38. When the divisor is a composite " number. RULE. Divide by one of the component parts, and the quotient by another, and (if there are more than two) this last quotient by a third, and so on till you have divided successively by all the component parts; the last quotient is the answer *. To estimate the remainder when the divisor consists of two component parts. RULE. Multiply the first divisor by the last remainder, and add the first remainder to the product; the result will be the remainder, which may be placed over the original divisor, as directed in the preceding rule, (Art. 37. B.) u A composite number is that which is the product of two or more num→ bers, each greater than unity, which latter are called its component parts. * Division being the converse of Multiplication, the truth of this rule will be evident from Art. 35. To make it still plainer, we observe, that in ex. 27. the dividend, which is required to be divided by 14, is divided by 2, and the quotient by 7. Now dividing by 2 gives the half, and dividing the half by 7 gives the seventh part of the half, which is evidently the fourteenth part of the whole; whence dividing successively by 2 and 7 is equivalent to dividing by 14. And the same may be shewn in all other cases. y The method of proof is thus accounted for from ex. 27. where every unit in the first quotient evidently contains 2 units of the dividend; consequently each unit in the remainder will likewise be equivalent to 2; and therefore the second remainder must be multiplied by 2, (to make it units of the dividend,) making 10, to which adding 1 (unit of the dividend), the remainder is 11. Likewise in ex. 29, every unit in the first quotient may be considered as containing 11 units of the dividend; every unit in the second, 11 units of the first; and every unit in the third, 11 units of the second; therefore by the foregoing ●bservations 7 x 11+7=84 the remainder, supposing the two last operations 39. If the divisor consist of several component parts, the remainder is found as follows; Multiply the last remainder by the preceding divisor; add the preceding remainder to the product, multiply this sum into the next preceding divisor, and add the next preceding remainder, and so on; the last sum is the remainder, and may be placed over the whole divisor as before. The method of proof is by changing the order of the divisors, that is, by performing the operation again, dividing by that number first which you divided by last in the preceding operation; if both quotients agree, the work is right. 27. Divide 824135 by 14, viz. by 2×7 and 7×2. OPERATIONS. First, 2)824135(1 Explanation. In the first operation, I divide first by 2, and the quotient by 7. I place each remainder to the right 7)412067(5 opposite its respective dividend; and then to find the true remainder, I multiply the first divisor 2 by the last remainder 5, (which make 10,) and add the first remainder 1, making 11 for the whole remainder, which is placed over the divisor at the right hand of the quotient. Quot. .588661 2×5+1=11 Rem. Second, 7)824135(4 In the second operation, I divide first by 7, then by 2)117733(1 2. To find the remainder, I multiply 7 by 1, and Quot. .5886611 add 4 to the product, which gives 11 for the remainder as before. 7x1+4=11 Rem. Both quotients agreeing, and likewise both remainders, the work is right. 28. Divide 813713 by 72 four ways; viz. 6 × 12 ... 12 × 6 ... only. Wherefore also 84 × 11+9=933 the remainder, supposing all three operations and the same may be shewn for four, or any number of successive operations. Another method of finding the remainder, is by multiplying the quotient by the divisor, and subtracting the product from the dividend; the result will be the true remainder. 29. Divide 904682 by 1331; viz. by 11 x 11 x 11. OPERATION. 11).62243(7 11) .7476(7 Quotient 679133T 933 Explanation. The divisors being all alike, the operation cannot be varied. To get the remainder, I multiply the lower 7 by the middle 11, and add the upper 7 to the product, which gives 84. I next multiply 84 by the upper 11, and add in the 9, which gives 9.33 for the remainder; thus 7 X 1! +7=84. Then 84 X 11+9=933. 30. Divide 316718 by 15, or 3 x 5 and 5×3. Quotient 21114, 31. Divide 801234 by 28, or 4 × 7 and 7 × 4. Quotient 28615 1-3. 32. Divide 814776 by 66, or 6×11 and 11×6. Quotient 12345 33. Divide 979488 by 96, or 12x8 and 8x 12. Quotient 10203. 40. When there are ciphers at the right hand of the divisor. RULE. Cut off the ciphers by a thin straight stroke drawn between them and the other figures; and cut off as many figures from the right hand of the dividend as there are ciphers cut off from the divisor; then divide the remaining figures in the dividend by the remaining figures in the divisor. If nothing remain after the operation, the figures cut off from the dividend will be the remainder; but if any thing remain, the figures cut off must be placed to the right of it for a remainder 2. The method of proof is by multiplying the quotient by the divisor, and adding in the remainder; or by dividing by the component parts of the divisor when it can be done. z Cutting off one figure from both terms is equivalent to dividing both by 10; cutting off two figures from cach is equivalent to dividing both by 100, &c. as is obvious from the method of Notation; and it is plain, that as often as the whole divisor is contained in the whole dividend, so often must any part of the divisor be contained in a like part of the dividend. Now the cutting off directed in the rule is nothing more than taking like parts of both, the figures cut off from the dividend being (from the nature of Notation) the right hand figures of the remainder; wherefore the rule is evident. This useful mode of contraction saves a multitude of ciphers, and shortens the work amazingly. 34. Divide 1571234 by 20. OPERATION. 2|0)1571234( Proof 20 1571234 Explanation. Here having cut off the cipher from the divisor, and the last figure 4 from the dividend, I divide the Quotient 78561 remaining figures in the latter by 2. Having divided the 3, I have I remaining, which is placed before the 4 cut off, making the remainder 14; this is placed over 20, the divisor, at the end of the quotient, as in the former rules. The proof arises by multiplying the quotient by the divisor 20, (Art. 30.) and adding in the remainder 14. For other proofs, divide by 4 × 5 and 5 X 4. Art. 39. 36. 3/00)100123 43 4000)791236|131( 50000)12345678( 35. 37. 24645538 12345678 Having cut off the cipher from the divisor, for the 21 I divide by 3X7 and 7 × 3, having first cut off the 7 from the dividend. For the remainder, in the first operation, I say 6 times 3 are 18, and place it before the 7, maksay twice 7 are 14 and 4 are 18; this which in both is placed over the divisor. Quotient 4113533. Quotient 17636. 39. Divide 2468123 by 60. 41. When the divisor is any number greater than 12. RULE I. Having placed the divisor on the left as before, find how many times it is contained in the least number of figures possible of those on the left of the dividend, and set the quotient figure to the right. II. Multiply the divisor by the quotient figure, and set the product under the forementioned left hand figures of the dividend. II. Subtract the said product from the figures above it, and to the right of the remainder bring down the next figure (viz. the next to those already used) in the dividend, and find how often the divisor is contained in this number. IV. Set the quotient figure to the right of the dividend as before; multiply the divisor by it; place the product under the last found number; subtract, bring down, divide, &c. as before, until all the figures in the dividend are brought downa. a This rule differs from the operation directed in Art. 37 B. only in one particular, namely, there the work is for the most part performed mentally, whereas here the whole process is put down. Learners usually experience some difficulty in finding how many times the divisor will go in the figures to be divided; the rule is, when the divisor is contained in its own number of figures, find how often the first figure of the divisor is contained in the first of those to be divided; but when the number of figures in the divisor is one less than the number of figures to be divided, find how often the first figure of the former will go in the two first of the latter. This method, if proper allowances be made for carrying, serves to determine the quotient figures very readily. Another method is, to multiply the divisor (previous to the operation) by 2, 3, 4, &c. to 9, and set the products in their order in a column under the divisor, and their respective multipliers opposite; then at every step of the work you have only to take that product which is nearest to, but not greater than, the number to be divided; place it under the said number, and put the opposite multiplier in the quotient. For example, to divide 42021687 by 23; thus, 1.... 23) 42021687 (1827029 2.. 46 23 3....69 190 4....92 184 5.. 115 6.. 138 62 7.. 161 46 8.. 184 161 9.. 207 161 68 46 227 207 20 Here the left hand column contains the multipliers of the divisor; the second column contains the products, each opposite its proper multiplier; to the right of this stands the operation. First then, to find how often 23 goes in 42, I look among the products, and find that 23 is the only one not greater than 42; I therefore put 23 under the 42, and its opposite multiplier 1 in the quotient. Having subtracted, and brought down the 0, the number next to be divided is 190; the nearest number not greater than this among the products is 184; this I put below 190, and its multiplier 8 in the quotient: the next number which arises to be divided is 62; the nearest product for this is 46, which I put below it, putting the corresponding multiplier 2 in the quotient : I proceed in this manner until the work is finished. The first method of proof is obvious; for it is plain, that if the dividend contain the divisor some number of times exactly, that number will be the quotient; and (this rule being the converse of Multiplication) the divisor multiplied by the quotient will produce the dividend exactly when the work is right. But if there be a remainder, it is plain that the dividend exceeds the aforementioned product by that remainder, which therefore must be added to the product in order to produce the dividend. The second method by casting out the nines will be demonstrated algebrai eally in its proper place. The third method of proof by adding, depends on this consideration, that the |