In practice, the name of the order of units may be omitted. Thus, in the operation we can say: 7 times 4 are 28; we write the 8, and add the 2 to the next product: 7 times 6 are 42, and 2 are 44; we write 4, and add 4 to the next product; 7 times 5 are 35, and 4 are 39; answer, 3948. 2. Let it be required to find the product of 736 by 206. OPERATION. Multiplicand, 736 Product, 151616 For convenience, we write the multiplier under the multiplicand, so that figures of the same order stand in the same column; and multiplying by the units, as in the preceding operation, we obtain 4416. There being 0 tens, we write a cipher in the order of tens underneath, and pass to the hundreds' figure of the multiplier. 2 hundreds are 2 hundred units, and 2 hundred times 6 units are 12 hundreds, or 1 thousand and 2 hundreds. We write the 2 hundreds, and reserve the 1 thousand to add to the next product. 2 hundred times 3 tens are 6 hundred tens or 6 thousands, and the 1 thousand added are 7 thousands, which we write. 2 hundred times 7 hundreds are 14 hundred-hundreds, equal 1 hundred-thousand, and 4 ten-thousands, which we write, and obtain 1472 hundreds, or, with the cipher on the right, 14720 tens. Adding the two partial products we have for the entire product 151616. Therefore, the product of 736 by 206 is 151616. RULE. Write the multiplier under the multiplicand, so that units may stand under units, tens under tens, etc. If the multiplier contains but one order of units, beginning at the right multiply each order of the multiplicand by it, riting the right-hand figure of each product underneath, adding the numbers expressed by the other figures, if any, to the next product, observing to write all the figures of the last product. If the multiplier contains more than one order of units, REVIEW QUESTIONS. What is a Sign? (38) The Sign of Addition? (38) Of Equality? (38) Of Subtraction? (43) Of Multiplication? (48) multiply by each of the orders, successively, writing the righthand figure of each partial product under the order used. The sum of the partial products will be the entire product. PROOF. Multiply the multiplier by the multiplicand; and, if the product is the same as that first obtained, the work is supposed to be correct. Or, Separate the multiplier into parts and make each of them a multiplier, and, if the sum of the products equals the first product, the work is supposed to be correct. 24. 8320 by 13. Ans. 108160. | 36. 81201 by 73. Ans. 5927673. 37. 75452 × 47 = how many ? 38. 54302 × 89 = how many ? Ans. 3546244. Ans. 4832878. Ans. 159152. 39. 784 X 203 = how many? 40. What is the product of 137 by 35? 46. Multiply 34618 by 259. 47. Multiply 80704 by 432. Ans. 4795. Ans. 61236. Ans. 4625. Ans. 8654042. Ans. 125874. Ans. 8966062. Ans. 34864128. 48. Multiply thirty-one thousand three hundred and eleven by one thousand two hundred and thirteen. 49. Multiply ninety-three thousand one hundred and eighty six by four thousand four hundred and fifty-five. Ans. 415143630. REVIEW QUESTIONS. What is the answer called in Addition? (37) In Subtractio? (42) In Multiplication? (47) 52. When there are ciphers between significant figures in the multiplier, the operation may be shortened by passing over each 0 of the multiplier. 53. When the multiplier consists of two significant figures, with or without intervening ciphers, and begins or ends with 1, we may consider the multiplicand as a product by the 1, and write the other partial product as many orders to the right or left as is required by the multiplier. In example 56, the one partial product is units and the other tens, and in example 57, the one partial product is thousands and the other units; and they are so written that, in each case, the sum of the partial products may be the required product. 55. Multiply 3403 by 501. Ans. 1704903. How may you multiply when there are ciphers between the significant figures of the multiplier? When either of the two significant figures of the multiplier is 1? 56. Multiply 5121 by 1002. Ans. 5131242. Ans. 42973403. 57. Multiply 61303 by 701. 54. When the multiplier is 10, 100, 1000, etc., the product may be obtained, at once, by annexing to the multiplicand as many ciphers as there are in the multiplier, and regarding the decimal point as removed an equal number of places to the right. For, the value expressed by figures is made tenfold by each removal of them an order to the left. (Art. 30.) Thus, 2 X 10 = 20, 2 × 100 =200, etc. 58. Multiply 619 by 100. 59. Multiply 11644 by 1000. 60. 45681000000= how many? Ans. 61900. Ans. 11644000. Ans. 4568000000. 55. When there are ciphers on the right of either or both of the factors, we may multiply without reference to them, and annex to the product as many ciphers as there are on the right of both factors. 61. Find the product of 2050 by 1300. The second operation is evidently the same as the first, except that the ciphers on the right are not written until the partial products are added. 62. Multiply 485 by 240. 63. Multiply 36500 by 730. Ans. 116400. Ans. 26645000. 64. Multiply six hundred seventy-four thousand and two hundred by two thousand one hundred and four. Ans. 1418516800. When the multiplier is 10, 100, 1000, etc.? How do you multiply when there are ciphers at the right of either or both of the factors? |