If, for instance, it were required to divide 64 into 4 equal parts; to find the value of these parts, it would be necessary to ascertain the number, that is contained 4 times in 64, and consequently to regard 64 as a product, having for its factors 4 and one of the required parts, which is here 16.

If it were asked how many parts, of 16 each, 64 is composed of, it would be necessary, in order to ascertain the number of these parts, to find how many times 64 contains 16, and consequently, 64 must be regarded as a product, of which one of the factors is 16, and the other the number sought, which is 4.

Whatever then may be the object in view, division consists in finding one of the factors of a given product, when the other is known.

37. The number to be divided is called the dividend, the factor, that is known, and by which we must divide, is called the divisor, the factor found by the division is called the quotient, and always shows how many times the divisor is contained in the dividend.

It follows then, from what has been said, that the divisor multiplied by the quotient ought to reproducé the dividend.

38. When the dividend can contain the divisor a great many times, it would be inconvenient in practice to make use of repeated subtraction for finding the quotient; it then becomes necessary to have recourse to an abbreviation analogous to that which is given for multiplication. If the dividend is not ten times larger than the divisor, which may be easily perceived by the inspection of the numbers, and if the divisor consists of only one figure, the quotient may be found by the table of Pythagoras, since that contains all the products of factors that consist of only one figure each. If it were asked, for instance, how many times 8 is contained in 56, it would be necessary to go down the 8th column, to the line in which 56 is found; the figure 7, at the beginning of this line, shows the second factor of the number 56, or how many times 8 is contained in this number.

We see by the same table, that there are numbers, which cannot be exactly divided by others. For instance, as the seventh line, which contains all the multiples of 7, has not 40 in it, it

follows that 40 is not divisible by 7; but as it comes between 35 and 42, we see that the greatest multiple of 7, it can contain, is 35, the factors of which are 5 and 7. By means of this elementary information, and the considerations, which will now be offered, any division whatever may be performed.

59. Let it be required, for example, to divide 1656 by 3; this question may be changed into another form, namely; To find such a number, that multiplying its units, tens, hundreds, &c. by 3, the product of these units, tens, hundreds, &c. may be the dvvidend, 1656.

It is plain, that this number will not have units of a higher order than thousands, for, if it had tens of thousands, there would be tens of thousands in the product, which is not the case. Neither can it have units of as high an order as thousands, for if it had but one of this order, the product would contain at least 3, which is not the case. It appears then, that the thousand in the dividend is a number reserved, when the hundreds of the quotient were multiplied by 3, the divisor.

This premised, the figure occupying the place of hundreds, in the required quotient, ought to be such, that, when multiplied by 3, its product may be 16, or the greatest multiple of 3 less than 16. This restriction is necessary, on account of the reserved numbers, which the other figures of the quotient may furnish, when multiplied by the divisor, and which should be united to the product of the hundreds.

The number, which fulfils this condition, is 5; but 5 hundreds, multiplied by 3, gives 15 hundreds, and the dividend, 1656, contains 16 hundreds; the difference, 1 hundred, must have come then from the reserved number, arising from the multiplication of the other figures of the quotient by the divisor. If we now subtract the partial product, 15 hundreds, or 1500, from the total product, 1656, the remainder, 156, will contain the product of the units and tens of the quotient by the divisor, and the question will be reduced to finding a number, which, multiplied by 3, gives 156, a question similar to that, which presented itself above. Thus when the first figure of the quotient shall have been found in this last question, as it was in the first, let it be multiplied by the divisor, then subtracting this partial product from the whole

product, the result will be a new dividend, which may be treated in the same manner as the preceding, and so on, until the original dividend is exhausted.

40. The operation just described is disposed of thus ;

The dividend and divisor are separated by a line, and another line is drawn under the divisor, to mark the place of the quotient. This being done, we take on the left of the dividend the part 16, capable of containing the divisor, 3, and dividing it by this number, we get 5 for the first figure of the quotient on the left; then taking the product of the divisor by the number just found, and subtracting it from 16, the partial dividend, we write, underneath, the remainder, 1, by the side of which we bring down the 5 tens of the dividend. Considering the number, as it now stands, a second partial dividend, we divide it also by the divisor, 3, and obtain 5 for the second figure of the quotient; we then take the product of this number by the divisor, and subtracting it from the partial dividend, get 0 for the remainder. We then bring down the last figure of the dividend, 6, and divide, this third partial dividend by the divisor, 3, and get 2 for the last figure of the quotient.

41. It is manifest that, if we find a partial dividend, which cannot contain the divisor, it must be because the quotient has no units of the order of that dividend, and that those which it contains arise from the products of the divisor by the units of the lower orders in the quotient; it is necessary, therefore, whenever this is the case, to put a 0 in the quotient, to occupy the place of the order of units that is wanting.

The division of the 15 hundreds of the dividend, by the divisor, leaving no remainder, the 3 tens, which form the second partial dividend, do not contain the divisor. Hence it appears, that the quotient ought to have no tens; consequently this place must be filled with a cipher, in order to give to the first figure of the quotient the value, it ought to have, compared with the others; then bringing down the last figure of the dividend, we form a third partial dividend, which, divided by 5, gives 7 for the units of the quotient, the whole of which is now 307.

42. The considerations, presented in article 40, apply equally to the case, in which the divisor consists of any number of figures.

If, for instance, it were required to divide 57981 by 251, it would easily be seen, that the quotient can have no figures of a higher order than hundreds, because, if it had thousands, the dividend would contain hundreds of thousands, which is not the case; further, the number of hundreds should be such, that, multiplied by 251, the product would be 579, or the multiple of 251 next less than 579; this restriction is necessary on account of the reserved numbers which may have been furnished by the multiplication of the other figures of the quotient by the divisor. The number, which answers to this condition, is 2; but 2 hundreds, multiplied by 251, give 502 hundreds, and the divisor contains 579; the difference, 77 hundreds, arises from the reserved numbers resulting from the multiplication of the units and tens of the quotient, by the divisor.

If we now subtract the partial product, 502 hundreds, or 50200, from the total product, 57981, the remainder, 7781, will contain the products of the units and tens of the quotient by the divisor,

and the operation will be reduced to finding a number, which, multiplied by 251, will give for a product 7781.

Thus, when the first figure of the quotient shall have been determined, it must be multiplied by the divisor, the product being subtracted from the whole dividend, a new dividend will be the result, which must be operated upon like the preceding; and so on, till the whole dividend is exhausted.

It is always necessary, for obtaining the first figure of the quotient, to separate, on the left of the dividend, so many figures, as, considered as simple units, will contain the divisor, and admit of this partial division.

43. Disposing of the operation as before, the calculation, just explained, is performed in the following order;

The 3 first figures, on the left of the dividend, are taken to form the partial dividend; they are divided by the divisor, and the number 2, thence resulting, is written in the quotient; the divisor is then multiplied by this number, and the product, 502, is written under the partial dividend, 579. Subtraction being performed, the 8 tens of the dividend are brought down to the side of the remainder, 77; this new partial dividend is then divided by the divisor, and 3 is obtained for the second figure of the quotient; the divisor is multiplied by this, the product subtracted from the corresponding partial dividend, and to the remainder, 25, is brought down the last figure of the dividend, 1; this last partial dividend, 251, being equal to the divisor, gives 1 for the units of the quotient.

44. When the divisor contains many figures, some difficulty may be found in ascertaining how many times it is contained in