CASE III. § 25. When the multiplier is 1 and any number of ciphers after it, as 10, 100, 1000, &c. Placing a cipher on the right of a number changes the units place into tens, the tens into hundreds, the hundreds into thousands, &c., and therefore increases the number ten times. Thus, 5 is increased ten times by making it 50. So the addition of two ciphers increases a number one hundred times; the addition of three ciphers, a thousand times, &c. Thus, 6 is increased a hundred times by making it 600, and 5 is increased a thousand times, by making it 5000. Hence, we have the following RULE. Place on the right of the multiplicand as many ciphers as there are in the multiplier, and the number so formed will be the required product. Q. If you place one cipher on the right of a number, what effect has it on its value? If you place two, what effect has it? If you place three? And for any number of ciphers, how much will each increase it? How do you multiply by 10, 100, 1000, &c.? § 26. When there are ciphers on the right hand of one or both of the factors. RULE. Neglect the ciphers and multiply the significant figures: then place as many ciphers to the right hand of the product, as there are in both of the factors. Q. When there are ciphers on the right hand of both the factors, how do you multiply? § 27. When the multiplier is a composite number. A composite number is one that may be produced by the multiplication of two or more numbers, which are called the components or factors. Thus, 2×3=6. Here 6 is the composite number, and 2 and 3 are the factors, or components. The number 16-8×2: here 16 is a composite number, and 8 and 2 are the factors; and since 4x4-16, we may also regard 4 and 4 as factors or components of 16. Q. What is a composite number? Is 6 a composite number? What are its components or factors? What are the factors of the composite number 16? What are the factors of the composite number 12 ? EXAMPLES. 1. Let it be required to multiply 8 by the composite number 6, in which the factors are 2 and 3. If we write 6 horizontal lines with 8 units in each, it is evident that the product of 8×6=48, the number of units in all the lines. But let us first connect the lines in sets of 2 each, as on the right; there will then be in each set 8x2=16; or 16 units in each set. But there are 3 sets; hence, 16x3=48, the number of units in all the sets. If we divide the lines into sets of 3 each, as on the left, the number of units in each set will be equal to 8x3=24, and there being 2 sets, 24×2=48, the whole number of units. As the same may be shown for all numbers we have the following RULE. When the multiplier is a composite number, multiply by each of the factors in succession, and the last product will be the entire product sought. EXAMPLES. 1. Multiply 327 by 12. The factors of 12, are 2 and 6, or they are 3 and 4, or they are 3, 2 and 2: for, 2×6=12, 3×4=12, and 2. Multiply 5709 by 48; the factors being 8 and 6, or 16 and 3. 3. Multiply 342516 by 56. 4. Multiply 209402 by 72. 5. Multiply 937387 by 54. 6. Multiply 91738 by 81. 7. Multiply 3842 by 144. APPLICATIONS. Ans. Ans. 19180896. Ans. 50618898. Ans. 553248. 1. There are ten bags of coffee, each containing 48 pounds: how much coffee is there in all the bags ? Ans. lbs. 2. There are 20 pieces of cloth each containing 37 yards; and 49 other pieces, each containing 75 yards: how many yards of cloth are there in all the pieces? Ans. 4415 yards. 3. There are 24 hours in a day, and 7 days in a week: many hours in a week? how Ans. hours. 4. A merchant buys a piece of cloth containing 97 yards, at 3 dollars a yard: what does the piece cost him? Ans. dollars. 5. A farmer bought a farm containing 10 fields; three of the fields contained 9 acres each; three other of the fields 12 acres each; and the remaining 4 fields, each 15 acres : how many acres were there in the farm, and how much did the whole cost at 18 dollars an acre? The farm contained 123 acres. Ans. {It cost 2214 dollars. 6. A merchant bought 49 hogsheads of molasses, each containing 63 gallons: how many gallons of molasses were there in the parcel ? Ans. gallons. 7. Suppose a man were to travel 32 miles a day: how far would he travel in 365 days? Ans. 11680 miles. 8. In a certain city, there are 3751 houses. If each house on an average contains 5 persons, how many inhabitants are there in the town? Ans. inhabitants. 9. When a person sells goods he generally gives with them a bill, showing the amount charged for them, and acknowledging the receipt of the money paid; such bills are usually called Bills of Parcels. BILLS OF PARCELS. James Johnson New-York, Oct. 1, 1838. 4 Chests of tea, of 45 pounds each, at 1 doll. a pound. 3 Firkins of butter at 17 dolls. per firkin 4 Boxes of raisins at 3 dolls. per box 36 Bags of coffee at 16 dolls. each 14 Hogsheads of molasses at 28 dolls. each. Received the amount in full, Amount 1211 dollars. W. Smith. Hartford, Nov. 1, 1837. James Hughes 27 Bags of coffee at 14 dolls. per bag 18 Chests of tea at 25 dolls. per chest 75 Barrels of shad at 9 dolls. per barrel 87 Barrels of mackerel at 8 dolls. per barrel 67 Cheeses at 2 dolls. each 59 Hogsheads of molasses at 29 dolls. per hogshead, Amount 4044 dollars. per James Cross. Received the amount in full, for W. Jones, DIVISION OF SIMPLE NUMBERS. § 28. Charles has 12 apples, and wishes to divide them equally between his four brothers. He gives one to each, which takes 4. Subtracting 4 from 12, 8 remains. He then gives another to each, which takes 4 more. Subtracting this 4 from 8 leaves OPERATION. 12 4 8 1st remain. 4 4 2d remain. 4 4. He then gives one more to each, which takes all his apples, and leaves nothing. He has then divided them equally, and found that 12 contains 4, three times, for he has three times subtracted 4 from 12. Suppose he had 28 apples and wished to divide them equally among 8 boys. Giving each one, would take 8 and leave 20. Giving each one, a second time, would take 8 and leave 12. Giving each one, a third time, would take 8 and leave 4. Hence, 8 is contained three times in 28, and there are 4 over. 0 3d remain. OPERATION. 28 8 20 1st remain. 8 12 2d remain. 8 By continued subtraction we can always find how many times one number is contained in another, and also, what is left when it is not contained an exact number of times. 4 3d remain. |