SECTION IV. MULTIPLICATION. ART. 33. Ex. 1. What will three lemcns cost, at 2 cents apiece? Analysis. Since 1 lemon costs 2 cents, 3 lemons will cost 3 times 2 cents; and 3 times 2 cents are 6 cents. Therefore, 3 lemons, at 2 cents apiece, will cost 6 cents. OBS. The preceding operation is a short method of finding how much 2 cents will amount to, when repeated or taken 3 times, and is called Multiplication. Thus, 2 cents + 2 cents + 2 cents are 6 cents. Hence, 34. MUTIPLICATION is the process of finding the amount of a number repeated or added to itself, a given number of times. The number to be repeated or multiplied, is called the multiplicand. The number by which we multiply, is cailed the multiplier, and shows how many times the multiplicand is to be repeated or taken. The answer, or number produced by multiplication, is called the product. Thus, when we say 5 times 7 are 35, 7 is the multipli cand, 5 the multiplier, and 35 the product. OBS. When the multiplicand denotes things of one kind, or denomination only, the operation is called Simple Multiplication. QUEST.-34. What is multiplication? What is the number to be repeated or multiplied called? What the number by which we multiply? What does the multiplier show? What is the answer called? When wo say 5 times 7 are 35, which is the multiplicand? Which the multiplier? Which the product? Obs. When the multiplicand denotes things of one denomination only, what is the operation called ? 38. The product cf any two numbers will be the same, whichever factor is taken for the multiplier. Thus, If a garden contains 3 rows of trees as represented by the number of horizontal rows of stars in the margin, and each row has 5 trees as represented by the number of stars in a row, it is evident, that the whole *** number of trees in the garden is equal either to the number of stars in a horizontal row, taken three times, or to the number of stars in a perpendicular row taken five times; that is, equal to 5 × 3, or 3×5. EXAMPLES. 39. When the multiplier contains but ONE figure. Ex. 1. What will 3 horses cost, at 123 dollars apiece? Analysis. Since 1 horse costs 123 dollars, 3 horses will cost 3 times 123 dollars. - Operation. 3 multiplier. Dolls. 369 product. Directions.-Write the multiplier under the multiplicand; then, beginning at the right hand, multiply each figure of the multiplicand by the multiplier. Thus, 3 times 3 units are 9 units, or we may simply say 3 times 3 are 9; set the 9 in units' place under the figure multiplied. 3 times 2 are 6; set the 6 in tens' place. 3 times 1 are 3; set the 3 in hundreds' place. Note. The pupil should be required to analyze every example, and to give the reasoning in full; otherwise the operation is liable to become mere guess-work, and a habit is formed, which is alike destructive to mental discipline and all substantial improvement. Solve the following examples in a similar manner : 10. What will 6 cows cost at 23 dollars apiece. Suggestion.In this example the product of the different figures of the multiplicand into the multiplier, exceeds 9; we must therefore write the units' figure under the figure multiplied, and carry the tens to the next product on the left, as in addition. Thus, begin Operation. 23 dolls. 6 ning at the right hand as before, 6 times 3 units are 18 units, or we may simply say 6 times 3 are 18. Now it requires two figures to express 18; we there- Ans. 138 dollars. fore set the 8 under the figure multi plied, and reserving the 1, carry it to the product of the next figure, as in addition. (Art. 23.) Next, 6 times 2 are 12, and 1 (to carry) makes 13. Since there are no more figures to be multiplied, we set down the 13 in full. The product is 138 dollars. Hence, 40. When the multiplier contains but one figure. Write the multiplier under the multiplicand, units un der units, and draw a line beneath them. Begin with the units, and multiply each figure of the multiplicand by the multiplier, setting down the result arıl carrying as in addition. (Art. 23.) Multiply the following numbers together. 25. What will 115 barrels of flour cost, at 6 dollars per barrel? 26. A man bought 460 pair of boots, at 5 dollars a pair: ow much did he pay for the whole? 27. What cost 196 acres of land, at 7 dollars per acre? 28. What cost 310 ploughs, at 8 dollars apiece? 29. What cost 691 hats, at 7 dollars apiece? 30. What cost 865 heifers, at 9 dollars per head? 31. What cost 968 cheeses, at 8 dollars apiece? 32. What cost 1260 sheep, at 7 dollars per head? 33. What cost 9 farms, at 2365 dollars apiece? 41. When the multiplier contains more than ONE figure. 34. A man sold 23 sleighs, at 54 dollars apiece: how much did he receive for them all? Suggestion.-Reasoning as before, if 1 sleigh costs 54 dollars, 23 sleighs will cost 23 times as much. Directions.-As it is not convenient to multiply by 23 at once, we first multiply by the 3 units, then by the 2 tens, and add-the two results together. Thus, 3 times 4 are 12, set the 2 under the Operation. 162 cost of 3 s. 108 66 "20 s. Dolls. 1242 66 "23 s. figure 3, by which we are multiplying, and carry the 1 as above. 3 times 5 are 15, and 1 (to carry) makes 16. Next, we multiply by the 2 tens thus: 20 times 4 units are 80 units or 8 tens; or we may simply say 2 times 4 are 8. Set the 8 under the figure 2 by which we are multiplying, that is, in tens' place, because it is tens. 2 times 5 are 10. Finally, adding these two products together as they stand, units to units, tens to tens, &c., we have 1242 dollars, which is the whole product required. Note. When the multiplier contains more than one figure, the several products of the multiplicand into the separate figures of the multiplier, are called partial products. 35. Multiply 45 by 36, and prove the operation. 36. What is the product of 234 multiplied by 165? Suggestion.-Proceed in the same man ner as when the multiplier contains but two figures, remembering to place the right hand figure of each partial product directly under the figure by which you multiply. Operation. 234 165 1170 1404 234 38610 Ans 37. What is the product of 326 multiplied by 205 ? Suggestion.-Since multiplying by a cipher produces nothing, in the operation we omit the 0 in the multiplier. Thus, having multiplied by the 5 units, we next multiply by the 2 hundreds, and place the first figure of this partial product under the figure by which we are multiplying. Operation. 326 205 1630 652 66830 Ans, |