We now pass to the multiplication of numbers expressed by many figures. MULTIPLICATION OF A NUMBER EXPRESSED IN MANY FIGURES BY A SINGLE FIGURE. 5. Write the multiplier, which in this case is supposed to ba single figure, under the unit figure of the multiplicand. Multiply the units of the multiplicand by the multiplier, and if the product be units, write it underneath; but if the product contain tens, write only the units underneath, and reserve the tens to be added, as so many units, with the product of the next multiplication. In the same manner, multiply the tens of the multiplicand, and to their product add the tens, if any, which were reserved from the multiplication of the units. Write under the place of tens in the multiplicand, only the units of this new product; and reserve the tens of it (which are hundreds) to be added as units to the product of the next multiplication. Continue to multiply according to the same rule, all the figures of the multiplicand; the series of figures written underneath will be the product. EXAMPLE. How many feet are there in 2864 fathoms? The fathom is 6 feet. And the question requires that 6 feet should be repeated 2864 times; or, (44) that 2864 feet should be taken 6 times. We therefore write 2864 for the multiplicand; 6 for the multiplier. and 17184 is the product. To perform this operation, commence with the units by saying, 6 times 4 make 24; write 4 under the place of units, and retain 2 for the two tens. 2dly say, 6 times 6 make 36; and 2 which were retained, make 38; put down 8 and retain 3. 3dly, 6 times 8 make 48, and 3 which were retained, make 51; put down 1 and retain 5. 4thly, 6 times 2 make 12, and 5 which were retained, make 17. This number is written down because the multiplication is finished. The number 17184 is the product required, or the number of feet in 2864 fathoins; because it contains the 4 units 6 times, the 6 tens 6 times, the 8 hundreds 6 times, and the 2 thousands 6 times, consequently 6 times the number 2864. MULTIPLICATION BY A NUMBER CONSISTING OF 51. When the multiplier is expressed by several figures, the same operation which has already been explained, is performed successively with each of its figures, commencing always at the right. Thus, all the figures of the multiplicand are first multiplied by the unit figure of the multiplier, then by the figure in the place of tens, and this second product is written under the first; but as the product of units multiplied by tens is tens, (44) the first figure of the second product is written under the place of tens in the first, in order to express its true value. The third product, which is obtained by multiplying by hundreds, is placed in like manner under the second, but advanced one place towards the left, because the product of units by hundreds is hundreds. The same rule is followed for all other figures in the multiplier. After multiplication is performed by all the figures in the multiplier, the several products are added together, and their sum is the whole product. 455658546 product. We first multiply 65487 by 8, the unit figure of the multiplier, and write successively under the line which was drawn below the multiplier, 593896, the figures of the product found by the rule given for the first case (50). The number 65487 is multiplied in like manner by 5, the second figure of the multiplier; and the product $27435 is written under the first product, but having its first figure 5 under the tens of the first product. Multiplying 65487 by the third figure 9, the product, 389383 is written under the preceding, but having its first figure 3 put in the place of hundreds, because 9, the multiplying figure, has the value of hundreds. Finally 65487 is multiplied by 6 the last figure of the multiplier, and 392922, the product, is written under the preceding product, but with its first figure in the place of thousands; because the multiplying figure expresses thousands. These several products are then added together, and 455658546 is the product of 65487 multiplied by 6958, or the value of 65487 taken 6958 times. Because the number 65487 was taken 8 times by the first operation, 50 times by the second, 900 times by the third, and 6000 times by the fourth. EXAMPLES FOR PRACTICE. 1. Multiply 847483567 by 768. Ans. 374387379456. 2. Multiply 264648436 by 3639604. Ans. 968215506259844. 3. What will 587 firkins of butter come to at 7 dollars per firkin? Ans. 4771 dollars. 4. What will be the worth of 924 tons of potash, if 1 ton sell for 95 dollars? Ans. 87780 dollars. 5. There is an orchard containing 9 rows of trees, and there are 57 trees in each row; how many trees are there in the orchard? Ans. 513. 6. A man bought 9 pieces of broadcloth, each piece containing 47 yards, at 6 dollars a yard; and 25 barrels of flour at 7 dollars a barrel; what did he give for the whole ? Aus. 2713 dolls. 52. When the multiplicand or the multiplier or both, are terminated by ciphers, the operation may be abridged by neglecting the ciphers in the multiplication, but placing all of them afterwards at the right of the product. EXAMPLE. It is proposed to multiply 6500 by 350 325 195 2275000 This operation is performed by neglecting the ciphers, multiplying 65 by 35, and placing at the right of their product as many ciphers as were contained in both the factors. The multiplicand, 6500, truly represents the value of 65 hundreds; therefore when 65 is multiplied, the product is understood to be hundreds. In like manner, the multiplier, 350, denotes 35 tens; when multiplication is made by 35, the product will be understood to be tens; the whole product will then have the value of tens of hundreds, that is, of thousands; three ciphers therefore must be added to the right. This reasoning applies to all cases, in which one, or both factors are terminated by ciphers. 53. When ciphers are found between the figures of the multiplier, we may dispense with writing them in the product, and pass to the next figure. But the product, which arises from multiplication by this next figure, must be advanced towards the left one place more than the number of ciphers in the multiplier, that is, two places if there is one cipher; three if there are two; and so on. 126408312 is the product. In this operation, after having multiplied by 6, and written down the product 252312, we next multiply by 3, because multiplication by a cipher gives only ciphers for the product; but the product 126156 must be written in such a manner as to express thousands; its first figure on the right will therefore be removed three places towards the left, that is, one place more than the number of ciphers interposed between the figures of the multiplier. EXAMPLES POR PRACTICE IN THE TWO LAST CASES. 1. Multiply 35012 by 100. Ans. 3501200. Ans. 319213500. Ans. 147553000. Ans. 820836000. Ans. 27489000. Ans. 32500000. MULTIPLICATION OF DECIMAL PARTS. 54. The same rule is observed for the multiplication decimal parts as for whole numbers. In the operaon the separatrix is neglected, but in the product as many figures are pointed off on the right, as there are decimal places in both the factors. It is proposed to multiply 54,23 by 8,3 16269 43384 450,109 We multiply 5433 by 83 and the product is 450109, but as there are two decimal places in the multiplicand and one in the multiplier, three figures are separated upon the right of the product, which becomes 450,109, such as it should be. The reason for this rule is easily understood by observing, that if the multiplier were 83, the product in decimals would be hundredths, because the multiplicand 54,23 would have been repeated 83 times, the decimals of which are hundredths. But as the multiplier is 8, 3, that is, (21,) ten times smaller than 83, the product must contain units ten times smaller than hundredths; the last figure in the product must then be thousandths. And there must be three decimal figures in the product, D |