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CONTRACTIONS IN MULTIPLICAT

94. The general rule is adequate to the solu ples that occur in multiplication. In many ins by the exercise of judgment in applying the pred the operation may be very much abridged.

95. Any number which may be produced by or more numbers together, is called a Composite Thus, 4, 15, 21, are composite numbers; for 5X3; 21=7X3.

OBS. 1. The factors which, being multiplied together, pro number, are sometimes called the component parts of the nu 2. The process of finding the factors of which a given nu is called resolving the number into factors.

Ex. 1. Resolve 9, 10, 14, 22, into their factors. 2. What are the factors of 35, 54, 56, 63? 3. What are the factors of 45, 72, 64, 81, 96? 96. Some numbers may be resolved into more tors; and also into different sets of factors. Thus, 1 also 12 4X3=6X2.

4. What are the different factors and sets of fac 18, 20, 24?

5. What are the different factors and sets of facto 36, 40, 48?

96. a. We have seen that the product of any tw the same, whichever factor is taken for the multiplie In like manner, it may be shown that the product of

QUEST.-95. What is a composite number? Obs. What are the factors w sometimes called? What is meant by resolving a number into factors? 9 ever composed of more than two factors? 96. a. When three or more fa multiplied together, does it make any difference in what order they are tak

more factors will be the same, in whatever order they are multiplied. For, the product of two factors may be considered as one number, and this may be taken either for the multiplicand, or the multiplier. Again, the product of three factors may be considered as one number, and be taken for the multiplicand, or the multiplier, &c. Thus, 24=3×2×2×2=6×2×2=12X2=6X4= 4X2X3=8X3.

CASE 1.-When the multiplier is a composite number.

6. What will 27 bureaus cost, at 31 dollars apiece?

Analysis. Since 27 is three times as much as 9; that is, 27=9 X3, it is manifest that 27 bureaus will cost three times as much as 9 bureaus.

Operation.

Dolls. 31 cost of 1 B.

9

Dolls. 279 cost of 9 B. 3

Dolls. 837 cost of 27 B.

Having resolved 27 into the factors 9 and 3, we find the cost of 9 bureaus, then multiplying that by 3, we have the cost of 27 bureaus.

7. What will 36 oxen cost, at 43 dollars per head? Solution.-36=9×4; and 43X9X4=1548 dolls. Ans. Or, 36=3×3×4; and 43×3×3×4=1548 dolls. Ans. Hence,

97. To multiply by a composite number.

Resolve the multiplier into two or more factors; multiply the multiplicand by one of these factors, and this product by another factor, and so on till you have multiplied by all the factors. The last product will be the answer required.

OBS. The factors into which a number may be resolved, must not be confounded with the parts into which it may be separated. (Art. 53.) The former have reference to multiplication, the latter to addition; that is, factors must be multiplied together, but parts must be added together to produce the given number. Thus, 56 may be resolved into two factors, 8 and 7; it may be separated into two parts, 5 tens or 50, and 6. Now, 8x7=56, and 50+6=56.

8. What will 24 horses cost, at 74 dollars a head?

QUEST.-97. When the multiplier is a composite number, how do you proceed? Obs. What is the difference between the factors into which a number may be resolved, and the parts into which it may be separated?

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9. What cost 45 hogsheads of tobacco, at 128 head?

10. What cost 54 acres of land, at 150 dollars 11. At 118 shillings per week, how much will to board 49 weeks?

12. If a man travels at the rate of 372 miles will he travel in 64 days?

13. At 163 dollars per ton, how much will 72 ton 14. What cost 81 pieces of broadcloth, at 245 sh 15. What cost 84 carriages, at 384 dollars apiece

CASE II. When the multiplier is 1 with ciphers 98. It is a fundamental principle of notation, moval of a figure one place towards the left, incre ten times; (Art. 36;) consequently, annexing a cipher will increase its value ten times, or multiply it by 1 two ciphers will increase its value a hundred times, by 100; annexing three ciphers will increase it a tho or multiply it by 1000, &c. Thus, 15 with a cipher a comes 150, and is the same as 15X10; 15 with two nexed, becomes 1500, and is the same as 15×100; 1 ciphers annexed, becomes 15000, and is the same as &c. Hence,

99. To multiply by 10, 100, 1000, &c.

Annex as many ciphers to the multiplicand as there in the multiplier, and the number thus formed will be required.

Note.-To annex means to place after, or at the right hand.

16. What will ten boxes of lemons cost, at 63 shi box? Ans. 630 shillings.

17. How many bushels of corn will 465 acres of land at 100 bushels per acre?

QUEST.-98. What is the effect of annexing a cipher to a number? T Three? Four? 99. How do you proceed when the multiplier is 10, 100, 1000, What is the meaning of the term annex?

18. Allowing 365 days for a year, how many days are there in

1000 years?

19. Multiply 153486 by 10000.
20. Multiply 3120467 by 100000.
21. Multiply 52690078 by 1000000.
22. Multiply 689063457 by 10000000.
23. Multiply 4946030506 by 100000000.
24. Multiply 87831206507 by 1000000000.
25. Multiply 67856005109 by 10000000000.

CASE III.-When the multiplier has ciphers on the right hand. 26. What will 30 wagons cost, at 45 dollars apiece?

Note. Any number with ciphers on its right hand, is obviously a composite number; the significant figure or figures being one factor, and 1, with the given ciphers annexed to it, the other factor. Thus, 30 may be resolved into the factors 3 and 10. We may therefore first multiply by 3 and then by 10, by annexing a cipher as above.

Solution.-45X3=135, and 135X10=1350 dolls. Ans.

27. How many acres of land are there in 3000 farms, if each farm contains 475 acres?

Analysis.-3000=3×1000. Now 475X 3=1425; and adding three ciphers to this product, multiplies it by 1000. (Art. 99.) Hence,

Operation.
475

3

Ans. 1425000 acres.

100. When there are ciphers on the right of the multiplier. Multiply the multiplicand by the significant figures of the multiplier, and to this product annex as many ciphers, as are found on the right of the multiplier.

OBS. It will be perceived that this case combines the principles of the two preceding cases; for, the multiplier is a composite number, and one of its factors is 1 with ciphers annexed to it.

28. How much will 50 hogs weigh, at 375 pounds apiece? 29. If 1 barrel of flour weighs 192 pounds, how much 500 barrels weigh?

30. Multiply 14376 by 25000.

QUEST.-100. When there are ciphers on the right of the multiplier, how do yo ceed? Obs. What principles does this case combine?

31. Multiply 350634 by 410000.

32. Multiply 4630425 by 6200000.

CASE IV. When the multiplicand has ciphers on the right hand.

33. What will 37 ships cost, at 29000 dollars apiece?

Analysis.-29000=29×1000. But the product of two or more factors is the same in whatever order they are multiplied. (Art 96. a.) We therefore multiply 29 by 37, and this product by 1000 by adding three ciphers to it.

Operation.
29000
37

203

87

Ans. 1073000 dolls.

PROOF.-29000×37=1073000, the same as before.

Hence,

101. When there are ciphers on the right of the multiplicand. Multiply the significant figures of the multiplicand by the multiplier, and to the product annex as many ciphers, as are found on the right of the multiplicand.

OBS. When both the multiplier and multiplicand have ciphers on the right, multiply the significant figures together as if there were no ciphers, and to their product annex as many ciphers, as are found on the right of both factors.

34. Multiply 2370000 by 52.

35. Multiply 48120000 by 48.
36. Multiply 356300000 by 74.
37. Multiply 1623000000 by 89.
38. Multiply 540000 by 700.
Analysis.-540000=54×10000, and
700=7×100; we therefore multiply the
significant figures, or the factors 54 and
7 together, (Art. 96. a,) and to this pro-
duct annex six ciphers. (Art. 99.)

39. Multiply 1563800 by 20000.
40. Multiply 31230000 by 120000.
41. Multiply 5310200 by 3400000.
42. Multiply 82065000 by 8100000.
43. Multiply 210909000 by 5100000.

Operation. 540000

700

Ans. 378000000

QUEST.-101. When there are ciphers on the right of the multiplicand, how proceed? Obs. How, when there are ciphers on the right both of the multiplier and multiplicand?

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