IS ses? By the last rule, 4 times 472=1,888. Thus we multiply by a larger number, by multiplying successively by two smaller ones. But the two smaller ones must be such that their product shall be just equal to the larger one ; that is, the two smaller numbers must be factors of the larger one. THE FINDING OF TWO OR MORE NUMBERS, WHOSB PRODUCT SHALL BE EQUAL TO A GIVEN NUMBER, CALLED RESOLVING THE GIVEN NUMBER INTO FACTORS. After thoroughly understanding the foregoing, let the pupil perform the following EXAMPĻES FOR PRACTICE. 3. At 21 dollars a hogshead, what cost 284 hogsheads of molas. Ans. 5,964 dolls. 4. What cost 28 bales of cotton at 75 dollars a bale ? Ans. 2,100 dolls. 5. If one cask of brandy cost 15 dolls., what cost 321 ? Ans. 4,815 dolls. 6. If 1 pound of coffee cost 27 cents, what cost 1,234 ? Ans. 333 dolls. 18 cts. 7. If 1 barrel of sugar cost 35 dolls., what cost 11,962 ? Ans. 418,670 dolls. 8. At 42 dollars an acre, what cost 825 acres of ground ? Ans. 34,650 dolls. 9. At 81 cents per pound, what cost 1,437 pounds of tea ? Ans. 1,163 dolls. 97 cts. 10. At 63 cents per pound, what cost 127 pounds of tobacco ? Ans. 80 dolls. 01 ct. 11. 275x18 Ans 4,950 | 18. 9,876X 49 Ans. 483,924 12. 164X 45 7,380 | 19. 21,378 X 36 769,608 13. 40,073X54 2,163,942 ) 20. 209,402 x 72 15,076,944 14. 764,131 X 48 36,678,288 21. 786,340 X 56 15. 91,738X81 7,430,778 22. 92,810,333 X 48 16. 3,002X28 84,056 23. 987,543,221 x 63 17. 4,381x32 140,192 | 24. 9,895,461,343X49 So far, our examples have been performed by dividing or resolving the multiplier into two factors. But, we may resolve it into three, four, or any number of factors, and proceed in the same way. For example, 25. How many quarts of grain in 42 sacks, each sack containing 64 quarts? We know that 3 times 7 are 21, and that 2 times 21 are 42. Therefore, 42 may be resolved into the factors 2, 3, and 7. 64 First, multiply 64 by 7, thus, 2688 Ans. This answer, viz. 2,688, is 42 times: 64, because twice 64 multiplied by 3, (that is, taken 3 times,) is 6 times 64; and 6 times 64 multiplied by 7, (that is, taken 7 times,) is 42 times 64. 26. Find how many gallons are contained in 16 casks, each hold. ing 58 gallone. times 2 are 4; 2 times 4 are 8; and 2 times 8 are 16. We may, therefore, resolve 16 into the 4 factors, 2, 2, 2, and 2. A. 928. When the above is thoroughly understood, let the pupil proceed to the following EXAMPLES FOR PRACTICE. 27. At 108 dollars an acre, what cost 34,462 acres of ground ? 108=3X4X9. Ans. 3,721,896. 28. At 144 dollars a pipe, what cost 615,243 pipes of wine ? 144-4X4X9. Ans. 88,594,992. 29. What cost 223,332 bales of cotton at 105 dollars a bale ? 105=7X3X5. Ans. 23,449,860. 30. At 64 dolls. a ton, what cost 87,543 tons of iron ? 64=4X4X4. Ans. 5,602,752. 31. At 72 cents per pound, what cost 507,832 pounds of aqua. fortis ? 72=3X3X8. Ans. 365,639 dolls. 04 cts. 32. At 56 cts. per pound, what cost 342,516 pounds of snuff? 56=2X4X7. Ans. 191,808 dolls. 96 cts. 33. At 1 dollar and 40 cents, that is, 140 cts. per pound, what cost 41,364 pounds of tea ? 140=2X2X7X5. Ans. 5,790,960 cts. 34. At 125 cents, or 1 dollar and 25 cts. per bushel, what cost 22,874 bushels of rye ? 125=5X5X5. Ans, 2,859,250 cts. 35. Multiply 78,946 by 84. 39. 72,863,452 X 72. 36. 839,675X96. 40. 81.998,473 X 64. 37. 2,847,333X54. 41. 91,339,478X81. 38. 9,999,888X56. 42, 637,853,962X108, The learner will soon find, that there are some numbers which he cannot resolve into factors. We shall teach him how to multiply by these hereafter. A NUMBER WHICH CAN BE RESOLVED INTO FACTORS, (that is, one which is the product of two or more other numbers,) IS CALLED A COMPOUND OR COMPOSITE NUMBER. The factors which produce the composite number are sometimes called its component parts. Hence, to multiply by a composite number, RESOLVE THE MULTIPLIER INTO FACTORS, AND MULTIPLY BY THESE FACTORS SUCCESSIVELY. [It may here be suggested, once for all, that the teacher should always require an explanation of the language employed. If the pupil hesitates, he should be referred to that part of the book which explains it.] § XIII. 1. What cost 13 tons of potash, at 139 dollars a ton ? Here 13, the multiplier, is more than 12, and is not a composite number. First find, then, what 6 tons cost, thus, 139 6 834 Then, what 7 tons cost, thús, 139 973 Now 6 tons and 7 tons are 13 tons. Therefore, add the two products together, thus, 834 973 1,807 dollars, is, therefore, the price of 13 tons. 1807 2. At 227 dollars an acre, what cost 23 acres of land ? A. 5,221. Then, when the multiplier is more than 12, and is not a composite number, we may SEPARATE THE MULTIPLIER INTO SUCH PARTS THAT NONE OF THEM SHALL EXCEED 12, MULTIPLY BY THOSE PARTS SEPARATELY, AND ADD TOGETHER THE SEVERAL PRODUCTS. EXAMPLES FOR PRACTICE. 3. What cost 123 barrels of flour at 17 dollars a barrel ? Ans. 2,091 dolls. 4. At 13 cents per pound, what cost 175 pounds of sugar ? Ans. 22 dolls. 75 cts. 5. At 19 cents a quart, what cost 245 quarts of gin ? Ans. 46 dolls. 55 cts. 6. At 34 cents per pound, what cost 141 pounds of chocolate ? Ans. 47 dolls 94 cents. 7. 206X39 Ans. 8,034 | 13. 57,895X31 8. 205 X23 4,715 14. 893,654X23 9. 876X29 25,404 | 15. 900,009X17 10. 9,893x31 16. 111,222 29 11. 7.640X37 17. 34,756,810X43 12. 63,994X41 18. 450,000,354X 47 XIV. The following mode, however, will be found more convenient. 1. At 134 dollars a pipe, what cost 31 pipes of brandy? Here 31, the multiplier, is not a composite number ; but 30 is, having the factors 3 and 10. Now, if we multiply by 3 and 10 successively, we shall obtain the price of 30 pipes; to this we can add the price of 1 pipe, and we shall then have the price of 31 pipes, the answer required. Multiply then, by 30, multiplying first by 3; 134 3 thus, Then by 10; thus, 402 10 This is the price of 30 pipes ; add the price of 1 more ; thus, 4020 134 Therefore, 4,154 dollars is the price of 31 pipes. 4151 2. In 47 boxes, each containing 348 candles, how many candles ? Here 47, the multiplier, is not a composite number : but 45 is, having the factors 9 and 5. Ans. 16,356. Hence, when the multiplier is not a composite number, I. TAKE A CONVENIENT COMPOSITE NUMBER LESS THAN THE MULTIPLIER, 11. MULTIPLY BY IT ACCORDING TO THE RULE. III. TO THE PRODUCT, ADD AS MANY TIMES THE MULTIPLICAND, AS THE COMPOSITE NUMBER CONTAINS UNITS LESS THAN THE MULTIPLIER. EXAMPLES FOR PRACTICE, 3. If 1 acre of ground cost 23 dollars, what cost 62,123 acres ? Ans. 1,428,829 dollars. 4. If 1 pipe of wine contain 1,008 pints; what will 19 pipes con. tain ? Ans. 19,152. 5. Jf, in digging a canal, 35,432 cubic yards are excavated in 1 day, how much will be excavated in 29 days ? Ans. 1,027,528 yds. 6. If33 dollars be paid a soldier, how much must be paid an army of 65,217 men ? Ans. 2,152,161. --+ 7. Multiply 151x29. Ans. 4,379 | 13. 28,933 241 14. 36,856X37 9. 7,512X17 127,704 15. 72,945 x 47 10. 2,931X23 67,413 | 16. 100,001 X 53 11. 2,004X31 62,124 17. 200,002X 43 12. 19,070 X13 247,910 | 18. 404,505X61 Ş XV. 1. At 10.cents a pound, what cost 53 pounds of butter ? Ilere 10 is the multiplier, By the first rule, 53 10 530 Here the product is the same as the multiplicand, with a cypher annexed. 2. At 10 cents apiece, what will 25 oranges cost? 25 250 Here, again, the product is the same as the multiplicand, with a cypher annexed. It seems, then, that annexing a cypher at the right of any number, multiplies it by 10. The reason of this is very plain, when we consider that a figure in the tens' place, is ten times as great as the same figure in the units' place; because numbers increase towards the left, in ten-fold proportion. Also, A figure in the hundreds' place, is ten times as great as the same figure in the tens' place. And, universally, |