In this case the two first figures of the dividend are taken, because the first does not contain the divisor. In 14, 8 is contained once, and 1 is written in the quotient. 8 is then multiplied by 1, and the product 8 subtracted from 14, and 6 is the remainder. By the side of 6, 4 the next figure in the dividend is brought down; and the operation continued by saying, in 64 how many times 8? It is contained 8 times; and 8 is written in the quotient. After multiplication, the product 64 is subtracted from 64, and 0 remains; by the side of which, 6 the fourth figure in the dividend is brought down. But as 6 does not contain 8, 0 is written in the quotient, and by the side of 6, the last figure 4 of the dividend is brought down; we then say, in 64 how many times 8? It is contained 8 times. After having written 8 in the quotient and subtracted 64, the product by multiplication, nothing remains. And as there is no remainder, we conclude that 14464 contains 8, 1808 times.

4. Divide 3728675 by 8.

466084,

3.

5. Divide 3656 dollars equally among 8 men.

6. How much beef at 8 cents per pound can be bought for $1,12?

Ans. 14 tbs.

Ans. 457 dolls. to each.

DIVISION BY A NUMBER COMPOSED OF SEVERAL

FIGURES.

62. When the divisor contains many figures, the operation is conducted in the following manner.

Take upon the left of the dividend as many figures as are necessary to contain the divisor. Then, instead of seeking, as before, how often this part of the dividend contains the entire divisor, seek how many times the first figure of the divisor is contained in the first figure of the dividend, or in the two first figures, if the first does not contain it, and place this quotient under the divisor as in the former case.

Multiply successively, according to the rule in article (50,) all the figures of the divisor by the quotient figurė, and place the product under the corresponding figures of the partial dividend. Subtract it therefrom, and by the side of the remainder bring down the next figure of the dividend, to continue the operation in the same manner.

Any embarrassment in the application of this rule will be removed by the following examples.

In commencing the operation, only the two first figures of the dividend are taken, because they contain the divisor; and instead of saying in 75 how many times

53, we seek how often the 7 tens of 75 contain the 5 tens of 53, that is, how often 7 contains 5. It is contained once; and 1 is written in the quotient. 53 is then multiplied by 1 and the product 53 written under 75. After subtraction there remains 22, by the side of which, 3, the next figure in the dividend is brought down, and the operation continued by saying, in 22 how many times 5 (instead of saying in 223 how many times 53.) It is contained 4 times; and 4 is written in the quotient. The two figures of the divisor are then multiplied successively by 4, and the product 212 is placed under 223, the partial dividend. After subtraction there is 11 for remainder; by the side of which, 4 the next figure of the dividend is brought down; we then say, as before, in 11 how many times 4? It is contained twice, and 2 is written in the quotient. We then multiply 53 by 2 and write 106, the product, under 114 the partial dividend. After subtraction there remains 8, by the side of which, the last figure 7 in the dividend is brought down; 87 is then divided as in the preceding manner, and 1 found for the quotient and 34 for remainder, which is written in the quotient as directed in article (60.)

63. When the operation is performed with exactness, we seek the number of times each partial dividend contains the divisor, but as this process often becomes tedious, it is sufficient, as has been seen, to seek how often the highest part of the dividend contains the highest part of the divisor. The quotient thus found is not always the true one; because, by taking parts, a near approximation only can be made. But this estimated value does not differ widely from the truth, and the ensuing multiplication serves to correct any error in the quotient. Thus, if the partial dividend really contain the divisor three times; but on making trial in the manner directed, if it is supposed to be contained 4 times; when the quotient is multiplied by 4, the product obtained will be greater than the dividend. And this multiplication detects the error, for the divisor has been taken more times than it is really contained in the dividend, consequently subtraction cannot be made. The quotient therefore must be diminished by one, two, three, &c. units, until the product can be subtracted.

On the contrary, if the divisor assumed

on trial be too small, the remainder after subtraction will be greater than the divisor; which proves that the divisor is contained again in the dividend; and consequently, that the quotient is too small.

A little practice will teach the learner how to augment or diminish the quotient assumed on the first trial.

This operation is commenced by taking the four first figures of the dividend, because the three first do not contain the divisor. We then say, in 18 only, how many times 3? It is in fact contained 6 times; but the divisor multiplied by 6 would give a product greater than 1894 the dividend; then 5 only is written in the quotient. We then multiply 375 by 5, and after subtracting the product from 1894, there is 19 for remainder. By the side of 19, the next figure 9 of the dividend is brought down; but as 199 does not contain $75, 0) is put in thequotient, and by the side of 199, the next figure 2 in the dividend is brought down, and 1992 is the next partial dividend; of which we take the two first figures only and say, in 19 how many times 3? It is contained 6 times; but for the same reason as before, 5 only is written in the quotient. After operating as before, 117 is found for remainder.

64. An observation may be here made, that will prevent many useless trials, which the learner is liable to make when the second figure of the divisor is considerably greater than the first. In this case, instead of seeking how often the first figure of the divisor is contained in the corresponding part of the dividend, it is better to seek how often the first figure augmented by unity, is contained in the corresponding part of the dividend. This

trial will give a quotient nearer approximated to truth than the first.

In this example, instead of saying in 18 how many times 2, we say in 18 how many times 3, because 288 is a number much nearer to 300 than to 200. And 6, found by this operation, is the true quotient; but if the other method had been followed, 9 would have been the quotient, and three useless trials must have been made.

EXAMPLES FOR PRACTICE.

1. Divide 2150596750 by 125. 2. Divide 49561776 by 5137.

Ans. 17204774.

Ans. 9648.

3. The price of a pair of shoes is 2 dollars. How many pair may be had for 56 dollars?

Ans. 28. 4. Fifty-four apples are to be divided equally between two boys. How many must each boy have? Ans. 27. 5. Divide 1030603615 by 3215. Ans. 320561. 6. Divide 4917968967 by 2359. Ans. 20847681 7. Divide 1474 dollars equally among 25 men, 30 women, 17 boys, and 27 girls; after taking out 75 dolars for the payment of a legacy. Ans. $141.

8. If 45 horses were sold in the West Indies for 9900 dollars. What was the average price of each ?

Ans. $220.

9. If a farmer, who has a plantation of 520 acres, buy an adjoining one of 375 acres, and divide the whole into five equal portions; how many acres will there be in each portion?

Ans. 179.

METHOD OF ABRIDGING DIVISION.

65. In order to render division more easy to be understood, the learner has been required to write under each partial dividend, the product found on multiplying