The four significant figures of this product result from the multiplication of 764 by 3, and are placed two places towards the left to admit the two ciphers, which terminate the multiplier. In general when the multiplier is terminated by a number of ciphers, first multiply the multiplicand by the significant figure of the multiplier, and place, after the product, as many ciphers as there are in the multiplier. Multiply Examples. 35012 by 100. Ans. 3501200. 638427 by 500. Ans. 319213500. 2107900 by 70. Ans. 147553000. 9120400 by 90.Ans. 820836000. 33. The preceding rules apply to the case, in which the mul tiplier is any number whatever, by considering separately each of the collections of units of which it is composed. To multiply, for instance, 793 by 345, or, which is the same thing, to repeat 793, 345 times, is to take 793, 5 times, added to 40 times, added to 300 times, and the operation to be performed is resolved into 3 others, in each of which the multipliers, 5, 40, and 300, have but one significant figure. To add the result of these three operations easily, the calculation is disposed thus; 793 345 3965 31720 237900 273585 The multiplicand is multiplied successively by the units, tens, hundreds, &c. of the multiplier, observing to place a cipher on the right of the partial product, given by the tens in the multiplier, and two on the right of the product given by hundreds, which advances the first of these products one place towards the left, and the second, two. The three partial products are then added together, to obtain the total product of the given numbers. As the ciphers placed at the end of these partial products, are of no value in the addition, we may dispense with writing them, provided we take care to put in its proper place the first figure of the product given by each significant figure of the multiplier; that is, to put in the place of tens the first figure of the product given by the tens in the multiplier; in the place of hundreds the first figure of the product given by the hundreds in the multiplier, and so on. 34. According to what has been said, the rule is as follows. To multiply any two numbers, one by the other, form successively (according to the rule in article 30,) the products of the multiplicand, by the different orders of units in the multiplier; observing to place the first figure of each partial product under the units of the same order with the figure of the multiplier, by which the product is given; and then add together all the partial products. 35. When the multiplicand is terminated by ciphers, they may at first be neglected, and all the partial multiplications begin with the first significant figure of the multiplicand; but afterwards, to put in their proper rank the figures of the total product, as many ciphers, as there are in the multiplicand, must be written on the right of this product. If the multiplier is terminated by ciphers, we may, according to the remark in article 31, neglect these also, provided we write an equal number on the right of the product. Hence it results that, when both multiplicand and multiplier are terminated by ciphers, these ciphers may at first be neglected, and after the other figures of the product are obtained, the same number may be written on the right of the product.) When there are ciphers between the significant figures of the multiplier, as they give no product, they may be passed over, observing to put in its proper place the unit of the product, given by the figure on the left of these ciphers. 526 Multiply 9648 by 5137. Ans. 49561776. 300 40 307 36. THE product of two numbers being formed by repeating one of these numbers as many times as there are units in the other, we can, from the product, find one of the factors, by ascertaining how many times it contains the other; subtraction alone is necessary for this. Thus, if it be required to ascertain the number of times 64 contains 16, we need only subtract 16 from 64 as many times as it can be done; and since, after 4 subtractions, nothing is left, we conclude, that 16 is contained 4 times in 64. This manner of decomposing one number by another, in order to know how many times the last is contained in the first, is called division, because it serves to divide, or portion out, a given number into equal parts, of which the number or value is given. 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 must reproduce 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. 39. 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 dividend, 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. Arith. 4 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. 16. 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 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 ; |