102. There are other methods of contracting the operations in multiplication, which, in certain cases, may be resorted to with advantage. Some of the most useful are the following. 44. How many gallons of water will a hydrant discharge in 13 hours, if it discharges 2325 gallons per hour? Operation. Multiplying by the 3 units, we set the 2325 X 13 first figure of the product one place to the 6975 right of the multiplicand. Now, since Ans. 30225 gallons. multiplying by 1 is taking the multiplicand once, (Art. 82,) we add together the multiplicand and the partial product already obtained, and the result is the answer. PROOF.-2325×13=30225 gallons, the same as above. Hence, 103. To multiply by 13, 14, 15, &c., or 1, with either of the other digits annexed to it. Multiply by the units' figure of the multiplier, and write each figure of the partial product one place to the right of that from which it arises; finally, add the partial product to the multiplicand, and the result will be the answer required. Note. This method is the same, in effect, as if we actually multiplied by the 1 ten, and placed the first figure of the partial product under the figure by which we multiply. (Art. 87. II.) 46. Multiply 4028 by 17. 45. Multiply 3251 by 14. 47. Multiply 25039 by 16. 48. Multiply 50389 by 18. 49. If 21 men can do a job of work in 365 days, how long will it take 1 man to do it? Operation. .365 X 21 730 Ans. 7665 days. PROOF.-365X21=7665 days, th same as above. We first multiply by the 2 tens, and set the first product figure in tens' place, then adding this partial product to the multiplicand, we have 7665, for the answer. Hence, 104. To multiply by 21, 31, 41, &c., or 1 with either of the other significant figures prefixed to it. Multiply by the tens' figure of the multiplier, and write the first figure of the partial product in tens' place; finally, add this partial product to the multiplicand, and the result will be the answer required. Note. The reason of this method of contraction is substantially the same as that of the preceding. Operation.. 50. Multiply 4275 by 31. 52. Multiply 38256 by 61. 54. How much will 99 carriages cost, Analysis. Since 1 carriage costs 235 dollars, 100 carriages will cost 100 times 23500 price of 100 C. as much, which is 23500 dollars. (Art. 235 of 1 C. 99.) But we wished to find the cost 23265 of of 99 carriages only. Now 99 is 1 less than 100; therefore, if we subtract the price of 1 carriage from the price of 100, it will give the price of 99 carriages. Hence, 66 99 C. 51. Multiply 7504 by 41. 53. Multiply 70267 by 81. at 235 dollars apiece? 105. To multiply by 9, 99, 999, or any number of 9s. Annex as many ciphers to the multiplicand as there are 9s in the multiplier; from the result subtract the given multiplicand, and the remainder will be the answer required. Note.-The reason of this method is obvious from the fact that annexing as many ciphers to the multiplicand as there are 9s in the multiplier, multiplies it by 100, or repeats it once more than is required; (Art. 99;) consequently, subtracting the multiplicand from the number thus produced, must give the true answer. 55. Multiply 4791 by 99. 57. Multiply 7301 by 999. 59. What is the product of 867 multiplied by 84? 56. Multiply 6034 by 999. 58. Multiply 463 by 9999. Analysis. We first multiply by 4 in the usual way. Now, since 8=4X2, it is plain, if the partial product of 4 is multiplied by 2, it will give the partial product of 8. But as 8 denotes tens, the first figure of its product will also be tens. (Art. 86.) The sum of the two partial products will be the answer required. Operation. 867 84 3468X2 6936 72828 Ans. Note. For the sake of convenience in multiplying, the factor 2 is placed at he right of the partial product of 4, with the sign X, between them. 60. What is the product of 987 by 486? Operation. 987 486 5922 X8 47376 479682 Ans. (61.) 2378 936 PROOF.-987486=479682, the same as above. Hence, 106. When part of the multiplier is a composite number of which the other figure is a factor. First multiply by the figure that is a factor; then multiply this partial product by the other factor, or factors, taking care to write the first figure of each partial product in its proper order, and their sum will be the answer required. (Art. 86.) 21402 OBS. When the figure in thousands, ten thousands, or any other column, is a factor of the other part, or parts of the multiplier, care must be taken to place the first figure of its product under the factor itself, and the first figure of each of the other partial products in its own order. (Art. 86.) Since 48=6X8, we multiply the partial product of 6 by 8, and set the first product figure in tens' place as before. (Art. 86.) X4 85608 2225808 Ans. Operation. 63 45 2835 Ans. 63. Multiply 665 by 82. 65. Multiply 876 by 396. 67. 324325X54426. 69. What is the product of 63 multiplied by 45? Note. By multiplying the figures which produce the same order, and adding the results mentally, we may obtain the answer without setting down the partial products. First, multiplying the units into units, we set down the result and carry as usual. Now, since the 6 tens into 5 units, and 3 units into 4 tens will both produce the same order, viz: tens, (Art. 86,) we multiply them and add their products men tally. Thus, 6×5=30, and 3×4=12; now, 30+12=42, and 1 (to carry) makes 43. Finally, 6X4=24, and 4 (to carry) make 28. PROOF.-63X45-2835, the same as before. Hence, 107. To multiply any two numbers together without setting down the partial products. First multiply the units together; then multiply the figures which produce tens, and adding the products mentally, set down the result and carry as usual. Next multiply the figures which produce hundreds, and add the products, &c., as before. In like manner, perform the multiplications which produce thousands, ten thousands, &c., adding the products of each order as you proceed, and thus continue the operation till all the figures are multiplied. 70. What is the product of 23456789 into 54321? Analytic Operation. 2 3 4 5 6 7 8 9 5 4 3 2 1 2X13X14X15X16X17X18X19X1 2X23X24X25X26×27X28X29X2 2X33X34X35X36X37X38×39X3 2X43X44X45X46X47X48×49×4 2X53X54X55X56X57X58X59X5 12 7 4 1 9 6 2 3 5 2 6 9 Explanation. Having multiplied by the first two figures of the multiplier, as in the last example, we perceive that there are three multiplications which will produce hundreds, viz: 7X1, 8X2, and 9x3; (Art. 86;) we therefore perform these multiplications, add their products mentally, and proceed to the next order. Again, there are four multiplications which will produce thousands, viz: 6×1, 7×2, 8×3, and 9×4. (Art. 86.) We perform these multiplications as before, and proceed in a similar manner through all the remaining orders. Ans. 1274196235269. Note.-1. In the solution above, the multiplications of the different figures are arranged in separate columns, that the various combinations which produce the same order, may be seen at a glance. In practice it is unnecessary to denote these multiplications. The principle being understood. the process of multiplying and adding may easily be carried on in the mind, while the final product only is set down. 2. When the factors contain but two or three figures each, this method is very simple and expeditious. A little practice will enable the student to apply it with facility when the factors contain six or eight figures each, and its application will afford an excellent discipline to the mind. It has sometimes been used when the factors contain twenty-four figures each; but it is doubtful whether the attempt to extend it so far, is profitable. 71. Multiply 25×25. 72. Multiply 54×54. 79. 42634 X 63. 80. 50035 X 56. 83. 80000X25000. 76. Multiply 1234X125. 108. By suitable attention, the critical student will discover various other methods of abbreviating the processes of multiplication. Solve the following examples, contracting the operations when practicable. 99. 12900X14000. 105. 5234 X 2435. 106. 48743000 × 637. 107. 31890420 × 85672. |