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3 6 2,81
5 nines contained in 5, the left-hand figure, we carry 6, and proeeed as usual. Proof. 575% = 5183: then 9,183 XÍ= 312 as before.
Examples. 1. 9,08X9=81,8; 5,6X3 =17; and 4,813X5=24,06 2. ,14 X 9 = 1,3; ,2617 X 9=2,356, and 3,3X3=10
The scholar will easily see, that when the multiplier is 9, it destroys the divisor 9, or denominator of the period; therefore, in the product, the repeater vanishes.
267. When the period contains more than one figure, we multiply as in integers; and, if the product of the last or lefthand figure of the period does not exceed 9, we write it underDeath; and having marked the period, we continue the operation as usual. But if the product of this left-hand figure exceeds 9, we not only carry the left-hand figure of this product forward, to be added to the product of the next figure, but we also carry it backward, to be added to the unit figure of the period in the product. Now, as multiplication is a species of addition, the student is referred for the reason of this to art. 264.
For example, 23,168 X 4=92,672; the multiplication being performed exactly as in whole numbers.
But, if we have 54,0987 to multiply by 8, having multiplied .the period and found 7896, as the 7 divided by 999 would give 7 for both quotient and remainder, we add 7 to the unit figure 6, which gives 903 for the period; and carrying the quotient 7, we proceed as usual. Wherefore,
54,0987 x 8=432,7903 Proof. 54,0987 X 8=548987 X = 1339676 =432,7903
Examples. 1. 3,142 X 6 =18,852, and 14,026 X 9 = 126,234 2. 4,0759 x8=32,6076, and ,00925 X 9=,08325 3. 35,6798 X7=249,7590, and ,0928 X 9=
When the multiplier contains several figures.
268. When the period in the multiplicand contains but one figure, we multiply by each figure of the multiplier, as in art. 266; and having found the several products, we continue the repeating figure in each to the unit place of the first partial product. Then having found the sum of the right-hand column, which is evidently a number of ninths, we divide this sum by 9, setting down the remainder as ninths, and carrying the quotient to the next column, after which we add as usual.
For example, if we have 2,354 to multiply by 125, having placed the numbers
125 11772 282533
294,305 and found the two partial products, in multiplying by 5 and by 12, as in art. 266, we repeat the figure 3 of the second product, so that it may stand under the repeater of the first; and having added the first column, we mark the sum 5 as a repeater; for it is evident that as both figures 3 and 2 repeat, their sum must also repeat.
269. A mixed decimal, as well as any other, may be multiplied by 10, 100, 1000, &c., by removing the comma, as usual, towards the right; but if it should extend into, or beyond the period, care must be taken to repeat the period till there are as many figures on the right of the comma as are contained in the period. If, in proceeding thus, the order of the figures in the period on the right of the comma appears to be changed, this is of no consequence; because they still repeat, in regular order, from the place where they begin, to infinity; and because the ratio of increase and diminution is, in integral and decimal numbers, the same. Thus: 2,354 X 1000=2354,4; also, 2,351 x 100 = 235,435
As the number 125 is the eighth part of 1000, if we divide 2354, À by 8, we shall have the product of 2,354 X 125. Thus :
8) 2354,444, &c.
where we repeat 4 in the dividend, till we find that the quotient figure 5 will repeat, which we mark accordingly. This may serve as proof of the example in art. 268.
To multiply 6,567 by 25, we place the numbers and multiply as before.
164,194 But as the sum of 5 and 8, which is 13, exceeds 9, we divide by 9; or rather, we say 1 and three is 4, which we write as a repeater underneath, and carrying the quotient 1, we proceed as usual.
4) 656,777, &c.
164,194 This last operation, which may serve as proof of the preceding, consists in multiplying by 100, and dividing by 4, because 25 is the fourth part of 100.
4. ,018798 X 2,9479=,0554172445 270. When there are several figures in the period of the multiplicand, we multiply by each figure of the multiplier, as in art. 267; and, having found the several partial products, we continue the repeating figures of each, in their proper order, to the right-hand or lowest place of the first partial product. Then, taking as many of the right-hand columns as there are figures in the period of the multiplicand, we consider them as belonging to the common period, and add them as in art. 264. The reason for operating thus will appear plain, if we consider that the period in the multiplicand is the numerator of a fraction, which does not change the denomination of its units in being
multiplied by any number; the products, therefore, must all terminate at the same place as this period; and we must have just the same number of repeating figures in the sum of these products, or total product of the given numbers, as in the multiplicand. For example, if we have 65,437 to multiply by 3462:
3462 130874 3926246 26174974 196312312
226544,408 As the product of the period by 2 does not exceed 999, the first product is found as in whole numbers; its figures 874 showing the limit of the period. The second product, 392624, is first found as in art. 267. If this stood alone, as the product of the multiplicand by 6, its period would be 624; but as the multiplier is 6 tens, this product must be ten times greater. Now, to multiply 392,624 by 10, we remove the comma one place towards the right; and, as this enters the period, we must (269) repeat the figure 6, which it passes, by placing it on the right of 4. The second product is therefore 3926,246. As the next multiplier is 400, we must, in the third product, suppose
the comma to be removed two places, and repeat the figures 74 of its period. As the last multiplier is 3000, we must, in the product, repeat the whole period 312. Thus, the products will always terminate at the right-hand place of the first product.
The work may be proved by multiplying 65437 by 3462, and reducing the product to decimals.
If the multiplier had been 34,62, the product would have been 2265,44108.
271. When there are repeating decimals in both factors, we may find the fractional equivalent of the multiplier; multiply by its numerator as in the preceding article, and divide the product by the denominator, taking care to repeat the period in the number we are dividing, till we obtain the quotient to a sufficient degree of exactness, or till its figures repeat, as may be thought proper.
Or, we may find the equivalents of both factors, multiply them as vulgar fractions, and reduce the fractional product to decimals. The scholar may operate by both methods, and see which he prefers; also, the one will be a proof of the other. For example, if we have 81,6275 to multiply by ,358, we may operate thus : ,358 x 10=399
,358 x 10=359 = 355, and 355 --10=35. Now, though this fraction might be reduced, it may be as well to let it remain as it is; then,
having multiplied by 5, we multiply the first product by 7, for 35. To divide by 990, we remove the comma one place to the left, which divides by 10, and then divide by 9 and by 11, which are the factors of the remaining number 99. Or thus : ,358 = 353 = 115, and 81,6275
5789575.89 = 29,27047704269 +, as before.
815 4 5 9