Ex. 1. We are to multiply 17059 by 45. We first multiply 17059 by 9. The product is 153531. But in using 9 as a multiplier instead of 45 we use a multiplier only onefifth large enough. Our product, therefore, is only one-fifth large enough, and to obtain the correct product we must multiply by 5. Multiplying 153531 by 5 we have as our final product 767655. Ex. 2. We are to divide 43275 by 56. We first divide 43275 by 8. Our quotient is 5409%. But in using 8 as a divisor instead of 56 we use a divisor only one-seventh large enough. Our quotient, therefore, is 7 times too large, and to obtain the correct quotient we must divide by 7. In dividing 5409% by 7 we have after the third division a remainder of 5%. 1 integer equals 8 eighths, 5 integers equal 40 eighths; and 40 eighths plus 3 eighths equal 43 eighths, or 43. 43 is an expression of division, 43 being the dividend of the expression and 8 the divisor. Therefore, in dividing 43 by 7 we have a dividend, 43, and two divisors, 8 and 7. 43 is divisible by neither divisor, therefore we retain our dividend unchanged and multiply together the divisors 8 and 7. The result thus obtained, which forms the fractional part of our quotient, is 3, and our total quotient is 77233. Ex. 3. Explain the solution of Ex. 3. Ex. 4. Explain the solution of Ex. 4. NOTE. It is evident that in Exs. 2 and 4 the numerators of the fractional part of each quotient might have been written as remainders. When the remainder is expressed as a fractional part of the quotient the fraction should be reduced to its lowest terms by dividing the numerator and denominator by their common factors. 37. To Multiply when there are O's at the Right of the Multiplier and Multiplicand. We are to multiply 2700 by 460. 162 Into what two convenient factors can 2700 be separated? 108 1242000 Into what two 460? What two factors do we first multiply together? What is the remaining factor of 2700 ? What is the remaining factor of 460? 100 multiplied by 10 gives what product? 1242 multiplied by 1000 gives what product? 2700 multiplied by 460 gives, then, what product? Give, then, a rule for multiplying when there are 0's at the right of the multiplier or multiplicand. Ex. 1. We are to multiply 9600 by 40. The factors of 9600 are 96 and 100, and the factors of 40, 4 and 10. We first multiply together the factors 96 and 4. The product thus obtained is 384. To multiply 384 by the two remaining factors we annex as many 0's as are contained in those factors. There are two 0's in 100 and one in 10. We therefore annex three 0's to 384, thus obtaining as our final product 384000. Ex. 2. We are to multiply 79.6 by 2700. The factors of 2700 are 27 and 100. We first multiply 79.6 by 27, thus obtaining as our product 2149.2. To multiply 2149.2 by 100, we remove the decimal point two places to the right. We thus obtain as our final product 214920. Ex. 3. Explain the solution of Ex. 3. 38. To Divide when there are O's at the Right of the Divisor. We are to divide 67865 by 3700. separate our divisor? 308 By which factor do we first divide? 296 How do we divide by 100? 12.65 What is our quotient? What is the next step of the division? What is the integral part of our quotient? The fractional part? By what must we multiply 37 to change it to 3700, the original divisor? By what, then, must we multiply 12.65 to obtain the true remainder? What, then, is the true remainder? Give, then, a rule for dividing when there are 0's at the right of the divisor. 19.56 The factors of 1800 are 18 and 100. We first To do this we remove the decimal point two divide 9736 by 100. places to the left. We next divide 97.36 by 18. Dividing we obtain 5 as the integral part of our quotient, and 7.36.18 as the fractional part. To change the divisor 18 to 1800, the original divisor, we multiply it by 100; therefore to obtain the true remainder 7.36 must be multiplied by 100. Multiplying 7.36 by 100 we have as our true remainder 736. Ex. 2. Explain the solution of Ex. 2. 39. To Multiply by 147, 125255, etc. We are to multiply a certain number by 147. We first multiply by 7 units. We have left 14 tens by which to multiply. How will the product by 14 compare with the product by 7? How, then, after obtaining the product by 7 can we obtain the product of the same multiplicand by 14? If the product by 7 is 28, what will be the product by 14? Where, then, should the right-hand figure of the product by the 14 tens be placed with reference to the right-hand figure of the product by the 7 units? Give, then, a special rule for multiplying by 147. We are to multiply a number by 125255. We first multiply by 5. After obtaining the product by 5, how can we obtain the product by 25? Where shall we place the right-hand figure of the product by the 25 tens? How will the product by 125 compare with the product by 25? How, then, after obtaining the product by 25, can we obtain the product by 125? How does the order of the 125 compare with the order of the 25? How, then, will the order of the product by 125 compare with the order of the product by 25? Where, then, should the right-hand figure of the product by 125 be placed with reference to the product by 25? Give, then, a special rule for multiplying by 125255. Where should the right-hand figure of a partial product be placed with reference to the right figure of a preceding partial product If the partial multiplier is four orders higher than the preceding partial multiplier ? If it is five orders higher? If it is six orders higher? Give then a special rule for multiplying a number By 426. By 642. By 366. By 217. By 721. By 600 150 306. Ry 625 125 255. By 256 128 328. By 210 105 155. By 3927 81. Ex. 1. We are to multiply 476589 by 625125255. We first multiply 476589 by 5. The product thus obtained is 2382945. 25 is 5 times 5. The product of 476589 by 25 will, therefore, be 5 times its product by 5, or 5 times 2382945, or 11914725. But the 25 is one order higher than the 5. The second product, therefore, will be one order higher than the first, and its right-hand figure must be written one order at the left of the right-hand figure of the first product. 125 is 5 times 25. The product of 476589 by 125 will, therefore, be 5 times its product by 25, or 5 times 11914725, or 59573625. But the 125 is two orders higher than the 25. The third product, therefore, will be two orders higher than the second, and its right-hand figure must be written two places at the left of the right-hand figure of the second product. 625 is 5 times 125. The product of 476589 by 625 will, therefore, be 5 times its product by 125, or 5 times 59573625, or 297868125. But the 625 is three orders higher than the 125. The fourth product, therefore, will be three orders higher than the third, and its right-hand figure must be written three places to the left of the right-hand figure of the third product. Adding the partial products we have as our total product 297927820155195. Explain the solutions of Exs. 2 and 3. Ex. 51. Multiply the first and each fifth multiplicand in Ex. 34 by 147; the second and each fifth by 125255 the third and each fifth by 252 63 9; the fourth and each fifth by 56872; the fifth and each fifth by 6 42 84 220. Multiply the first and each third multiplicand in Ex. 36 by a multiplier similar to 426; the second and each third by a multiplier similar to 642; the third and each third by a multiplier similar to 42 6 48. |