In this example the third partial division being impossible, O is made to occupy the place of the deficient figure; also O is annexed to the last significant figure of the quotient. Restoring the detailed process, as explained in Example 1, the reason becomes manifest, or it may be inferred from the considerations following: had a remainder been left from the preceding subtraction, this remainder, with the 0 of the dividend annexed, must have been divisible by 8; another quotient figure must have been obtained, and the 6 made to occupy (its proper place) the place of tens. 86. When the divisor is a factor contained in the multiplication table, the successive quotient figures may be found, and the partial products formed and subtracted, mentally. In this manner the calculation is made to appear in a very abridged form; the results only, or results and partial remainders, being registered thus: 87. Lastly, let D and d be any numbers whatever, with this limitation, however, that D is greater than d. Since D=dxq, the dividend is composed of the partial products of the divisor by the figures of the quotient taken singly, and with regard to their relative values (Art. 53). The relative values of these figures may be determined from the principles established in Articles 73 and 85; and when the absolute values also have been found, the products by the divisor may be subtracted from the dividend and remainders, as has been done in the case of Example 1, Art. 84. When the divisor is less than 10, the quotient figures are obtained individually from the multiplication table; but when greater than 10, the manner of discovering them is not at first so obvious. It is certain, however, that the significant part of the partial product of the divisor by any figure of the quotient must be the product of the divisor by some number in the series 1, 2, 3, 4, 5, 6, 7, 8, 9. Whence, if the products of the divisor by these numbers are computed and arranged in a tabular form, it will be as easy from this table to determine the quotient figures in succession as it is, in the case of a divisor less than 10, to find the quotient figures by means of the multiplication table. 88. The process of dividing by a divisor greater than 10 being thus rendered identical with that of dividing by a divisor less than 10, the calculation in the former case may be abridged as it has been in the latter. (See Article 85.) Provided always, that attention is given to the explanations and directions contained in Article 85. Example:-Divide 640791247 by 74693. The product of 74693 by each of the nine digits may be formed by repeated additions, as follows: 74693 x1= 74693 74693 74693 X2=149386 74693 Detail of the operation : 64079 +74693; this division gives no whole number to the quotient. -640791+74693=8*+43247 remainder. 432472+74693-5 +59007 remainder. 590074+74693=7 +67223 remainder. 672237+74693=9 + O remainder. 89. The labour of computing such a table for every example more than counterbalances the advantage to be derived from its use. In practice the successive quotient figures are obtained by trial, and a little experience renders it easy to select the proper figures. Usually the first, or first and second figures of the dividend are divided by the first figure of the divisor. The result (given by the multiplication table) is taken as a first quotient figure. If the product of the divisor by this quotient figure is greater than the partial dividend from which it is to be subtracted, the figure next lower in the series, 1, 2, 3, 9 must be taken, and tried in the same manner; on the other hand, if the remainder is greater than the divisor, the figure next higher is to be taken. The subtraction being possible, and the remainder less than the divisor, the quotient figure chosen is the proper one. To apply these remarks, let it be required to divide 2577922208 by * The product of the first quotient figure by the units of the divisor falls under the thousands of the dividend. Whence the quotient must contain four figures. Detail of the operation : 2+5; the division cannot be made. Whence the first partial dividend contains 7 figures, and the quotient 4 figures (Art. 85). 25+5=5, and 538976 × 5=2694880. This product is greater than the partial dividend, 2577922; the quotient figure 5 is therefore too great. Trying 4, the next lower figure, 538976×4=2155904, and 257792-2155904=422018; the remainder, 422018, being less than the divisor, 4 is the proper quotient figure. Annexing 2, the next figure of the dividend, to the remainder 422018, the second partial dividend is 4220182, 42+5=8, and 538976 × 8=4311808. This number exceeds the dividend; the quotient figure 8 is therefore too great. Trying 7, 538976 x7=3772832, and 4220182-3772832=447350; whence 7 is the proper quotient figure, and 447350 the remainder. Annexing to this remainder the next figure of the dividend, which is 0, 4473500 is obtained for the third partial dividend; 44+5=8, and 538976×8=4311808, a number less than the partial dividend; 4473500-4311808=161692; 8 is the proper quotient, and 161692 the remainder. Annexing to this remainder 8, the last figure of the dividend, 1616928 is the fourth partial dividend. 16+5=3, and 538976×3=1616928, a number equal to the partial dividend. Whence the division is exact, and the quotient=4783. In the first and second partial divisions it is found, that, although 25+5=5, and 42+5=8, yet 5 and 8 are not the proper quotient figures. This arises from the circumstance, that when the product of the remaining part of the divisor (namely, 38976) by 5 or 8 is added to the product of 500000 by 5 or 8, the sum of both exceeds the partial dividend. 90. After finding a quotient figure in the manner prescribed, if its product by the first and second figures of the divisor is formed mentally, and compared with the corresponding figures of the partial dividend, it will in general be possible, without further trial, to ascertain whether the figure found is or is not too great. In the preceding instances 53×5, or 265, is greater than 257; and 53 × 8, or 424, is greater than 422. The preceding remarks will be found applicable to the following example: Divide 303374185950 by 4985607. 4985607)303374185950(60850 29913642 42377659 39884856 24928035 24928035 It does not seem necessary to go at length through the details of the process in this example also. Why zeros are written in the quotient in the places of thousands and simple units is explained in Article 85. 91. An abridged form of division in the case of a divisor less than 10 has been noticed in Article 86. When the divisor is greater than 10, the successive multiplications and subtractions may, in like manner, be made mentally, and the quotient figures and remainders alone recorded. In division by this method the divisor seems most conveniently placed over the figures composing the first partial dividend. As this abridged method may be most intelligibly explained through the medium of an example, let it be required to divide 945945 by 2457. The calculation is represented thus, 2457 945945 (385 20884 12285 0000 And the details of the process are, 1st. To find the highest quotient figure, 9+2=4. But since 24 × 4=96, and 96 is greater than 94, this result is too great. The first quotient figure cannot, therefore, be greater than 3; and since 24x3=72, a number considerably less than 94, it may be presumed that 3 is the proper figure. 2d. To find the remainder left by the subtraction of 3 times the divisor from the first partial dividend. In the ordinary process the divisor, 2457, is multiplied by 3, and the product subtracted from the partial dividend 9459; but in this abridged method the multiplication of each figure of the divisor by 3, and the subtraction of this product from the corresponding figure of the partial dividend, are performed simultaneously thus, The last figure of the first partial dividend is 9, and 7×3=21; therefore 21 is to be taken from 9. To render this subtraction possible 2 is borrowed from 5, the next higher figure of the dividend; regard being had to the relative value of the 5 as compared with that of the 9, the 2 borrowed are equal to 20 units of the order of the 9. Whence 20+9=29, and 29-21=8. This 8 is written under the figure 9 of the dividend, and the 2 borrowed is either taken from the next higher figure of the dividend, 5, or reserved for combination with the product of the tens of the divisor by the quotient figure 3. The latter course, which is found the more convenient, is generally adopted in practice. Proceeding to the second figures of the divisor and partial dividend, 5x3=15, and 15+2 borrowed=17. 5-17, subtraction not possible: borrow 20, 20+5=25, and 25-17=8. The remainder 8 is written, and the two tens borrowed are carried, as before. Third figures of the divisor and partial dividend, 4x8-12, and 12+2 borrowed=14. 4-14, subtraction not possible: borrow 10, The O is written, and the 10 carried, as before. Fourth figures of the divisor and partial quotient, 2x3=6, and 6+1 borrowed=7. 9-7=2. The 2 being written in its proper place, it is found that the first figure of the quotient is 3, and the first remainder 2088 annexing the next figure of the dividend to this remainder, 20884 is obtained as the second partial dividend. The quotient evidently contains two figures in addition to that already found. For each the process is the same as that which has been developed at so great length. E To find the second figure of the quotient. 20+2=10. If the quotient really is 10, the preceding quotient figure has been taken too low, for it is evident that this 10 in the second place is equal to 1 in the first place, and that the 3 must become 3+1, or 4. Applying the practical test of Article 90, 24× 10=240: but the corresponding number of the dividend is only 208 ; therefore the quotient is less than 10. In the same manner it is found to be less than 9; to be 8. 8 is written in the quotient; and the multiplication, by 8, of the figures of the divisor, and the subtraction of the uncombined products continued as before, 7x8=56. To render the subtraction possible 60 is borrowed, 60+4 64, and 64-56-8. The 8 remainder is written under the 4 of the partial dividend, and 6 carried for the 60 borrowed, 5 x 8=40, and 40+6=46. 8-46. To render the subtraction possible 40 is borrowed, The 2 remainder is written under the proper figure, and 4 is carried. 8-36. Borrow 30, Write 2 remainder, and carry 3 for 30 borrowed. 2x8=16, and 16+3=19. 20-19-1. Writing 1 under the 0, the second remainder is found to be 1228. If to this remainder the last figure of the dividend is annexed, the third partial dividend is 12285. To find the last figure of the quotient, and the remainder, if any. 12+2=6; but since 24×6=144, and the corresponding figures of the partial dividend are 122, the quotient figure must be less than 6. 24×5=120, a number less than 122; therefore 5 may be tried. Writing 5 in the quotient, and proceeding as before, 7x5=35. 5-35: to render the subtraction possible borrow 30, 30+5=35, and 35-35=0. Write 0 under 5 of the partial dividend, and carry 3, 5×5=25, and 25+3 borrowed=28. 8-28. Borrow 20. Write O under 8, and carry 2. 4×5=20, and 20+2 borrowed=22. 2-22. Borrow 20. 20+2=22, and 22-22=0. Write O under 2, and carry 2. 2x5=10, and 10+2 borrowed=12. 12-12=0. Whence the quotient=385, and the remainder=0. Second example of calculation by this method. 7584 3548728(467 51512 60088 |