Thus, 21 is a composite number: the factors are 7 and 3, because 7 multiplied by 3 produces 21. So, also, 12 is a com posite number; 6X2=12; or, 3X4=12; or, 2X2X3=12. 1. What will 15 oranges cost, at 8 cents each? Since 5X3-15, it follows that 15 is a composite number of which the factors are 5 and 3. Cost of 1 orange, 8 cents. 5 Cost of 5 oranges, 40 cents ANALYSIS.-Since 15 are 3 times 5, 15 oranges will cost 3 times as much as 5 oranges. Therefore, instead of multiplying 8 by 15, first find the cost of 5 oranges, by multiplying 8 cents by 5; then take 3 times that product for the cost of 15 oranges. PROOF.-8×15=120. 3 Cost of 15 oranges, 120 cents. Rule for Case I.-Separate the multiplier into two or more factors. 2. Multiply the multiplicand by one of the factors, and this product by another factor, till every factor is used; the last product will be the one required. REM. Do not confound the factors of a number with the parts into which it may be separated. Thus, the factors of 15 are 5 and 3, while the parts into which 15 may be separated, are any numbers whose sum equals 15; as, 14 and 1; or, 2, 9, and 4. 4 EXAMPLES FOR PRACTICE. 2. What will 24 acres of land cost, at $124 an acre ? Ans. $2976. 3. How far will a ship sail in 56 weeks, at the rate of 2395 miles per week? Ans. 134120 miles. 4. How many pounds of iron are there in 54 loads, each weighing 2873 pounds? Ans. 155142 pounds. 5. How many gallons of wine in 63 vats, each containing 1673 gallons? 6. Multiply 2874 by 72. 7. Multiply 8074 by 108. Ans. 105399 galls. Ans. 206928. Ans. 871992, Exp CASE II. ART. 34. When the Multiplier is 1 with ciphers annexed to it; as, 10, 100, 1000, &c. Placing one cipher on the right of a number, (Art. 8), changes the units into tens, the tens into hundreds, and so on, and, therefore, multiplies the number by 10. Annexing two ciphers, changes units into hundreds, tens into thousands, &c., and multiplies the number by 100. Annex one cipher to the right of 25, and it becomes 250, the product arising from multiplying it by 10. Annex two ciphers to 25, and it becomes 2500, the product of 25X100. Hence the Rule for Case II.-Annex as many ciphers to the Multiplicand as there are ciphers in the multiplier, and the number thus formed will be the product required. ART 35. When there are ciphers at the right hand of one or both of the factors. Find the product of 2300 × 170. ANALYSIS. The number 2300 may be regarded as a composite number, of which the factors are 23 and 100; and 170 as a composite number, of which the factors are 17 and 10. 2300 170 161 23 391000 By Art. 33, the product of 2300 by 170 will be found by multiplying 23 by 17, and this product by 100, and the resulting product by 10; that is, by finding the product of 23 multiplied by 17, and then annexing to the product 3 ciphers, as there are 3 ciphers at the right of both factors. Hence the Rule for Case III.—Multiply without regarding the ciphers on the right of the factors; then annex to the product as many riphers as are at the right of both factors. 10. Multiply 904000 by 10200. For additional problems, see Ray's Test Examples. Ans. 9220800000. V. DIVISION. ART. 36. 1. If you divide 6 apples equally between 2 boys, how many will each have? ANALYSIS.-It will require two apples to give each boy 1; hence, each boy will have as many apples as there are times 2 apples in 6 apples, that is 3. How many times 2 in 6?—Ans. 3. Why? times 2 are 6. Because 3 2. If you divide 8 peaches equally between 2 boys, how many will each have? Ans. 4. Why? 3. How many times 2 in 10? Why?" REVIEW.-33. What is a composite number? What are its component parts or factors? Give an example. ber, Rule? Iilustrate this method. How multiply by a composite numREM. In what respect do the factors of a number differ from its parts? Give an example. 84. If one cipher is placed on the right of a number, how are the orders changed? If two ciphers? How multiply by 10, 100, 1000, &c.? CASE II. ART. 34. When the Multiplier is 1 with ciphers annexed to it; as, 10, 100, 1000, &c. Placing one cipher on the right of a number, (Art. 8), changes the units into tens, the tens into hundreds, and so on, and, therefore, multiplies the number by 10. Annexing two ciphers, changes units into hundreds, tens into thousands, &c., and multiplies the number by 100. Annex one cipher to the right of 25, and it becomes 250, the product arising from multiplying it by 10. Annex two ciphers to 25, and it becomes 2500, the duct of 25X100. Hence the pro Rule for Case II.-Annex as many ciphers to the Multiplicand as there are ciphers in the multiplier, and the number thus formed will be the product required. |