102. Contracted Multiplication.—The partial products are sometimes omitted; the process saves time in writing, but is not easy. The principle is that, e.g. (a·102 + b·10 + c)(p·102 + q10 + r) = ap·104 + (aq + bp)·103 + (ar + bq + cp)·102 + (br + cq)·10 + cr. Hence the digits are multiplied in pairs, and grouped according to the power of 10 which each product contains. A method of performing the process is shown here for the case of 162·427. The principle is that 162·427 = 100·427 + 60·427 + 2·427 = 1·42700 + 6·4270 + 2·427; but, instead of writing down the separate products, we (in effect) write 42700, 4270, and 427 in separate rows, with the multipliers 1, 6, 2 in the margin, and then multiply each number in each column by the corresponding multiplier in the margin, making allowance for any figures to be “carried.” Thus the second figure (from the right) is given by 1 + 2·2 + 6·7 = 47, the 1 being carried.
103. Aliquot Parts.—For multiplication by a proper fraction or a decimal, it is sometimes convenient, especially when we are dealing with mixed quantities, to convert the multiplier into the sum or difference of a number of fractions, each of which has 1 as its numerator. Such fractions are called aliquot parts (from Lat. aliquot, some, several). This can usually be done in a good many ways. Thus 5⁄6 = 1 − 1⁄6, and also = ½ + 1⁄3; and 15% = .15 = 1⁄10 + 1⁄20 = 1⁄6 − 1⁄60 = 1⁄8 + 1⁄40. The fractions should generally be chosen so that each part of the product may be obtained from an earlier part by a comparatively simple division. Thus ½ + 1⁄20 − 1⁄60 is a simpler expression for 8⁄15 than ½ + 1⁄30.
The process may sometimes by applied two or three times in succession; thus 8⁄15 = 4⁄5·2⁄3 = (1 − 1⁄5)(1 − 1⁄3), and 33⁄40 = ¾·11⁄10 = (1 − ¼)(1 + 1⁄10).
104. Practice.—The above is a particular case of the method called practice, but the nomenclature of the method is confusing. There are two kinds of practice, simple practice and compound practice, but the latter is the simpler of the two. To find the cost of 2 ℔ 8 oz. of butter at 1s. 2d. a ℔, we multiply 1s. 2d. by 28⁄16 = 2½. This straightforward process is called “compound” practice. “Simple” practice involves an application of the commutative law. To find the cost of n articles at £a, bs, cd. each, we express £a, bs, cd. in the form £(a + f), where f is a fraction (or the sum of several fractions); we then say that the cost, being n × £(a + f), is equal to (a + f) × £n, and apply the method of compound practice, i.e. the method of aliquot parts.
105. Multiplication of a Mixed Number.—When a mixed quantity or a mixed number has to be multiplied by a large number, it is sometimes convenient to express the former in terms of one only of its denominations. Thus, to multiply £7, 13s. 6d. by 469, we may express the former in any of the ways £7.675, 307⁄40 of £1, 153½s., 153.5s., 307 sixpences, or 1842 pence. Expression in £ and decimals of £1 is usually recommended, but it depends on circumstances whether some other method may not be simpler.
A sum of money cannot be expressed exactly as a decimal of £1 unless it is a multiple of ¾d. A rule for approximate conversion is that 1s. = .05 of £1, and that 2½d.= .01 of £1. For accurate conversion we write .1£ for each 2s., and .001£ for each farthing beyond 2s., their number being first increased by one twenty-fourth.
106. Division. Of the two kinds of division, although the idea of partition is perhaps the more elementary, the process of measuring is the easier to perform, since it is equivalent to a series of subtractions. Starting from the dividend, we in theory keep on subtracting the unit, and count the number of subtractions that have to be performed until nothing is left. In actual practice, of course, we subtract large multiples at a time. Thus, to divide 987063 by 427, we reverse the procedure of § 101, but with intermediate stages. We first construct the multiple-table C, and then subtract successively 200 times, 30 times and 1 times; these numbers being the partial quotients. The theory of the process is shown fully in F. Treating x as the unknown quotient corresponding to the original dividend, we obtain successive dividends corresponding to quotients x − 200, x − 230 and x − 231. The original dividend is written as 0987063, since its initial figures are greater than those of the divisor; if the dividend had commenced with (e.g.) 3 ... it would not have been necessary to insert the initial 0. At each stage of the division the number of digits in the reduced dividend is decreased by one. The final dividend being 0000, we have x − 231 = 0, and therefore x = 231.