another number, as 5, the new product will be 5 times 3 times 17, which is evidently 15 times 17. Hence, 17 X 3 X 5 = 17 x 15; the same reasoning would extend to three or more multipliers. 2d. Since 5 times 3 is equal to 3 times 5, (82, I), it follows that 17 multiplied by 5 times 3 is the same as 17 multiplied by 3 times 5; or 17 x 3 x 5= 17 X 5 X 3. Hence, the product is not changed by changing the orders of the factors. These principles may be stated as follows: I. If a given number be multiplied by several factors in continued multiplication, the result will be the same as if the given number were multiplied by the product of the several multipliers. II. The product of several factors in continued multiplication will be the same, in whatever order the factors are taken. CONTRACTIONS IN MULTIPLICATION. CASE I. = 18; or, 96. When the multiplier is a composite number. A Composite Number is one that may be produced by multiplying together two or more numbers. Thus, 18 is a composite number, since 6 x 3 9 X 2 18; or, 3 x 3 x 2= 18. 97. The Component Factors of a number are the several numbers which, multiplied together, produce the given number; thus, the component factors of 20 are 10 and 2 (10 x 2 = 20); or, 4 and 5 (4 x 5= 20); or, 2 and 2 and 5 (2 x 2 x 5= 20). Note.—The pupil must not confound the factors with the parts of a number. Thus, the fuctors of which 12 is composed, are 4 and 3 (4 X 3 = 12); while the parts of which 12 is composed are 8 and 4 (8+4=12); or 10 and 2 (10+2=12). The factors are multiplied, while the parts are added, to produce the number. 98. 1. Multiply 327 by 35. OPERATION. 7 2289 5 ANALYSIS. The factors of 35 are 7 and 5. We multiply 327 by 7, and this result by 5, and obtain 11445, which must be the same as the product of 327 by 5 times 7, or 35. (95, I). Hence we have the following 11445 RULE. I. Separate the composite number into two or more factors. II. Multiply the multiplicand by one of these factors, and that product by another, and so on until all the factors have been used buccessively; the last product will be the product required. Note.—The factors may be used in any order that is most convenient, (95, II). EXAMPLES FOR PRACTICE. 1. Multiply 736 by 24. Ans. 17664. 2. Multiply 538 by 56. Ans. 30128. 3. Multiply 27865 by 84. 4. Multiply 7856 by 144. Ans. 1131264. 5. What will 56 horses cost at 185 each ? 6. If a river discharge 17740872 cubic feet of water in one hour, how much will it discharge in 96 hours ? Ans 1703123712 cubic feet. CASE II. 99. When the multiplier is a unit of any order. If we annex a cipher to the multiplicand, each figure is removed one place toward the left, and consequently the value of the whole number is increased tenfoid, (57, III). If two ciphers are annexed, each figure is removed two places toward the left, and the value of the number is increased one hundred fold; and every additional cipher increases the value tenfold. Hence, the RULE Annex as many ciphers to the multiplicand as there are ciphers in the multiplier. EXAMPLES FOR PRACTICE. 1. Multiply 364 by 100. Ans. 36400. 2. Multiply 248 by 1000. Ans. 248000. 3. What cost 1000 head of cattle at 50 dollars each? 4. Multiply one million by one hundred thousand ? 5. How many letters will there be on 100 sheets, if each sheet have 100 lines, and each line 100 letters ? Ans. 1000000. CASE III. OPERATION. 100. When there are ciphers at the right hand of one or both of the factors. 1. Multiply 7200 by 40. Analysis. The multiplicand, factored, is 7200 equal to 72 x 100; the multiplier, factored, is 40 equal to 4 x 10; and as these factors taken in 288000 any order will give the same product, (95, II), we first multiply 72 by 4, then this product by 100 by annexing two ciphers, and this product by 10 by annexing one cipher. Hence, the following RULE. Multiply the significant figures of the multiplicand by those of the multiplier, and to the product annex as many ciphers as there are ciphers on the right of both factors. EXAMPLES FOR PRACTICE. Ans. 222000. Ans. 8640000. 1. Multiply 740 by 300. 2. Multiply 36000 by 240. 3. Multiply 20700 by 500. 4. Multiply 4007000 by 3002. 5. Multiply 300200 by 610. Ans. 12029014000. CASE IV. OPERATION. 101. When one part of the multiplier is a factor of another part. 1. Multiply 4739 by 357. ANALYSIS. In this example, 7, one 4739 part of the multiplier, is a factor of 357 35, the other part. We first find, in the usual manner, the product of the 33173 Prod. by 7 units. multiplicand by the 7 units; multi165865 Prod. by 35 tens plying this product by 5, and writing 1691823 Ans. the first figure of the result in tens' place, we obtain the product of the multiplicand by 7 x 5 x 10 = 35 tens; and the sum of these two par tial products must be the whole product required. OPERATION. 2. Multiply 58327 by 21318. ANALYSIS. In this exam 58327 ple, the 3 hundreds is a factor 21318 of 18, the part on the right of it, and also of 21, the part on 174981 Prod. by 3 hundreds. the left of it. We first mul1049886 Prod. by 18 units. tiply by 3, writing the first 1224867 Prod. by 21 thousands. figure in hundreds' place; 1213414986 Ans. multiplying this product by 6, and writing the first figure in units' place, we obtain the product of the multiplicand by 3 X 6 = 18 units; multiplying the first partial product by 7, and writing the first figure in thousands' place, we obtain the product of the multiplicand by 7 X 3 X 1000 = 21 thousands, and the sum of these three partial products must be the entire product required. Note.— The product obtained by multiplying any partial product is called a derived product. 102. From these illustrations we have the following RULE. I. Find the product of the multiplicand by some figure of the multiplier which is a factor of one or more parts of the multiplier. II. Multiply this product by that factor which, taken with the figure of the multiplier first used, will produce other parts of the multiplier, and write the first figure of each result under the first figure of the part of the multiplier thus used. III. In like manner, find the product, either direct or derived, for every figure or part of the multiplier; the sum of all the products will be the whole procluct required. 8. Multiply 3578426785 by 64532164. 9. Multiply 2703605 by 4249784. 10. What is the product of 9462108 multiplied by 16824? Ars. 159,190,504,992. EXAMPLES COMBINING THE PRECEDING RULES. 1. A man bought two farms, one containing 175 acres at $28 per acre, and the other containing 320 acres at $37 per acre; what was the cost of both ? Ans. $16,740. 2. If a man receive $1209 salary, and pay $364 for board, $275 for clothing, $150 for books, and $187 for other expenses, how much can he save in 5 years ? Ans. $1,120. 3. Two persons start from the same point, and travel in opposite directions; one travels 29 miles a day, and the other 32 miles. How far apart will they be in 17 days? Ans. 1,037 miles. 4. A drover bought 127 head of cattle at $34 a head, and 97 head at $47 a head, and sold the whole lot at $40 a head; what was his entire profit or loss? Ans. $83 profit. 5. Multiply 675—(77 + 56) by (3 x 156)—(214—28). Ans. 152844. 6. Multiply 98 + 6 x (37 + 50) by (64 - 50) x 5 – 10. Ans. 37200. 7. What is the product of (1+ x 25) — (9 X 36) + 4324 x (280 — 112) + (376 + 42) 4? Ans. 8,004,000. 8. In 1850 South Carolina cultivated 29967 farms and plantations, containing an average of 541 acres each, at an average value of $2751 for each farm ; New Jersey cultivated 23905 farms, containing an average of 115 acres each, at an average value of $5030 per farm. How much more were the farming lands of the latter valued at, than those of the former ? 9. There are in the United States 1922890880 acres of land; of this there were reported under cultivation, in 1850, 1419075 farms, each embracing an average of 203 acres. How many acres were still uncultivated ? 10. Each of the above farms in the United States was valued at an average of $2258, and upon each farm there was an average |