To indicate the manner, let m be put for any compound multiplicand; and since this method is most advantageous when the multiplier is a large number, let it be required to multiply m by 57684. 57684-50000+7000+600+80+4. ... mx 57684=mx 50000+mx7000+m × 600+m × 80+mx 4. Form of calculation. Giving to m the particular value 5£. 9'sh. 73 d., and retaining the same multiplier, 57684, the preceding calculation is replaced by the following. In this calculation 50000 repetitions of the multiplicand are obtained by multiplying consecutively by 10, 10, 10, 10, 5. 7000 repetitions are obtained by multiplying the partial product 1000m by 7. 600 repetitions are obtained by multiplying the partial product 100m by 6. 80 repetitions, by multiplying the partial product 10m by 8. d. When the multiplicand is a compound number, and the multiplier a fraction, if the multiplicand is reduced to units of the lowest denomination contained in it, then multiplied by the fractional multiplier, and the product reduced to a compound number, the result is the product of the given compound multiplicand and fractional multiplier. Otherwise, since the product of the multiplicand by a fraction must contain some part or parts of the multiplicand as often as the multiplier contains the same part or parts of unity, the product may be found by resolving the multiplicand into as many equal parts as the denominator of the multiplier contains simple units, and taking of these equal parts the number expressed by the numerator. To obtain the product of a compound number by a fraction, without reduction of the former to a single denomination, it is therefore necessary to divide a compound number by a whole number. e. In like manner it follows from Article 205, that to divide a compound number by a fraction, it is necessary to multiply the dividend by the denominator, and to divide the product by the numerator of the fractional divisor. Consequently, to complete the principles for either multiplication or division of compound numbers, the principles of both must be developed. DIVISION OF COMPOUND NUMBERS. 321. The process of dividing a compound quantity into any number of equal parts, for the purpose of obtaining one of these parts, differs from that of dividing one whole number by another in respect only of the multipliers employed to reduce the successive remainders from any denomination to the denomination which is next lower. In whole numbers, 1 of any order of units is equal to 10 of the next lower, and by annexing a figure of the dividend to a remainder for the purpose of forming a new partial dividend, the multiplication of the remainder by 10 and the addition of the units of the next lower denomination are both accomplished. In compound numbers, the multiplication of the remainder, and addition. of the units of the next lower denomination to the product, cannot be effected by mere annexation of these units. But the principle of reduction is the same; the remainder is multiplied by a factor which expresses that number of units of the next lower denomination which make one unit of the remainder, and the units of that lower denomination are added to the product. Examples of division of compound numbers. 1st. Divide 27£. 14sh. 83d. by 6. (The quotient is obtained by dividing the number in the highest denomination by 6, reducing the compound number composed of the remainder and the number in the second denomination to units of the second denomination, dividing this result by 6, &c.) 1st. To find the pounds of the quotient. 11qrs.+6=1&qrs. quotient. Whence the complete quotient is 4 £. 12 sh. 5 d. 13 qrs. The calculation may be represented as in Article 86 thus, £ sh. d. qrs. 6)27 14 8 3 4 12 5 1 The manner of passing from any denomination to the denomination next lower is the same for divisors either not greater than 12 or greater than 12; but in the latter case the calculation must be made as in Article 88 or Article 91. 2d Example. Divide 576 miles 4 furlongs 24 poles 4 yards by 168. Mi. fur. po. yd. Mi. fur. po. yd. ft. 168)576 4 24 4 (3 3 18 1 1 504 72 8 168)580(3 furlongs. 504 76 40 168)3064(18 poles. 168 1384 1344 40 54 20 204 168)224(1 yard. 168 56 3 168)168(1 foot. Performing the multiplications and subtractions mentally, and recording the remainders only, as in Article 91, the preceding calculation is reduced Since 168=4×6×7, the quotient may be obtained by dividing successively by these factors of 168 (Art. 95). Mi. fur. po. yd. ft. 1686)144 1 6 1 0 732 3 3 18 1 1 Nothing in the division by 4 or in that by 6 seems to require explanation. But the division by 7 presents a partial dividend which, although not the last, contains a fraction. The process of the division by 7 seems therefore to require development. Detail of the division of 24 miles 0 furlongs 7 poles 3 yards 2 feet by 7. 24 miles+7=3 miles quotient +3 miles remainder. 3 miles remainder+0 furlongs dividend=24 furlongs. The complete quotient (as before) is 3 miles 3 furlongs 18 poles 1 yard 1 foot. 322. It has been already explained (Art. 320 d), that to multiply a compound number by a fraction, it is necessary to multiply the compound number by the numerator of the multiplier, and to divide the product by its denominator. This is the process to be followed in the case of a decimal as well as in that of a vulgar fraction. In most instances of multiplication by the former some advantage may be gained by writing the decimal as a vulgar fraction and reducing it to the lowest terms. When the multiplier is a mixed number, the products of the multiplicand by the integral and fractional parts of the multiplier may be formed separately, and the results added together. 323. Division by a fraction being reduced to multiplication by the reciprocal of the fraction, these principles for the multiplication serve also for the division of a compound number by a fraction. When the divisor is a mixed number it must be reduced to an improper fraction. The reciprocal is, of course, a proper fraction. Examples of the multiplication and division of compound numbers by fractional expressions. 1st. Multiply 104 £. 7 sh. 64 d. by §. To obtain the required product it is necessary to multiply 104 £. 7 sh. 63 d. by 5, and to divide the result by 8. Wherefore (104£. 7 sh. 64 d.) × ğ=65 £. 4 sh. 834 d. or 65 £. 4 sh. 8 d. 1qrs. 2d Example. Multiply 227 £. 9sh. 114 d. by 25-6. 25-6=25-253. What is required, therefore, is to find the partial product of the multiplicand, first by 25, and second by ; and to take the sum of these partial products for the complete product of the multiplicand by 25 5823 18 4 3 product of multiplicand by 25.6. 3d Example. Divide 24 £. 15 sh. 813 d. by 4. To obtain the required quotient it is necessary to multiply 24 £. 15 sh, 813 d. by the reciprocal of 3, that is, by 4. £. sh. 24 15 3)99 d. 818 4 2 114 33 0 113=quotient required. 4th Example. Divide 12 miles 4 furlongs 37 poles 3 yards by 4:05. The reciprocal of $ dividend by 20, and to 405-4-4= is . It is therefore necessary to multiply the divide the result by 81. Mi. fur. po. yds. 4 12 4 37 50 9)252 81 { 9)28 3 30 133 2 31 5 4 product of dividend by 20. 0 12 2 3 0 36 54 quotient required. ... 12 mi. 4 fur. 37 po. 38 yds.÷4·05=3 mi. 36 po. 54 yds. 324. Rules for the multiplication and division of compound numbers. a. To multiply a compound number by a whole number. 1st. When the multiplier does not exceed the greatest factor contained in the multiplication table. Multiply the number in the lowest denomination of the multiplicand by the multiplier. Reduce the product to a compound number expressed in terms of the denomination multiplied and of that which is next higher (the 2d). Write the units of the lowest denomination in their proper place, and reserve the units of the next higher denomination. Multiply the number in the next higher denomination by the multiplier; to the product add the units of this denomination reserved from the product of the lowest denomination; reduce the sum to a compound number expressed in terms of this and the next higher denomination; write the units of this (the 2d) denomination in their proper place, and reserve the units of the next higher denomination. Proceed in this manner to multiply, reduce, and carry from one denomination to the next higher, until all the denominations of the compound number are multiplied by the multiplier. The result is the product required. |