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more figures to be multiplied, set down the 14 in full as in addition. (Art. 53. Obs. 1.) The product is 14046 dollars.
83. The product of any two numbers will be the same, whichever factor is taken for the multiplier. Thus,
If an orchard contains 5 rows of trees, and each row has 7 trees, as represented by the Etars in the margin, it is evident the whole * * * number of trees is equal either to the number of stars in a horizontal row repeated five times, or to the number of stars in a perpendicular row repeated seven times, viz: 35. For, 7X5=35, also 5X7=35.
Obs. 1. It is more convenient and therefore customary to place the larger number for the multiplicand, and the smaller for the multiplier. Thus, it is easier to multiply 8468946 by 3, than it is to multiply 3 by 8468946, but the product would be the same.
Ex. 4. What will 237 coaches cost, at 675 dollars apiece ?
Since it is not convenient to multi- Operation. ply by 237 at once, we multiply first 675 Multiplicand. by the 7 units, next by the 3 tens, 237 Multiplier. then by the 2 hundreds, and place 4725 cost 7. coaches. each result in a separate line, with 2025* cost 30 the first figure of each line directly 1350** cost 200 under that by which we multiply. 159975 cost 237 Finally, adding these results together, units to units, &c., we have 159975 dollars, which is the whole product required. (Ax, 11.)
Note.—When the multiplier contains more than one figure, the several products of the multiplicand into the separate figures of the multiplier, are called partial products.
Obs. 2. The reason for placing the first figure of the several partial products under the figure by which we multiply, is to bring the same orders under each other, and thus prevent mistakes in adding them together. (Art. 51.)
3. The several partial products are added together for the obvious purpose of finding the whole product or answer required. (Ax. 11.)
Quest.-83. Does it make any difference with the result, which of the given numbers 1s taken for the multiplier ? Obs. Which is usually taken? Why?
84. The principle of carrying the tens in multiplication is the same as in addition, and may be illustrated in a similar manner. (Art. 53.) Thus, Ex. 5.
9382 Mult'd. Or, separating the multiplicand into
7 Mult'r. the orders of which it is composed, 14=units,
and 9000 X 7=63000 21**=hunds.
300X7= = 2100 63***=thou.
80X7= 560 65674 Product.
2 X 7= 14 Adding these results together, we have 65674 Ans. OBs. The reason for always beginning to multiply at the right hand of the multiplicand, is that we may carry the tens as we proceed in the operation.
85. From this illustration it will be observed that units multiplied into units produce units ; tens into units, or units into tens, produce tens ; (Art. 83 ;) hundreds into units, or units into hundreds, produce hundreds, &c. Hence,
86. When units are multiplied into any order whatever, the product will always be of the same order as the other figure.
And universally, the product of any two integers is of the order next less than that denoted by the sum of the orders of the two given figures. Thus, hundreds into tens produce thousands, or the 4th order, which is one less than the sum of the two given orders.
Oes. When the multiplier contains more than one figure, it is customary to begin to multiply with its units' figure. The result however will be the same, if we begin with its hundreds or any other order of the multiplier, and place the first figure of the partial products, so that the same orders shall stand under each other. First Operation.
Second Operation. 1357
5428 4849918. Prod.
QUEST.-85. What do units into units produce ? Units into tens, or tens into units ?
Ex. 6. What is the product of 5690 into 3008 ?
After multiplying by the 8 units, we next Operation. multiply by the 3 thousands, since there are no 5690 tens nor hundreds in the multiplier, and place 3008 the first figure of this partial product under the 45520 figure 3 by which we are multiplying.
17115520 Ans. 87. From the preceding illustrations and principles we derive the following
GENERAL RULE FOR MULTIPLICATION. I. When the multiplier contains but one figure.
Write the multiplier under the multiplicand, units under units, tens under tens, &c. (Art. 83. Obs.)
Begin at the right hand and multiply each figure of the multiplicand by the multiplier, setting down the result and carrying as in addition. (Art. 84. Obs.)
II. When the multiplier contains more than one figure.
Multiply each figure of the multiplicand by each figure of the multiplier separately, beginning with the units, and write the partial products in separate lines, placing the first figure of each line directly under the figure by which you multiply. (Art. 83. Obs. 2.)
Finally, add the several partial products together, and the sum will be the whole product. (Art. 83. Obs. 3.)
88. Proof.—Multiply the multiplier by the multiplicand, and if the product thus obtained is the same as the other product, the work is supposed to be right.
OBs. This method of proof depends upon the principle, that the product of any two numbers is the same, whichever is taken for the multiplier. (Art. 83.)
89. Second Method.-Add the multiplicand to itself as many QUEST.–86. When units are multiplied into any order, what order is the product ? When any two integers are multiplied together, of what order is the product ? 87. How
you write the numbers for multiplication ? When the multiplier contains but one fig ure, how proceed? Why begin at the right hand of the multiplicand ? When the multi plier contains more than one figure, how proceed? What is meant by partial products ? Why place the first figure of each partial product under the figure by which yon multipły? What is to be done with the partial products? Why add the several partial products together? Why should this give the whole product? 88. How is multiplication proved ? Obs. On what principle does this proof depend ?
times as there are units in the multiplier, and if the amount obtained is equal to the product, the work is right. Note.--When the multiplier is small, this is a very convenient mode of proof.
90. Third Method.--Cast the 9s out of the multiplicand and multiplier ; multiply their remainders together, and casting the 9s out of their product, set down the excess; then cast the Is out of the answer obtained, and if this excess be the same as that obtained from the multiplier and multiplicand, the work may be considered right. Ex. 7. Multiply 565 by 356. Operation.
1695 Prod. 201140. The excess of 9s in the Ans. is also 8. 91. Fourth Method.—Divide the product by one of the factors, and if the quotient thus arising is equal to the other factor, the work is right.
Note. This method of proof supposes the learner to be acquainted with division before he commences this work. (Art. 57. Note.) It is simply reversing the operation, and must obviously lead us back to the number with which we started : for, if a number is both multiplied and divided by the same number, its value will not be altered. (Ax. 9.)
92. Fifth Method.*_First, cast the 11s out of the multiplicand and multiplier; multiply their remainders together, cast the 11s out of the product, and set down the excess; then cast the 11s out of the answer obtained, and if the excess is the same as that obtained from the multiplier and multiplicand, the work is right.
Note.-1. This method depends on a peculiar property of the number 11. For its further development and illustration, see Art. 161. Prop. 18.
2. To cast the 11s out of a number, begin at the right hand, mark the alternate figures; then from the sum of the figures marked, increased by 11 if necessary, take the sum of those not marked, and the remainder will be the excess required. Thus to cast the 11s out of 39475025, mark the alternate figures, beginning at the right hand, 39475025, then the sum of 5+0+7+9=21. Again, the sum of the others, viz: 2+5+4+3=14. Now 21–1457, the excess of 11s.
QUEST.-Can multiplication be proved by any other methods?
* Leslie's Philosophy of Arithmetic.
Or, as soon as the sum is 11 or over, we may drop the 11, and add the remainder to the next digit. Thus, 5 and 7 are 12; dropping the 11, 1 and 9 are 10. Again, 2 and 5 are 7, and 4 are 11; drop the 11, and there are 3 left. Now, 10—3=7, the same excess as before. Ex. 8. Multiply 237956 by 3728. Operation.
Proof. 237956 Excess of 11s is 4. / Now, 4X10=40; the excess of 11s 3728
10.) in 40 is 7. Ans. 887099968 Excess of 11s in the answer is also 7.
EXAMPLES FOR PRACTICE.
93. Ex. 1. What will 435 acres of land cost, at 57 dollars
per acre ?
2. What cost 573 oxen, at 63 dollars
head ? 3. What cost 1260 tons of iron, at 45 dollars per ton ?
4. If a man can travel 248 miles in a day, how far can he travel in 365 days ?
5. If an army consume 645 pounds of meat in a day, how much will they consume in 115 days ?
6. If 1250 men can build a fort in 298 days, how long would it take 1 man to do it ?
7. How many rods is it across the Atlantic Ocean, allowing 320 rods to a mile, and the distance to be 3000 miles ?
8. What is the product of 463 X 45.?
12. What is the product of 7198 X 216 ? 13. 31416X175.
22. 8320900 X 1328. 14. 8862X189.
23. 17500 X 732. 15. 7071X556.
24. 15607X 3094. 16. 93186X4455.
25. 7422153 X 468. 17. 40930x779.
26. 9264397 x 9584. 18. 12345X686.
27. 4687319 X 1987. 19. 46481 x 936,
28. 9507340 X 7071. 20. 16734 X 708.
29. 39948123 X 6007. 21. 7575 X 7575.
30. 73885246 X 6079.