divisor remain unchanged, the quotient will be only half as large as before, or 2 ; as (32 • 2) = 8 2; and if the divisor be multiplied by 2, and the dividend remain unchanged, the quotient will be, likewise, only half as large, or 2 ; as 32 (8 X 2) = 2. 83. If the dividend and divisor be both multiplied, or both divided, by the same number, the quotient will not be changed. Thus, if the dividend be 16 and the divisor 4, the quotient will be 4. Now, if we multiply the dividend and divisor by some *number, as 2, their relative values are not changed, and we obtain 32 and 8 respectively, and 32 = 8 = 4, the same as the original quotient. Also, if we divide the dividend and quotient by some number, as 2, their relative values are not changed, and we obtain 16 and 2 respectively, and 16 = 2 = 8, the same quotient as before. 84. If a factor in any number is rejected or cancelled, the number is divided by that factor. Thus, if 24 is the dividend and 6 the divisor, the quotient will be 4. Now, since the divisor and quotient are the two factors which, being multiplied together, produce the dividend (Art. 72), it follows, if we reject or cancel the factor 6, the remaining 4 is the quotient; and, by the operation, the dividend 24 has been divided by 4. CANCELLATION. 85. CANCELLATION is the method of abbreviating arithmetical operations by rejecting any factor or factors common to the divisor and dividend. OPERATION Ex. 1. Sold 19 thousand shingles at 4 dollars a thousand, and received pay in wood at 4 dollars a cord; how many cords of wood was received ? Ans. 19 cords. Having indicated by signs Dividend 4 X 19 the multiplication and di = 19 Quotient. vision required by the quesDivisor 4 tion, then, since dividing both dividend and divisor by the same number will not change the quotient (Art. 83), we divide them by the common factor 4, by cancelling it in both, and obtain 25 for the quotient. 2. Divide the product of 15, 3, 28, and 13, by the product of 7, 30, and 4. OPERATION Dividend 1 $ x 3 x 28 x 13 39 = 194 Quotient. Divisor 7 X3 X4 2 2 The product of the 7 and 4 in the divisor equals the 28 in the dividend; we therefore cancel all these numbers. Finding 15 in the dividend to be a factor of 30 in the divisor, we cancel both of the numbers, and use the remaining factor 2 in place of the 30. There now being no factor common to both dividend and divisor uncancelled, we multiply together the remaining factors in the dividend, and divide the product by the remaining factor in the divisor, and obtain the quotient 191. RULE. Cancel the factor or factors common to the dividend and divisor, and then divide the product of the factors remaining in the dividend by the product of those remaining in the divisor. NOTE. 1. In arranging the numbers for cancellation, the dividend may be written above the divisor, with a horizontal line between them, as in division (Art. 67); or, as some prefer, the dividend may be written on the right of the divisor, with a vertical line between them. NotE. -- 2. Cancelling a factor does not leave 0, but the quotient 1, to take its place, since rejecting a factor is the same as dividing by that factor (Art. 84). Therefore, for every factor cancelled, either in the dividend or divisor, the factor 1 remains. EXAMPLES. 3. Multiply 24 by 16, and divide the product by 12. Ans. 32. 4. Divide 48 by 16, and multiply the quotient by 8. Ans. 24. 5. Divide the product of 7, 10, 12, and 5, by the product of 14, 18, and 6. 6. If 15 be multiplied by 7, 27, and 40, and the product divided by 54 multiplied by 14, 10, and 2, what will be the result ? 7. Divide the product of 13, 15, 20, and 5, by the product of 26, 10, 2, and 3. Ans. 123. 8. Divide the product of 28, 27, 21, 15, and 18, by the product of 7, 54, 7, 3, and 9. 9. How many pounds of butter at 28 cents a pound will be required to pay for 56 pounds of sugar at 11 cents a pound? Ans. 22 pounds. Ans. 71. 10. A. Holmes sold 14 boxes of soap, each containing 24 pounds, at 9 cents a pound, and received for pay 63 barrels of ashes, each containing 3 bushels. What was allowed a bushel for the ashes ? Ans. 16 cents. 11. M. Gardner sold 5 piles of brick, each containing 12 thousand, at 7 dollars a thousand, and was paid in wood, 3 ranges, at 4 dollars a cord. How many cords in each range? Ans. 35 cords. 12. A merchant exchanged 8 cases of shoes, each containing 60 pairs, at 75 cents a pair, for a certain number of casks of molasses, each containing 90 gallons, at 40 cents a gallon. How many casks did he get ? Ans. 10 casks. CONTRACTIONS IN MULTIPLICATION. SECOND OPERATION. 86. A CONTRACTION is the process of shortening any operation. 87. When the multiplier is 13, 14, etc., or 1 with a significant figure annexed. Ex. 1. Multiply 3126 by 14. Ans. 43764. FIRST OPERATION. In the first operation, 31 2 6 X 14 3 1 2 6 we multiply the multi1 2 5 0 4 14 plicand by the 4 units of the multiplier, and 4 3 7 6 4 Ans. 437 6 4 Ans. write the product un der the multiplicand one place to the right. To this partial product we add the multiplicand, since, as it stands, it represents the product of the multiplicand by the 1 ten of the multiplier; and obtain 43764, the answer required. In the second operation, we add in the multiplicand taken as the product by the 1 ten of the multiplier, as we multiply by the 4 units; thus, 6 X 4 24, of which we write down the 4 and carry the 2. 2 x 4 8, + 2 (carried) + 6 (from the multiplicand) 16, of which we write down the 6 and carry the 1. 1 X 4 = 4, +1+ 2= 7, which we write down. 3 x 4 : 12,+1= 13 of which we write down the 3 and carry the 1. 3+1= 4, which we write down; and have as the entire result 43764 as before. RULE. — Write the product by the units' figure of the multiplier under the multiplicand, one place to the right, and add them together. Or, Multiply each figure of the multiplicand by the units' figure of the multiplier, and, after the units' place, and in the preceding figure of the multiplicand. EXAMPLES. Ans. 1071221. 2. Multiply 63013 by 17. Ans. 513084992. OPERATION. 88. When the multiplier is 101, 102, etc., or 1 with one oj more ciphers and a significant figure annexed. Ex. 1. Multiply 8107 by 103. Ans. 835021. We multiply by 3, the units' figure of the 81 07 x 103 multiplicand, and write the product under 2 4 3 2 1 the multiplicand, two places to the right, so that the multiplicand, as it stands over 8 3 5 0 21 Ans. this partial product, will represent the pro: duct of the multiplicand by the 1 hundred of the multiplier; and, adding these, we obtain 835021, the result required. For the reason given, if the multiplier had been such as to have contained one more intervening cipher, we should have written the product by the units' figure three places to the right, and so on, one place farther to the right for every additional intervening cipher. Rule. --- Write the product by the units' figure of the multiplier under the multiplicand, as many places to the right as there are in the multiplier intervening ciphers plus 1; and add them together. EXAMPLES. 2. Multiply 6651 by 108. Ans. 718308. Ans. 11567192. OPERATION. 89. When the multiplier is 21, 31, etc., or 1 with a significant figure prefixed. Ex. 1. Multiply 3113 by 41. Ans. 127633. We multiply by 4, the tens' figure of the 3113 X 41 multiplier, and write the product under the 1 2 4 5 2 multiplicand, one place to the left, so that the multiplicand, as it stands over this par1 2 7 6 3 3 Ans. tial product, will represent the product of the multiplicand by the 1 unit of the multiplier; and, adding these, obtain the result required. RULE. -- Write the product by the tens' figure of the multiplier under the multiplicand, one place to the left, and add them together. EXAMPLES Ans. 679167. 2. Multiply 13317 by 51. 3. Multiply 71389 by 21. 4. Multiply 12062 by 91. Ans. 1097642. OPERATION. 90. When the multiplier is 201, 301, etc., or 1 with one or more ciphers and a significant figure prefixed. Ex. 1. Multiply 14118 by 601. Ans. 8484918. We multiply by 6, the hundreds 1 4118 x 60 1. figure of the multiplier, and write the 8 47 08 product under the multiplicand, two places to the left, so that the multipli8 48 4918 Ans. cand, as it stands over this partial pro duct, will represent the product of the multiplicand by the 1 unit of the multiplier; and adding these we have the answer required. RULE. – Write the product by the hundreds' figure of the multiplier under the multiplicand, as many places to the left as there are in the multiplier intervening ciphers plus 1; and add them together. EXAMPLES. 2. Multiply 8360 by 7001. 3. Multiply 10613 by 801. 4. Multiply 91603 by 2001 Ans. 58528360. 91. When the multiplier or multiplicand has a fraction annexed. Ex. 1. Multiply 426 by 11. By 83. Ans. 3124; 3692. OPERATION. OPERATION. 4 2 6 4 2 6 74 83 2 982 Product by 1. 3 408 Product by 8. 1 4 2 Product by s. 2 84 Product by . 31 2 4 Product ky 71. 3 6 9 2 Product by 8. In multiplying 426 by 7$, we first obtain the product of 426 by 7, and then the product of 426 by }, and, adding these two partial products together, have 3124, the product by 7%. In multiplying by $, we take one third of the multiplicand, by dividing it by 3; thus, 426 x š=426 3.3= 142. In multiplying 426 by 83, we proceed as in the other cae, except in obtaining the product by the fraction; |