2°. By the multiplication table find the greatest number of times the divisor is contained in the first figure of the dividend, or, if necessary, the first two or first three figures of the dividend, and place the figure denoting the times directly under the figure divided, or under the last figure if more than one have been taken, and carry what is over, that is, regard the figure which is over to be prefixed to the following figure of the dividend.

3. Divide this number by the divisor, set down the result as the next figure of the quotient, carry the remainder to the next figure of the dividend; and if the divisor is not contained in any figure of the dividend, place a cypher in the quotient and prefix this figure to the next one of the dividend as if it were a remainder, and proceed in the same manner till all the figures of the dividend are exhausted. The number thus found is the quotient.

Ex. 1. Divide 25602 by 3.

EXAMPLES.

3)25602 8534

Placing the dividend and divisor (3) as in the margin, we proceed thus:3 is contained in 2, no times, so that nothing is to be placed under the 2; 3 is contained in 25, i.e., 25 thousands, 8 times (i.e., 8 thousands) and I over, 8 and carry 1; this regarded as prefixed to the 6, gives the number 16, .e., 16 hundreds, 3 is in 16, 5 times and I over; 3 is in 10 (i.e., 1 hundred) 3 times and 1 over; 3 is in 12, 4 times. Therefore, the quotient is 8534; and this is the complete quotient, as there is no remainder.

5)7804623

Ex. 2. Divide 7804623 by 5.

We say, 5 in 7, 1 and 2 over; 5 in 28, 5 and 3 over; 5 in 30, 6; 5 in 4, o and 4 over; 5 in 46, 9 and 1 over; 5 in 12, 2 and 2 over; 5 in 23, 4 and 3 over. As there is here a remainder, we annex it, with the divisor 5 under it, to the figures of the quotient, and call 1560924 the complete quotient.

1560924

Ex. 3. Divide 84111648 by 12.

12)84111648

We say, 12 in 8, no times, so that nothing is to be placed under the 8; 12 in 84 goes 7 times and o over; 12 in 1, no times and I over, put down o under the 1; 12 in 11, o times and 11 over; this 11 regarded as prefixed to the next figure of the dividend gives the number 111; therefore say, 12 in 111, 9 times and 3 over; 12 in 36, 3 times and o over; 12 in 4, o times and 4 over; 12 in 48, 4 times.

EXAMPLES FOR PRACTICE.

7009304

8. 2470263075 ÷ 9

4. 2982447605

109513717012

12. 59437055312 12

Divide each of the following dividends six times successively by each of the following divisors-2, 3, 4, 5, 6, 7, 8, 9, 11, 12.

1. 6154778

2. 7000000

3. 4086791

4. 8821000

42. If the divisor be greater than 12.

RULE VIII.

1°. Place the divisor and dividend in the same line, separated by a small curve line, and on the right of the dividend draw another line of the same kind; thus,— divisor) dividend (quotient

2°. Mark off from the left-hand side of the dividend a numbor of figures equal in number to those of the divisor, or one more, if necessary, and find the greatest number of times the divisor is contained in this number; place the figure which denotes this number on the right as the first figure of the quotient.

3°. Multiply the divisor by this number and place the product under the number marked off at the left of the dividend. Subtract the said product from that part of the dividend under which it stands.

4. Bring down the next figure of the dividend and place it to the right of the remainder, and if the number thus formed be greater than the divisor, find the greatest number of times the divisor is contained in it, and write this number as the second figure of the quotient; but if this number be less than the divisor bring down the next figure of the dividend, or more, until a number not less than the divisor, is formed, remembering to place a cypher in the quotient for every figure brought down, except the last; find how often the divisor is contained in this number; then multiply, subtract, and bring down, &c., as before, till all the figures of the dividend are exhausted.

The number thus obtained is the quotient required.

NOTE.-When the divisor is large, the learner will find assistance in determining the quotient figure, by finding how many times the first one or two figures on the left hand of the divisor is contained in the first one or more of those of the dividend. This will give pretty nearly the right figure. Some allowance must, however, be made for carrying from the product of the other figures of the divisor, to the product of the first into the quotient figure. If any product be greater than the number which stands above it, the last figure in the quotient must be changed for one of smaller value; but if any remainder be greater than the divisor, or equal to it, the last figure of the quotient must be changed for a greater.

2422

1423

1384

394

346

48

Looking at the leading figure of the divisor, and also at that 346)256434(741 quotient. of the dividend, with the view of seeing whether the latter contains the former, which it does not, 3 being greater than 2; we therefore commence with the number 25, formed by the first two figures of the dividend, and seeing that 3 is contained in 25 8 times, we should put 8 for the first quotient figure; but bearing in mind that when the whole divisor is multiplied by this 8 we must attend to the carryings; we perceive that 8 is too great, we therefore try 7, and find 7 times 346 to be 2422, a number less than 2564 above it, so that we can subtract; the remainder is 142, which, when the next figure of the dividend is brought down, becomes 1423. We now take this as a dividend, and looking at the leading figures in this new dividend and the divisor, we see that the latter will go 4 times, we therefore put 4 for the second quotient figure, and multiplying and subtracting we get 39 for the second remainder, and, by bringing down another figure we get 394 for a new dividend; the divisor goes into this once, so that the quotient is 741, and the final remainder 48; this remainder must be annexed with the divisor underneath to the quotient figures, so that the complete quotient is 741, which is the 346th part of 256434.

It must be noticed that if any dividend formed by a remainder and a figure brought down should be less than the divisor, that the divisor will go no times in that dividend; so that a o will be the corresponding quotient figure; and that, then, a second figure must be brought down as in the operation annexed. The steps marked are inserted merely to show the principle. In practice we simply put down the two noughts in the quotient, and go at once to 32214 for the divisor.

Ex. 3. Divide 6421284 by 642.

642)6421284(10002 642

1284
1284

642 goes once into 642, and leaves no remainder. Bring down the next figure (1) of the dividend, then 642 is in 1 no times, put o in the quotient after the 1. The next figure of the dividend (2) being brought down to the right of the I forms the dividend 12, then 642 is in 12 no times, put a cypher as the next quotient figure; bring down the next figure of the dividend (8), then 642 in 128 goes no times, write o as the corresponding quotient figure; but the next figure brought down (4) makes the quantity brought down 1284, which contains the divisor twice, and gives no remainder.

29. Divide 1000........) with as many noughts added as may be necessary to give ten places of figures in the quotient) by 2302585093.

43. When the divisor is a composite number, that is, can be separated into two or more factors, the division can be effected by the following rule :RULE IX.

1o. Divide by one factor, setting down the quotient and remainder.

2°. Divide the quotient thus obtained by another factor, setting down the quotient and remainder, and so on, till all the factors are employed. The last quotient will be the answer required.

3°. The proper remainder is found, when the divisor is resolved into but two factors, by multiplying the second remainder by the first divisor, and to the product adding the first remainder, but when more than two factors are employed by multiplying the remainders after every line by all the divisors except their own, and adding the results.

Ex. 1. Divide 569736869 by 15.

EXAMPLES.

Here, since 15 is the product of 3 and 5, it is obvious that the quotient may be obtained from successive division by 3 and 5.

3569736869

15

189912289....2

37982457....4

Here, the remainder 2 in the first quotient is 2 units of the upper line; but the remainder 4 in the second line consists of 4 units of the second line; and as each unit in the second line is three times as great as each unit in the upper line, the remainder 4 is equal to 3 X 4 units of the upper line, i.e., is equal to 12 ordinary units, hence the whole remainder is 2 +12, or is 14. Ex. 2. Divide 8327965 by 72.

Ex. 3.

Divide 8327965 by 99.

115666....I

84120....9

To deduce the remainders which would have been left had the divisions been performed by 72 and 99 in the usual way, we may observe that the first partial remainder 5 (Ex. 2) must be units; but the second remainder must be regarded as so many collections of 8 units each, and that the first partial remainder 4 (Ex. 3) must be units; but the second dividend being as so many collections of 9 units each, the second remainder must, in this case, be regarded as so many collections of 9 units each; hence the true remainders in these examples are respectively 1 × 8+5=13, and 9 × 9 + 4 = 85.

13916....5X6X4+6=126

Dividing by 4 the remainder is 2, dividing by 6 the remainder is 1, dividing by 8 the remainder is 5, which gives a total remainder 126, found thus: the remainder 2 is two units, the second remainder is one 4 and 2, making 6, the third remainder 5 is five times 24 (4 x 6) = 120, because having divided by 4 x 6 = 24, the quotient 111333 are twenty-fours, so that any remainder must consist of twenty-fours; 120 added to the previous remainder 6, gives 126. Thus the answer is quotient 13916, remainder 126.

F

44. Division may also be abridged where the divisor is terminated by a cypher or cyphers; we proceed as follows:

RULE X.

1o. Cut off the cyphers from the divisor, and as many figures from the right-hand of the dividend as there are cyphers so cut off at the right-hand end of the divisor, then proceed with the remaining figures in the usual manner (Rule VII or VIII), and if there are anything remaining after the division annex those figures which are cut off from the dividend; otherwise, the figures cut off will be the remainder. NOTE.-The same rule applies when the divisor and dividend both terminate with cyphers.

Ex. I. Divide 3704196 by 20.

2,0)370419,6

185209 18

EXAMPLES.

In the first of these examples you mark off with a comma the cypher or o in the divisor, and the first figure 6 to the right in the dividend; this is equivalent to dividing both divisor and dividend by 10. You next divide the remaining figures 370419, to the left in the dividend, by the divisor 2, according to Rule VII; thus is obtained the quotient 185209, and remainder 1; to this remainder you annex the figure 6, which was cut off, and you have the complete remainder 16. The quotient may now be correctly represented thus, 18520918.

Ex. 3. Divide 271830 by 30. 3,0)27183,0

Ex. 2. Divide 31086901 by 7100.
71,00)310869,01(4378 188
284

268

213

556

497

599

568

31

In the second example you follow the same rule; that is, you cut off two cyphers in the divisor and two figures in the dividend, and obtain the quotient in the usual way, which is 4378, and remainder 31; to this 31 annex the two figures cut off from the dividend, and you have the complete remainder 3101.

1st. Cut off a cypher from the divisor, and also one from the dividend. 2nd. Divide 27183, the remainder of the dividend, by 3, the remainder 9061 Ans. of the divisor: the quotient 9061 is the answer.

45. Verification of Division.-(1.) Multiply the quotient by the divisor, or the divisor by the quotient, and to the product add the remainder, if there be one. The result ought to be the same as the dividend; because we are only adding the divisor the same number of times, as it was subtracted in the operation of division.