contain the divisor. Divide this remainder (thus in creased) in the same manner as before; and proceed in this manner until all the figures in the dividend are brought down and used.

PROOF.

Multiply the quotient by the divisor, and to the product add the last remainder, if there be any; if the work is right, the sum will be equal to the dividend.

29

27

26

24

27

27

18

18

In this example, I find that 3, the divisor, can not be contained in the first figure of the dividend; therefore I take two figures, viz: 14, and inquire how often 3 is contained therein,which! I find to be 4 times, and put 4 in the quotient.Then multiplying the divisor by it, I set the product under the 14, in the dividend, and find by subtracting that there is a remainder of two. To this 2, I bring down the next figure in the dividend, viz: 3, which increases the remainder to 23. I then seek how often 3 is contained in 23, and proceed as before. When I bring down the one that is in the dividend, I find that three can not be contained in it, and therefore place a cypher in the quotient and bring down the 8, which makes 18. Finding that 3 is contained 6 times in 18, and that there is no

Remainder.

remainder, I bring down the 2; but as 3 can not be con tained in it, I place a cypher in the quotient, and let 2 stand as the last remainder. In proving the sum by Multiplication, the 2 is added. This mode of operation is called LONG DIVISION.

1. Divide 87654 by 58 Quo. 1511 Rem. 16

2.

456789 679

672

501

3.

3875642

7898

490

5622

Note. When there is one cypher, or more, at the right hand of the divisor, it may be cut off; but when this is done, the same number of figures must be cut off from the right hand of the dividend; and the figures thus cut off, must be placed at the right hand of the remainder.

Note. In dividing by 10, 100, or 1000, &c.. when you cut off as many figures from the dividend as there are cyphers in the divisor, the sum is done; for the figures cut off at the right hand are the remainder, and those at the left are the quotient, as in the following

When the divisor does not exceed 12, seek how often it is contained in the first figure or figures of the dividend, and place the result in the quotient. Then multiply in your mind the divisor by the figure placed in the quotient, subtract the product from the figure under which it would properly stand in the former case of division and conceive the remainder, if there be any, to be prefixed to the next figure. See how often the divisor is contained in these, and proceed, as before, till

the whole is divided. This operation is called SHORT DIVISION.

EXAMPLES.

1.

4) 987654321

II.

8) 123456789

Quo. 2 4 6 913580-1 Quo. 154 3 2 0 9 8-5

In the first example, I find that 4 is contained twice in 9, and that 1 remains. The 1, I conceive as prefixed to the next figure, which is 8, and they become 18. In 18, I find 4 is contained 4 times, and 2 remain. By prefixing the 2 to the following figure, which is 7, they make 27. In this manner I proceed, setting the result of each calculation in the lower line which is the quotient. In the second example, as 8 can not be contained in 1, take two figures, and proceed as in the first.

Note. When the divisor is of such a number that two figures being multiplied together will produce it, divide the dividend by one of those figures, the quotient thence arising, by the other figure, and it will give the quotient required. As it sometimes happens that there is a remainder to each of the quotients, and neither of them the true one, it may be found thus:-Multiply the first divisor by the last remainder, and to the product add the first remainder, which will give the true one.