Detail of the operation, units units units units tens units. 4x7=28=20+8=2+8; the 8 units are written under the units of the multiplicand, and the 2 tens reserved for addition to the product of the tens; tens tens tens tens tens tens tens hds. tens. 5×7=35, and 35+2 carried = 37=30+7=3+7; the 7 tens are written under the tens of the multiplicand, and the 3 hundreds reserved for addition to the product of the hundreds ; hds. hds. hds. hds. hds. hds. hds. ths. hds. 8x7=56, and 56+3 carried = 59=50+9=5+9; the 9 hundreds are written under the hundreds of the multiplicand, and the 5 thousands reserved for addition to the product of the thousands; there being no higher unit to the product of which by 7 the 2 ten thousands can be carried, the 6 is written under the thousands of the multiplicand, and the 2, expressing ten thousands, in the next place to the left. The result of this detail is that 3854×7=26978. 58. 4th Example. Multiply 8762 by 47. The multiplicand and multiplier being both greater than 10, this example falls under Article 51. Whence, resolving the multiplier into 7 units and 4 tens, it follows that 8762 x7 units+8762 x 4 tens product required. tens units Proceeding as in Example 3, 8762×7=61334=product of multiplicand by units of multiplier, and 8762 × 4=ten times 8762×4 (Art. 51)=87620 × 4 =350480 product by the tens of the multiplier. Taking the sum of the partial products, 8762 × 7 = 61334 8762 × 40 = 350480 .. 8762 × 47 = 411814 5th Example. Multiply 684 by 35742. Employing 35742 as multiplier, five partial products must be found; whereas if 684 be taken as multiplier, the number of partial products to be found is only three. Now (Art. 56) 684 × 35742=35742 × 684; therefore it is permitted to employ 684 instead of 35742 as multiplier; and since 684 is composed of 4 units, 8 tens, and 6 hundreds (Art. 51).' 35742×4 units+35742×8 tens+35742×6 hundreds=product required. 35742 x 4 units 35742 x 8 tens =142968 =357420 × 8 units=2859360 35742 × 6 hundreds=3574200×6 units=21445200 Multiplicand= = 35742 684 142968 2859360 21445200 It seems unnecessary to repeat the details of the multiplication. 59. The calculation of the last example is exhibited in two forms, slightly differing the one from the other. In the first the zeros arising from the multiplication of the multiplicand by the tens, the hundreds, &c. of the multiplier, are retained, and in the second omitted. This omission may always be made, provided care is taken to place the units of the lowest order in each partial product directly under that figure by which this partial product arises. The reason of so arranging the partial products is obvious, for the product of simple units by simple units is simple units, of simple units by tens it is tens, &c. 60. When a multiplier contains the same digit oftener than once, the significant figures of the partial product must, for every repetition, be the same, but their relative values different. It is only necessary, therefore, for every such repetition, to write the partial product, already found, taking care to annex in each case the proper number of zeros, or to place the lowest figure under the partial multiplier. 7th Example. Multiply 13256 by 6080. 13256 6080 1060480 795360 80596480 This example affords the occasion to observe, that when the last figure of a multiplier is zero, or when zeros are interposed between significant figures, the preceding remarks with regard to placing the last figures of the partial products under those figures of the multiplicand to which they respectively belong, apply as well to a multiplier such as this, as to one wholly composed of significant figures. 61. Any necessity, except as an exercise, of finding the product of two very large numbers, can seldom occur. If, however, it were required to find the product of a number expressed by, ex. gr., 20 digits, by another containing as many, the labour might be considerably abridged by the following means. How many soever the number of digits in the multiplier, it is certain that there can be only 1, 2, 3,. . . . . 9 distinct figures; and that the excess, at the very least, of the whole number of digits over nine is composed of repetitions of some of the figures. Whence, if once, twice, three times, nine times the multiplicand be computed, all the partial products will be known; and it will further be only necessary to write in its proper place the product of the multiplicand by each digit of the multiplier, and to add these partial products together, in order to obtain the result sought. ..... The product of the multiplicand by 1, 2, 3, . . . . . 9, is readily found by repeated additions. Let the multiplicand=73854980671325487963 73854980671325487963 Multiplicand×2=147709961342650975926 73854980671325487963 Multiplicand×3=221564942013976463889 73854980671325487963 Multiplicand×4=295419922685301951852 The table being completed to multiplicand 9, the partial products are copied from it in their proper order, and added together: their sum, thus obtained, is the product of the given numbers. 62. The multiplication of a large number by a single figure is performed from right to left, to facilitate the combination of the tens produced by the multiplication of units of any order with the units of the product of the order which is next higher; and in the multiplication of one large number by another the partial products are, for the same reason, found by following the same course. The digits of the multiplier may, however, be taken from right to left, left to right, or in any order whatever; for, since the sum of any given numbers is independent of the order in which they are added together, the sum of the same partial products, in whatever order taken, must give the same result. In the following example the figures of the multiplier are first taken from left to right, and second from right to left: The order of the partial products in the first operation is the reverse of that in the second. Their sum is, of course, the same in both. Although the partial products are found most conveniently by taking the figures of the multiplicand from right to left, it is not impossible to perform the multiplication, in respect of both factors, from left to right, as the following instance may show : Upon the whole, the usual practice, by which the figures as well of the multiplier as of the multiplicand are taken from right to left, seems that which is most convenient. 63. As the quantity repeatedly added to itself may be of any kind, so the number expressing it (that is, the multiplicand) may be concrete or abstract. But the multiplier, expressing only times of repetition, must be always an abstract number. The product consists of units of the same kind as the multiplicand. In performing the multiplication, it is usual, on account that the factors are sometimes interchanged, to regard both as abstract numbers, and afterwards to restore to the product the denomination of the multiplicand. 64. To find the product of any two numbers, General Rule. Write the greater of the two numbers as multiplicand, and the less as multiplier, placing the units of each order in the latter under the corresponding units of the former. Take as a partial factor the figure expressing the units of the first order, or simple units of the multiplier; multiply the units of the first order in the multiplicand by this figure; write the units of the product in the same vertical column with the partial factor, and reserve the tens (if there be any) for combination with the product of the units of the second order of the multiplicand by this partial factor. Multiply the units of the second order of the multiplicand by the partial factor, and to the product add, as units of this order, the tens reserved from the product of the simple units; write the units of the sum on the immediate left of that figure of the product which is already found, and reserve the tens, as before, for combination with the product of the partial factor into the units of the third order of the multiplicand. Proceed in the same manner with the units of the 3d, the 4th, the nth order, so as to form the partial product of the multiplicand by the simple units of the multiplier. Next, take as partial factor the figure expressing the units of the second order, or tens of the multiplier; multiply the units of the first order of the multiplicand by this figure; write the units of the partial product in the same vertical column with this partial factor, and reserve the tens for combination with the product of the units of the second order of the multiplicand by the partial factor. Continue to multiply, one after another, the remaining figures of the multiplicand by the figure expressing the tens of the multiplier, and dispose of the units and tens of the successive products as in the case of the partial factor expressing units of the first order; the result is the product of the multiplicand by the tens of the multiplier. Find, in like manner, the product of the multiplicand by the units of the 3d, the 4th, . . . the mth order of the multiplier, and write the successive products, one below another, in horizontal lines, placing always the lowest figure of each partial product in the same column with that partial factor from the multiplication of the multiplicand by which it has been obtained. Then add together all these partial products. Their sum is the product of the given numbers. 65. Examples in the multiplication of whole numbers. Required the products of the following numbers? 1st. 765498 and 4? 5th. 153879864 and 8? 6th. 123456789 and 9? 7th. 987654321 and 9? 8th. 742 and 68 ? 10th. 6730 and 89?. 23d. 944784 and 972? 24th. 46090521 and 6789?....... ...Ans. 3061992. ..Ans. 1752430. ..Ans. 502773612. .Ans. 3357795252. ..Ans. 1231038912. .Ans. 1111111101. ...Ans. 8888888889. ....Ans. 50456. ...Ans. 614160. ...Ans. 598970. .Ans. 31394452. .Ans. 47272052. .Ans. 330098048. Ans. 5266392900. ..Ans. 83810205. .Ans. 87407959104. Ans. 1084294992. .Ans. 3450743782950. .Ans. 27778480. ...Ans. 918330048. .......Ans. 312908547069. 25th. 594786043 and 64074958 ?...... ..Ans. 38110890724211194. 26th. 3529987504 and 7985463 ?.........Ans. 28188584603654352. 27th. 179865380024 and 90730856?. Ans. 16319339894342820544. 28th. 405970382 and 470098435 ?........Ans. 190846041234552170. 66. The quantities a, b, may be considered as general expressions of number. Making them the factors of a product, that product is indicated in any of the three forms following: axb; a.b; ab; each of which is a general expression of the product of two factors. It is in the first form only that the product of two numerical factors can be expressed; 5x7 signifies 5 multiplied by 7; but 5'7 is not distinguishable from the expression of a decimal fraction; or 57 from the number fifty seven. If one factor is a particular, and the other a general expression of number, as 5 and a, the product is written in any of the forms 5×a; 5.a; 5 a. 67. Hitherto attention has been given to the formation of such products only as result from the combination of two factors. But it is often necessary to find the product of more than two numbers. To afford a distinct notion of this subject, it may be of use to take a particular case requiring the multiplication of more than two factors. One pound sterling contains 20 shillings; and one shilling contains 12 pence: let it be required to find the number of pence in three pounds. To find the number of pence in 3£, it is obviously necessary to determine the product of the three factors 12, 20, and 3. Now, granting that the whole sum is given in penny-pieces, the number of these may be determined as follows, by a recurrence to the first principles of multiplication :— |