &c. Hence,

QUEST -130. What is the effect of annexing a cipher to a number? What is the effect of removing a cipher from the right of a number? How does this appear?

131. To divide by 10, 100, 1000, &c.

Cut off as many figures from the right hand of the dividend as there are ciphers in the divisor. The remaining figures of the dividend will be the quotient, and those cut off the remainder.

8. In one dime there are 10 cents: how many dimes are there in 200 cents? In 340 cents? In 560 cents?

9. In one dollar there are 100 cents: how many dollars are there in 65000 cents? In 765000 cents? In 4320000 cents? 10. Divide 26750000 by 100000.

11. Divide 144360791 by 1000000.

12. Divide 582367180309 by 100000000.

CASE III.-When the divisor has ciphers on the right hand.

13. How many hogsheads of molasses, at 30 dollars apiece, can you buy for 9643 dollars?

OBS. The divisor 30, is a composite number, the factors of which are 3 and 10. (Arts. 95, 96.) We may, therefore, divide first by one factor and the quotient thence arising by the other. (Art. 129.) Now cutting off the right hand figure of the dividend, divides it by ten; (Art. 131;) consequently dividing the remaining figures of the dividend by 3, the other factor of the divisor, will give the quotient.

Operation. 310)96413

321 13 Ans.

Now as the 3 cut

nex it to the 1.

We first cut off the cipher on the right of the divisor, and also cut off the right hand figure of the dividend; then dividing 964 by 3, we have 1 remainder. off, is part of the remainder, we therefore Ans. 32113 hogsheads. Hence,

132. When there are ciphers on the right hand of the divisor. Cut off the ciphers, also cut off as many figures from the right of the dividend. Then divide the other figures of the dividend by the remaining figures of the divisor, and annex the figures cut off from the dividend to the remainder.

14. How many buggies, at 70 dollars apiece, can you buy for 7350 dollars?

QUEST. 131. How proceed when the divisor is 10, 100, 1000, &c.? 12. When there are eiphers on the right hand of the divisor, how proceed? What is to be done with figures cut off from the dividend?

15. How many barrels will it take to pack 36800 pounds of pork, allowing 200 pounds to a barrel ?

16. Divide 3360000 by 17000.

133. Operations in Long Division may be shortened by subtracting the product of the respective figures in the divisor into each quotient figure as we proceed in the operation, setting down the remainders only. This is called the Italian Method.

17. How many times is 21 contained in 4998 ?

Operation. 21)4998(238

79

168

This method, it will be seen, requires a much smaller number of figures than the ordinary process.

.18. Divide 1188 by 33.

19. Divide 2516 by 37.

20. Divide 3128 by 86.

21. Divide 7125 by 95.

22. A merchant laid out 873 dollars in flour, at 5 dollars a barrel how many barrels did he get?

Operation.
873

2

1|0)17416
174 Ans.

We first double the dividend, and then divide the product by 10, which is done by cutting off the right hand figure. (Art. 131.) But since we multiplied the dividend by 2, it is plain that the 6 cut off, is 2 times too large for the remainder; we therefore divide it by Hence,

2, and we have 3 for the true remainder.

134. When the divisor is 5.

Multiply the dividend by 2, and divide the product by 10. (Art. 131.)

Note.-1. When the figure cut off is a significant figure, it must be divided by 2 for the true remainder.

2. This contraction depends upon the principle that any given divisor is contained in any given dividend, just as many times as twice that divisor is contained in twice that dividend, three times that divisor in three times that dividend, &c. For a further illustration of this principle see General Principles in Division.

24. Divide 8450 by 5.

23. Divide 6035 by 5.

25. Divide 32561 by 5.

26. Divide 43270 ly 5.

135. When the divisor is 15, 35, 45, or 55.

Double the dividend, and divide the product by 30, 70, 90, or 110, as the case may be. (Art. 132.)

Note.-This method is simply doubling both the divisor and dividend. We must therefore divide the remainder, if any, by 2, for the true remainder.

27. Divide 1256 by 15.

29. Divide 3507 by 45.

136. When the divisor is 25.

28. Divide 2673 by 35.

30. Divide 7853 by 55.

Multiply the dividend by 4, and divide the product by 100. (Art. 131.)

Note.-This is obviously the same as multiplying both the dividend and divisor by 4. (Art. 134. Note 2.) Hence, we must divide the remainder, if any thus found, by 4, for the true remainder.

31. Divide 2350 by 25.

33. Divide 42340 by 25.

137. To divide by 125.

32. Divide 4860 by 25.

34. Divide 94880 by 25.

Multiply the dividend by 8, and divide the product by 1000. (Art. 131.)

Note. This contraction is multiplying both the dividend and divisor by 8. For the true remainder, therefore, we must divide the remainder, if any, by 8. 35. Divide 8375 by 125. 36. Divide 25426 by 125.

138. To divide by 75, 175, 225, or 275.

Multiply the dividend by 4, and divide the product by 300, 700 900, or 1100, as the case may be. (Art. 132.)

Note.-For the true remainder, divide the remainder, if any thus found, by 4

37. Divide 1125 by 75. 39. Divide 3825 by 225.

38. Divide 2876 by 175. 40. Divide 8250 by 275.

139. The preceding are among the most frequent and useful modes of contracting operations in division. Various other methods might be added, but they will naturally suggest themselves to the inventive student, as opportunities occur for their application.

41. How long would it take a vessel sailing 100 miles per day to circumnavigate the earth, whose circumference is 25000 miles?

42. The distance of the Earth from the Sun is 95,000,000 of miles: how long would it take a balloon going at the rate of 100,000 miles a year, to reach the sun?

43. The debts of the several States of the Union, in 1840, amounted to 171,000,000 of dollars, and the number of inhabitants was 17,000,000: how much must each individual have been taxed to pay the debt?

44. The national debt of Holland is 800,000,000 of dollars, and the number of inhabitants 2,800,000: what is the amount of indebtedness of each individual ?

45. The national debt of Spain is 467,000,000 of dollars, and the number of inhabitants 11,900,000: what is the amount of indebtedness of each individual ?

46. The national debt of Russia is 150,000,000 of dollars, and the number of inhabitants 51,100,000: what is the amount of indebtedness of each individual?

47. The national debt of Austria is 380,000,000 of dollars, and the number of inhabitants 34,100,000: what is the amount of indebtedness of each individual?

48. The national debt of France is 1,800,000,000 of dollars, and the number of inhabitants 33,300,000: what is the amount of indebtedness of each individual?

49. The national debt of Great Britain is 5,556,000,000 of dollars, and the number of inhabitants 25,300,000: what is the amount of indebtedness of each individual?

50. Divide 467000000000 by 25000000000.