place it in the quotient, and multiply the divisor by it, and place the product exactly under the figures of the dividend that you sought how often your divisor was contained in; then, draw a line underneath it, and subtract it from said figures; (here you must observe this last product; for, it must not be greater than the part of the dividend made use of; likewise, your remainder must not be greater than your divisor. If either be as here stated, the work cannot be right, you must observe this through the operation;) now, bring down the next figure in the dividend to the right of the remainder; after which, you must seek how often the divisor is contained in this number; place the number thus found in the quotient; then multiply and subtract as before; after this, bring down the next figure; so proceed on till all the figures of the dividend are brought down. Note, when it happens that the divisor is more than so many figures of the dividend, you must take one more figure in the dividend than you have in the divisor, and proceed as before, seeking how often the first figure of the divisor is contained in the two first figures of the dividend, making allowance for what you carry from the right hand. Also, you must mind that if in any case, the remainder be so small, that when the figure of the dividend is joined, or brought down to it, that it is still a less sum than the divisor; then a cipher is to be placed in the quotient and another figure brought down; if it be still less than the divisor, place another cipher in the quotient, and bring down another figure; so continue to do, till the number is as great, or greater, than the divisor, and then proceed as before.

PROOF. Observe the same as given under Short Division.

first figure of the divisor 2 was contained in 16, the first two figures of the dividend, which I found (after allowance was made for what I carried,) to be 6 times; this I placed in the quotient, and then multiplied the divisor by it; saying 6 times 4 are 24, set down 4 and carry 2; then 6 times 2 are 12; and 2 that I carry are 14; this, you will see comes under the 163, the three left hand figures of the dividend. Here I noted whether the under figures (that is the 144,) were greater than the figures immediately over them; that is the 163, which were not; I then proceeded to subtract the 144 from the 163; saying, 4 from 3 I cannot take, 4 from 13, nine; set down 9, and carry 1 to 4 are 5, then 5 from 6, one; then 1 from 1, nothing; here I had 19 remainder, which being less than the divisor, I proceeded to bring down the next figure of the dividend which was 4, and annexed it to the last remainder 19; this made 194; again, I said, how often is 24 contained in 194? I found by the same operation as before that it was 8 times; I placed this in the quotient to the right of the last figure, and proceeded to multiply the divisor by it; saying, 8 times 4 are 32, set down 2 and carry 3; then, 8 times 2 are 16 and 3 are 19; having set this down, I had 192 to subtract from 194, which I did by saying, 2 from 4, two; 9 from 9, nothing; 1 from 1, nothing; I now brought down the next figure of the dividend, and annexed it to 2, the last remainder and went on, saying, how often is 24, the divisor, contained in 21? I found that it was not contained in it; so I set a cipher in the quotient, and brought down the next figure of the dividend, annexing it to the last brought down, and proceeded to ascertain how many times 24 were contained in 219; which I found to be 9 times; this I placed in the quotient, saying 9 times 4 are 36; set down 6 and carry 3, nine times 2 are 18 and 3 are 21; here, I had 216 to subtract from 219; I went on, saying, 6 from 9, three; then, 1 from 1, nothing; 2 from 2, nothing; here, I had 3 remainder; I now brought down the last figure of the dividend, and annexed it to this remainder as before, and sought how often 24 was contained in 35, which being once, I set this in the quotient, and said once 4 is 4; once 2 is 2; I subtracted this, saying 4 from 5, one; two from 3, one; here, I had 11 for my last remainder; as all the figures of the dividend had been brought down, the sum was done.

I proceeded to prove it by multiplying the quotient by the divisor, taking care to add the last remainder to the product. I found that this product exactly corresponded with the dividend, so I concluded that the work was right.

It will be best for the learner to work the last three examples at length; by so doing, and strictly examining the rule, he may find the quotient or answer to the following questions.

4 Divide 1798467 by 24, Answer 74936 and 3 Remainder. Divisor. Dividend. Quotient.

To be placed thus, 24)1798467(74936 & 3 Remainder.

When the divisor has a cipher or ciphers on the right hand.

RULE.

Cut off the cipher or ciphers in the divisor, and cut off as many figures on the right hand of the dividend; divide as before figures of the divisor that are left when the ciphers when the division is ended, the figures that were

cut off in the dividend must be placed to the right of the remainder, (if any,) and the figures cut off in the divisor restored to their own places again, as in the

EXAMPLE FOR ILLUSTRATION.

62|000)127854|786(2062 Here, according to rule, I cut off

124

385

372

134

124

10786

62000

the ciphers in the divisor; and, then, cut off as many figures on the right of the dividend; you will see that I divided by 62; after I had brought down all the figures in the dividend (rejecting those cut off,) I placed those cut off to the right of the last remainder, under which I placed the divisor, remembering to restore the ciphers cut off. This operation brings the same answer and remainder that would have been brought had I divided with

out cutting off, but this method makes the work shorter.

When the divisor is such a number that any two figures in the multiplication table being multiplied together will produce said divisor.

RULE.

Divide the given dividend by one of those figures, and that quotient by the other, the last division will give the quotient required.

NOTE--The total remainder (or the remainder that would have been brought by dividing by the whole number at once,) is found by multiplying the last remainder by the first divisor, and adding in the first remainder.

EXAMPLE FOR ILLUSTRATION.

Divide 69483 by 48.

6)69483

To find the quotient or answer to this, I sought what two numbers being multiplied together would

8)11580-3 make, or produce 48; by the table I found that 6

times 8 produced 48; so, according to rule, I divi1447-4 ded first by 6, and then by 8; here you will see, that when I divided by 6 I had 3 remainder, and when by 8, I had 4; and, to find the total remainder I proceeded according to note under the rule; thus I multiplied 4, the last remainder, by 6 the first divisor; this produced 24, which I added to 3 the first remainder; this brought 27 the total remainder.

EXAMPLES.

2 Divide 897456 by 56 Ans. 16026 and 0 Total Remainder. You will find by the table that 7 times 8 are 56.

3 Divide 548369 by 35 Ans. 15667 and 24 Total Rem.

See that you endeavour to understand every Rule; read it over with care, and try to get the meaning of it; by so doing you will get along with less difficulty.

COMPOUND ADDITION teacheth how to add several sums or numbers together, having different names or denominations; as pounds, shillings, and pence; dollars and cents; tons, hundreds, quarters; &c.

RULE.

1 Place the numbers so that those of the same name or denomination may stand directly under each other; that is, place pounds under pounds; shillings under shillings; pence under pence; farthings under farthings, &c.

2 Begin to add at the lowest name or denomination; that is, begin at the right hand column, as in whole numbers, and after you have added it, divide the amount of the column by as many of the same name as makes one of the next greater, setting do the remainder under the column added and carry the

the next greater or superior denomination; then unt of that, and divide it by as many of the same