found by multiplying the multiplicand by the multiplier, the answer may be supposed correct. Illustration. The foregoing method of proving multiplication, is founded on the following proposition, that the product of two numbers will always be the same number which way soever they are multiplied. If we multiply 9 by 6, or 6 by 9, we obtain the same result, viz. 54. To express the same by adding, let 9 be separated into two parts, and added together 6 times. 6+3=9 6+3=9 6+3=9 6+3=9 6+3=9 6+3=9 By adding the two left hand columns, we, in reality, add 6 nine times which is the same as though we had multiplied 6 by 9. When we add the right hand column, we do the same thing as when we multiplied 9 by 6 ; and we obtain the same number now as before. 36 18 54 18 54 To show that the products of any two figures, when multiplied both ways, will be equal, let us make 5 rows of points, each row consisting of nine points, as in the parallelogram À. Now if we set down the number of points in one row, 5 times, and add them, the sum 45 would be the number of points in all the rows; but if we multiply the number in one row by 5, the number of rows, we shall obtain the same number, for the product contains the multiplicand as many times as the multiplier contains an unit. Next let us make 5 rows of points, each row consisting of 5, as seen in parallelogram B C; then sap. pose 4 points from each row in the parallelogram A, to be removed from the left and added to the right of the parallelogram B C; there will be points enough to make the 4 rows a, b, c, d ; then we have 9 rows, with 5 points in each, and by multiplying 5, the number of points in one row by 9, the number of rows, we shall obtain the number of points in the 9 rows, which is 45, the same number that we found in the parallelogram A, when we multiplied 9 by 5. We may suppose the parallelogram A to be increased by more rows, and each row to consist of more points, but by removing a part of them as before, and making rows the other way, we hould always obtain the same product. Let the scholar prove the following examples: 44. 9631 X 341. 45. 84213 x 919. 46. 705684 x 1763. 47. 123456 X 54321. 48. 9753201 X 98765. 49. 9897969 X 191817., 50. What is the product of one hundred sixtyseven thousand, five hundred and twenty-seven, into forty-five thousand, nine hundred and sixty-eight? 51. A merchant has a bale of bombazetts, consisting of 4 pieces, each piece 31 yards; and he gave 24 cents a yard. What did the 4 pieces cost him? Ans. $29,76. CONTRACTIONS. The operations in multiplication may often be contracted, or shortened. Contraction 1.-When ciphers stand at the right hand of the multiplier, set the multiplier so that the right hand significant figure may stand directly under units' place in the multiplicand. By bringing down the ciphers, the products of the other figures are brought into their proper places. 52. An army consisting of twenty-one thousand, three hundred and seventeen men, plundered so much money from a certain city, that each man received $50. How much money was taken? Operation. Nole 3.-By the illus5 0 tration of the proof, it is immaterial which of the two numbers is made the Ans. $1 '06 5' 8.50 multiplier; but it is gene nerally more convenient to make the less number the multiplier. 53. 107821 X 11000. 54. 107821 X 990000. The sum of the two last products is 107'928'821'000. 55. An army has 150 waggons loaded with beef ; each has six barrels, and each barrel contains 200 pounds. What is the whole number of pounds ? Ans. 180'000. 56. What number contains six thousand, seven hundred and sixty-five, nine thousand times? Ans, 60'885'000. 57. If each window in a house, having 20 windows, contains 24 panes of glass, how many panes in all the windows ? Ans. 480. 58. Suppose the number of inhabitants in the United States to be twelve million, and each person to be worth five hundred dollars ; what are they all worth. Ans. 6'000'000'000 dolls. Contraction 2.-If the multiplier be 1, with ci. phers at the right hand, we have only to annex the ciphers in the multiplier to the right hand of the multiplicand, for multiplying by the 1 does not alter the figures in the multiplicand, and annexing the ciphers, removes every figure in the multiplicand into its proper place, in the same manner that it would, were we to go through the process of multiplying. 59. How many bushels of salt in 161 casks, each containing 10 bushels ? Ans. 1'610. 60. One dollar contains 100 cents; how many cents is 65 dollars? Ans. 6500. 61. One thousand mills being equal to one dollar, how many mills in 65 dollars ? Ans. 65'000. 62. If 12029 were set down ten thousand times, and added up, what would be the sum ? Ans. 120'290'000. QUESTIONS TO BE WORKED BY THE ADDITION AND MULTIPLICATION OF SIMPLE NUMBERS. 1. Charles bought 29 oranges, and John 31, at 5 cents a piece. What did all the oranges cost? Operation 1. Operation 2. 2 9+3 1=60 5 1 4 5+1 5 5=$3,00 $3,00 Explanation.--The last question may be performed two ways, either by multiplying each quantity separately by 5, or adding the quantities together, and then multiplying. But had one boy given more a piece for his oranges than the other, the price of the whole would have been found as in the first ope. ration, for 5 times 29, and 7 times 31 makes more than 300. 2. A man bought 101 bushels of rye for 45 cents a bushel; 65 pounds of tea for 45 cents a pound; what is the cost of the rye and tea? Ans. $74,70. 3. A farmer sells 405 pounds of beef, at 6 cents a pound, and 5 tons of hay, at $15,00 a ton; what does he receive for both ? Ans. $99,30. 4. What is the amount of the following articles, viz. : 69 gallons of wine, at 81 cents a gallon; 59 pounds of sugar, at 13 cents a pound; 17 yards of riband, at 17 cents a yard; and 9 plates, at 7 cents a piece ? Ans. $67,08. 5. The cost of constructing 25 miles of a certain road, is twenty thousand dollars; for the remaining part, which is 14 miles, the cost is fifteen thousand and twenty-four dollars a mile. What was the whole cost of the road ? Ans. $230'336. 6. Multiply the sum of the following numbers by 203 ; viz, 1045, 69721, 621 and 10001. Ans. $16'521'764. 7. A man gave his estate to 8 sons; to 5 of them he gave nine hundred and eleven dollars each; to the other 3, seven hundred and one dollars each ; what was his estate? Ans. $6659. 8. Suppose 1002 men possess equal properties, and the property of each man to be as follows; 6000 acres of wild land, at 4 dollars an acre; buildings worth 4011 dollars ; 999 sheep, at 2 dollars each; 3 horses, each 95 dollars; money at interest, 3 times the value of his buildings; 10 shares in a bank, 500 dollars each; and other property, the value of which is equal to the product of 25 into the value of his bank stock and wild land. What would be the property of all the men ? Ans. $773'871'654. Remark. When one man sells another several articles, they are generally set down upon a picce of paper ; the price of each article found separately, and all the prices added up. If the buyer pays for the articles, the seller writes, received payment at the bottom of the bill, and signs his name; but if payment is not made, the several prices are only added up. |