ply the multiplicand separately by the parts of the multiplier, and add the products, we will have the same result; thus, 30684 40 30684 3 92052 1227360 Here you will perceive we multiply by the parts of our first multiplier; first our multiplicand by 3, and then our multiplicand by 40, which is the local value of the left hand figure of our multiplier; and multiplying the multiplicand by 4, increases it 4 times; then bringing a cipher at the right, increases it 10 times; which evidently makes 40 times repeated. Then, when we add the products, we obtain the same result; but in our first example, giving the product of our left hand figure its local place, was the same in effect. Had we placed the product of the left hand figure of our multiplier, directly under the product of the first, and then have added, it would have only been repeating the multiplicand 7 times; for when we multiply by 3, (the first figure of the multiplier,) we repeat the multiplicand 3 times; then if we multiply by 4, placing the product directly under the product of the first figure of the multiplier, we repeat it 4 times; and 3 times repeated, and 4 times repeated, make 7 times repeated; as the following work shows, Thus, you see the same result is produced. 1 3 6 4 in each bale 8. A merchant bought 481 bales of linen; there were 36 pieces; and in each piece, 24 yards; how many pieces, and how many yards were there? Ans. 17316 pieces, 415584 yards. 9. Three hundred and forty seven men shared equally a prize, and each received forty nine dollars; what was the whole prize? Ans. $17003. 10. If four bushels of wheat make a barrel of flour, and the price of wheat, one dollar a bushel; what will 375 barrels come to ? Ans. $1500. CASE III-When ciphers occur between the significant figures of the multiplier, omit them in the operation. RULE.-Place the product of each significant figure, directly under that, by which you multiply; then add the products together, and their sum will be the total product. EXAMPLES. 1st. Multiply 384 by 203. Here, when we multiply by 2, 38 4 203 1152 768 7 7 9 5 2 Ans. 2. 3604 the left hand figure of the multiplier, we commence the product directly under itself, which brings it in the place of hundreds; and it should stand in the place of hundreds; for it is repeating the multiplicand by 200: the multiplying figure, (2,) standing in the place of hundreds. 102 4005 3568059 7 8 8 6 4 Ans. 6 2768 087412 Ans. 7. What is the product of 365432 multiplied by 7608? Ans. 2780206656. 8. Multiply eight thousand five hundred and sixteen, by seventy-six thousand and two. 9. Multiply 120345 by 9004. 10. What is the product of 24393 multiplied by 402? Ans. 647233032. Ans. 1083586380. Ans. 9805986. CASE IV. When one or both of the factors have ciphers at the right hand. RULE.-Multiply the significant figures the same as if there were no ciphers at the right, and add their products; then, join to the right hand of the total product of the significant figures, as many ciphers as there are at the right hand of both the factors. EXAMPLES. 1. Multiply 4 6 3 by 4 0 40 18520 DEM.-Here, we first multiply by 4, which repeats the multiplicand 4 times; but we wish to have our multiplicand repeated forty times, and to effect this, we bring the cipher of our multiplier at the right of the product, which gives the product of the 4 its local value, which is ten times more than its simple value; and ten times the simple value of 4, is forty; consequently we have our multiplicand forty times repeated. 2. Multiply 3460 by 4200. 3460 4200 692 1384 14532000. DEM.-Here, we first multiply our significant figures, as our rule directs. And since we multiply only by 42, the hundredth part of our multiplier, our product is only a hundredth part of what it should be; we then increase it a hundred times, by joining two ciphers at the right hand of our product. We now have repeated the significant figures of our multiplicand 4200 times; but still, our product is only one tenth of what it should be; because we have only multiplied a tenth part of our multiplicand; then to give our product its proper value, we increase it ten times, by joining another cipher at the right hand ; which is for the cipher at the right of our multiplicand. 3. Multiply 21200 by 70. Ans. 1484000. Ans. 4560020000. how many shil 4. Multiply 340300 by 13400. 5. Twenty shillings make one pound: lings in four hundred and fifty pounds? Ans. 9000 s. 6. In eighty pounds, how many shillings? 7. One hundred and sixty square rods make one acre; Ans. 51200 rods., 8. Six hundred and 40 acres make one square mile; how square miles ? how many square rods in 320 acres? many acres in three hundred 9. Multiply 480000 by 12000. Ans. 192000 acres. Ans. 5760000000. CASE V. To multiply by 10, 100, 1000, &c. RULE. Add as many ciphers to your multiplicand, as there are ciphers in the multiplier; and the multiplying is performed. D EXAMPLES. 1. Multiply 146 by 10. Ans. 1460. DEM.-We join a cipher to the multiplicand, and our work is done, because it increases the figures of our multiplicand ten times, by giving units the place of tens; and tens, the place of hundreds; and hundreds, the place of thousands; &c. 2. Multiply 846 by 100. Ans 84600. DEM.-Here, we join two ciphers to the multiplicand; and the figures of our multiplicand are increased 100 times because 6, the first figure of our multiplicand, after the ciphers are joined, becomes hundreds; and the next left hand figure 4, becomes thousands, one hundred times the first value. 3. Multiply 8324 by 1000. 4. Multiply 30460 by 10000. Ans. 8324000. Ans. 304600000. CASE VI. To multiply by 9, 99, 999, &c. RULE. Join to the right of the multiplicand, as many ciphers, as your multiplier contains nines; and from the sum produced, subtract the multiplicand. Or you may simply multiply the multiplicand by the number of nines in the multiplier. The first method, frequently shortens the work. EXAMPLES. 1. Multiply 464 by 9. DEM.-Joining a cipher to the 4640 right of our multiplicand, is repeating our multipli464 cand ten times, and by subtracting our multiplicand once, must evidently leave it nine times repeated; as appears, from the same example multiplied by 9. 4 6 4 9 Ans 4176 DEM.-Here, it is plain, by joining two ciphers to the multiplicand, we increase it 100 times; and our multiplicand being once taken away, or subtracted, must leave it 99 times repeated. 3. Multiply 3472 by 999. 4. Multiply 34672 by 9999.. Ans. 3468528. Ans. 346685328. CASE VII. To multiply by a composite number; that is, when the multiplier can be produced by the multiplying of any two figures in the multiplication table. RULE.-Multiply the multiplicand first by one of the figures, and that product by the other, and this last product will be the total product required. EXAMPLES. 1. Multiply 243 by 35. 24 3 7 17 0 1 5 Ans. 8 5 0 5 In this example, 7 and 5 are called the component parts of 35; because 7 times 5 are 33; and it will make no difference which we multiply by first. DEM. We first multiply by 7, which repeats our multiplicand 7 times; and then we multiply this product by 5, which is repeating, 5 times, a number 7 times as great as our first multiplicand, consequently, our multiplicand is 35 times repeated, because 5 times 7 are 35. Our work perhaps may appear more clear to the young student, by multiplying the same multiplicand directly by 35; thus, the same result, you perceive, is produced. 3 5 243 NOTE. This method of work, will not be found of much service, in simple multiplication; but it is sometimes found very convenient in multiplying a compound quantity. 1 2 1 5 729 8.5 05 Ans. 17280. 2. Multiply 480 by 36. 3. Multiply 1324 by 45. Ans. 59580. Ans. 265248. Ans. 253044.. Ans. 445248. Ans. 148960. 4. Multiply 3684 by 72. 5. Multiply 3124 by 81. 6. Multiply 4638 by 96. 7. Multiply 3040 by 49. PROMISCUOUS EXAMPLES. 1. What will a year's board come to, at $2 a week; allowing fifty two weeks to the year? Ans. $104. 2. Suppose 80 seamen were concerned in taking a prize at sea, and on dividing their prize money, each seaman received $150; what was the amount of the prize? Ans. $12000. 3. A gentleman bought 307 horses for shipping, at the rate of $105 each; what did he pay for the whole? Ans. $32235. 4. A gentleman dividing his fortune among seven sons, found that the portion of each, was $1050; what was the gentleman's fortune? Ans. $7350. 5. There are 320 rods in a mile; now suppose the distance between the United States and Europe, to be 3000 miles ; |