the result at the right hand of the dividend, for the first figure of the quotient.

3. Multiply the divisor by that quotient figure, and place the product under the dividual, for a subtrahend.

4. Subtract the subtrahend from the dividual, and bring down the next figure of the dividend, and annex it to the remainder, for a new dividual. Then find another quotient figure as before; and so proceed until the work is finished.

Note 1.-If it be necessary to bring down more than one figure to any remainder, to make it as large as the divisor, then a cipher must be put in the quotient for every figure so brought down more than one.

Note 2.-When the divisor is a large number, each quotient figure must be found by trial; and, in making these trials, the learner should always bear in mind, that the subtrahend must not be greater than the dividual, nor so much less as to leave a remainder as great as the divisor.

Note 3.-When division is performed by the rule for Case first, it is called Short Division; and when performed by the rule for Case second, Long Division.

and perfect answer to the question; but when there is a remainder, it
may be observed, that it goes so much towards another time as it ap
-proaches the divisor: thus, if the remainder be equal to half the divisor,
it will go half a time more, and so on. In order, therefore, to complete
the quotient, put the last remainder to the end of it, above a line, and
the divisor below it. Hence the origin of vulgar fractions, which are
treated of hereafter. -[The preceding Note has been mostly copied from
Pike's Arithmetic ]
E

In performing the work of example 1st, I find that the first dividual, viz. 33, contains the divisor twice; and therefore I set down 2 for the first quotient figure. I then multiply the divisor by this quotient figure, and the product is 28, which I place under the dividual, for a subtrahend. then subtract the subtrahend from the dividual, and the remainder is 5, to which I annex the next quotient figure, and then I have 57, for a new dividual. This dividual contains the divisor 4 times, and therefore I set down 4 for the second quotient figure. Then I make out a subtrahend, &c. as before, and continue the operation in like manner until I have brought down all the figures of the dividend.

Note. When a figure is brought down from the dividend a dot should be made under it, to show that the figure has been brought down; as in the foregoing example.

4. If 645678 be divided by 13, what will the quotient be? Ans. 496673. 5. If 71640 be divided by 72, what will the quotient be?

Ans. 995.

6. How many times is 422 contained in 253622 ?

Ans. 601 times.

7. How many times is 456 contained in 378480?

Ans. 830.

As learners sometimes find it difficult, when the divisor is a large number, to find the number of times each dividual contains the divisor, I would advise them, in such cases, to

use the following method of finding the quotient figures. After having made out a dividual, begin at the right hand figure of it, and count off as many figures lacking two as there are in the divisor; then find, by trial, how many times the part of the dividual remaining to the left, contains the number expressed by the two left hand figures of the divisor; and the result will be either the required quotient figure, or the next greater figure. Then, by making out a subtrahend, as usual, you may know whether the quotient figure thus found is the right one or not; and if it should prove to be too great, the next less figure will be the one required.

(8)

157496)53277693607(338279 Quot.

472488

602889

472488

1304013

1259968

440456

314992

1254640

1102472

1521687

1417464

104223

In this example, the divisor contains six figures; therefore, in seeking each quotient figure, I reject the four right hand figures of the dividual, and find how many times the number expressed by the remaining figures contains the number expressed by the two left hand figures of the divisor, viz. 15. Thus, the first dividual is 532776; but I reject the four right hand figures, and the remaining part is 53, which contains 15, (the two left hand figures of the divisor,) 3 times. I set down 3 for the first quotient figure; and then by making out a subtrahend as usual, I find that 3

is the right quotient figure. I then make out the 2d dividual; and by rejecting the four right hand figures of it, there remains 60, which contains the two left hand figures of the divisor 4 times. But, by making out a subtrahend, I find that the whole dividual does not contain the whole divisor 4 times; and therefore, instead of 4, I put down in the quotient the next less figure, viz. 3, which proves to be the right quotient figure. In like manner the other quotient figures are found.

CONTRACTIONS IN SIMPLE DIVISION.

I. When there are ciphers at the right hand of the divisor; cut off those ciphers from the divisor, and cut off the same number of figures from the right hand of the dividend; then divide the remaining part of the dividend by the remaining part of the divisor, as usual. If there is a remainder after the division, place the figures thus cut off from the dividend, to the right hand of it, and the whole will be the true remainder; otherwise, the figures cut off will be the remainder.*

*The reason of this method of contraction is easy to conceive; for cuttling off the same number of figures from each, is the same as dividing by 10, or 100, or 1000, &c according to the number of figures cut off; and it is evident, that as often as the whole divisor is contained in the whole dividend, so often must any part of the divisor, be contained in the like part of the dividend. This method is used to avoid a needless repetition of ciphers, which would happen in the common way, as may be seen by working one of the examples in the common way, without cutting off the ciphers.

EXAMPLES.

1. Divide 48205 by 300. 2. Divide 7052708 by 25000.

Note. To divide any number by 10, or 100, or 1000, &c. you have only to cut off as many of the right hand figures of the dividend, as there are ciphers in the divisor; then the remaining part of the dividend will be the quotient, and the figures cut off, the remainder; which remainder being placed over the given divisor, and annexed to the quotient, will make it complete.

So, 764+10=76.

And 1800-100-18.

And 21507+1000-21,5.07.

II. When the divisor exceeds 12, and is the exact product of two or more of the small numbers not greater than 12; then, you may divide the dividend by one of those small numbers, or component parts of the given divisor; and the quotient thence arising, by another of them, and so on to the last; and the last quotient will be the one required.*

* This follows from the second contraction in Multiplication, being only the converse of it; for the half of the third part of any thing, is evidently the same as the sixth part of the whole; and so of any other numbers. The reason of