RULE SECOND. Commence at the right, and number the figures of one factor, with the indices 1, 2, 3, &c., and the figures of the other factor, with the indices 0, 1, 2, 3, &c. Indicate the product of any two of the figures, by the sum of their indices. Write the units' figure of product 1, for the right hand figure of the true product, and carry the tens to THE SUM of products 2. The units' figure of this sum is the second product figure, and the tens are carried to THE SUM of products 3. The units' figure of this product is the third figure sought, and the remaining figures are determined in a similar manner. EXAMPLE FOR THE BOARD. Set Multiply in one line, 268.1475 by 93.074. According to the rule, we say 4×5= 20. Set down 0 and carry 2. 4×7+ 5×7 63, and 2 to carry make 65. down 5 and carry 6. 4×4+7x7+5x 0+6=71. Set down 1 and carry 7. 4x1+7x4+7×0+5×3+7=54. 4X 8+7×1+4×0+7×3+5×9+5=110. 7 6 5 4 3 2 1 268.1475 4 3 2 10 9 3.074 11 10 9 8 7 6 5 4 3 2 1 24957.5 604150 3×4+9×7+11=166. 4×2+7×6+0×8+3×1+9×4+16= 105. 7×2+0×6+3×8+9×1+10=57. 0×2+3×6+9×8+ 5=95. 3×2+9×6+9=69. 9×2+6=24. This rule may be readily demonstrated, by performing the multiplication in the usual manner, and observing what figures are added together,, to obtain each figure of the entire product. The multiplication in one line will be found a useful exercise, and is particularly valuable for the facility it gives in addition, as well as for the exercise of the memory. 1. 49863 x .06 x 1.3=? 2. Multiply 7441.08 by 236. 3. Find .06 of 284931.75. 4. Find .0725 of 164.25 × 1000. 5. 96.8 × 12000 × 45.93 = ? 6. Multiply 4489.076 by 2604000. 7. Multiply 596000 by 70490000. 8. What is the product of 36 × 24 × 12 × 15 ? 9. A merchant borrowed $9965.73, and paid .1845 of the amount for the use of the money. How much did he pay in the whole? 10. Multiply 909871265000 by 470.0368. CHAPTER V. DIVISION. DIVISION is the process by which we find how many times one number or part of a number is contained in, or may be subtracted from, another. The number to be divided is the dividend. The number to divide by, is the divisor. The number of times the dividend contains the divisor, is the quotient. The divisor and quotient may also be regarded as factors of the dividend. The number left, (if any,) after the operation, is the remainder. 4 7 The sign ÷ (divided by) signifies that the former of two numbers is to be divided by the latter, as; 4÷2=2; 16÷4=8÷2. Division may also be expressed by writing the divisor under the dividend; as, 2, 46, 3, 11; which are read 42 or 4 halves; 16÷4 or 16 fourths; 7÷8 or 7 eighths; 11-12 or 11 twelfths, &c. Numbers written in this manner are called fractions, the number above the line, or the dividend, being the numerator, and the number below the line, or the divisor, the denominator. The remainder in any division may always be written as the numerator of a fraction, whose denominator will be the divisor. RULE. Write the divisor at the left of the dividend, and if necessary, annex decimal zeroes to the dividend, until it has as many decimal places as the divisor. From the left of the dividend, take as many figures as will contain the divisor one or more times, for a first partial dividend. Find how many times the partial dividend will contain the divisor, and write the result as the first quotient figure. Multiply the divisor by this figure, and subtract the product from the first partial dividend. To the remainder annex the next figure of the dividend for a second partial dividend, and divide as before. Thus continue until the division is complete, and point off as many decimals in the quotient, as there are in the dividend more than in the divisor, prefixing zeroes, if necessary, to make the required number of decimals. PROOF. Add the remainder to the product of the divisor by the quotient, and will obtain the dividend. you To divide any number by 10, 100, 1000, &c., remove the decimal point as many places to the right as there are zeroes in the divisor. When there are zeroes at the right hand of the divisor, cut them off, and remove the decimal point of the dividend as many places to the left. After dividing the integers, remove the decimal point from the remainder, and you will have the true remainder. When the divisor can be resolved into factors, we may either employ the whole divisor, or each of its factors in succession. When the quotient will contain a number of figures, it is often convenient to multiply the divisor by each of the nine digits, and write the products on the slate, before commencing the division. It will then be easy to determine the value of each quotient figure, and the product to be subtracted from the partial dividend. EXAMPLE FOR THE BOARD. Divide 199760 by 371. 1855 1426 1113 3130 2968 The divisor is between 300 and 400. 3 371)199760(538 hundreds are contained in 19 hundreds, 6 times, and 4 hundreds are contained 4 times. 4 and 6 are therefore the limits of the true quotient figure, which must be either 4, 5, or 6. Now the true quotient figure must be contained in 1997, at least 371 times, as its product by 371, is to be subtracted from 1997. Mentally dividing 1997 by 6, I find 6 is in 19, 3 times, 6 in 19, 3 times, which is smaller than the second figure of 371,-6 is therefore too large. Trying 5, I say, 5 in 19, 3 times, 5 in 49, 9 times, which is larger than the second figure of 371, and is therefore correct. The limits of the second quotient figure are 3 and 4. Dividing as before, 4 in 14, 3 times, 4 in 22, 5 times, which is too small. 3 is therefore the true quotient figure. The limits of the third quotient figure 162 are 7 and 10, or rather 7 and 9, as no figure can be greater than 9, and by mental division, 8 is found to be the true figure. Hence the following RULE FOR OBTAINING THE TRUE QUOTIENT FIGURE. Employ the first figure of the divisor, and a number, one larger than the first divisor figure, as trial divisors, to determine the limits within which the true quotient figure must be found. Mentally divide the partial dividend by each of the possible quotient figures, until you find one that will give a quotient as large as, or larger than, the divisor, If there is but one such figure, it will be the one sought; if there is more than one, the larger will be the true quotient figure. 1. Divide .0497 by 368000. 2. What is the quotient of 3.995 by 2×3×4? 3. 8.9×.3× 14÷27 x .04×13= ? 4. If the product of two factors is 81.45, and one of the factors is 18.1, what is the other? 5. If the product of four factors is 2520, and three of its factors are 8, 5, and 7, what is the fourth? 6. Divide 4×7× 9 × 2 by 9×7×2. By 9×4, By 4× 2 x 9. 7. Divide .00984 by .02613×7. 8. What is the quotient of .087 × .003 by 19000 × 70 ? 9. 27.3 is .16 of what number? .0095 is .84 of what number? 76.125 is 1.25 times what number? 841.21 is 1.34 of what number? 44 is .13 of what number? THE LEAST COMMON MULTIPLE, AND THE GREATEST COMMON DIVISOR. One number is called a multiple of another, when the former can be divided by the latter without any remainder. A number that can be exactly divided by two or more other numbers, is a common multiple of those numbers. The divisors are called sub-multiples, or aliquot parts. Thus 24 is a common multiple of 2, 4, 6, 8, and 12,—and these latter numbers are aliquot parts of 24. Any number that contains all the factors of a number, will evidently contain the number itself. Thus 24, which is equal to 2×2× 2× 3, contains 2×2 or 4, 2×3 or 6, 2× 2 × 2 or 8, and 2× 2 × 3 or 12. Then the least common multiple of any series of numbers, is the least number which contains all the factors of the given numbers, and may be found by the following RULE. Arrange the numbers in a horizontal line, and divide successively by the prime numbers 2, 3, 5, 7, 11, &c., employing each divisor as often as it will divide one or more of the numbers without a remainder, writing the quotients and undivided numbers, beneath. Continue this division until the last quotients are all 1, and you will have obtained all the prime factors of the given numbers. The product of these factors, is the least common multiple. 7)766 166 166 1766 166 1 66 166 166 166 1 By the table of prime factors, the least common multiple can be found much more readily, in the following manner. Form the product of all the prime factors of the given numbers, employing each factor the largest number of times it is used in either number. Referring to the numbers in the foregoing example, we find they are respectively equal to 2×7, 2×3×3, 3×3×3, 3×7, 2×2×7, and 2×3×3×7. The only prime numbers used, are 2, 3, and 7. 2 is employed 2 times in the 5th number, 3 is employed 3 times in the 3d number, and 7 is employed but once in either number. The least common multiple is then, 2×2×3× 3x3x7=756, as before. |