SECOND METHOD. OPERATION. 2:12 70 36 15 € 35 129. This method consists in taking out the prime fac tors of the last common multiple and finding their product. 1. Find the least common multiple of 12, 30, and 70. SOLUTION.-Placing the numbers one beside another and dividing by 2, we see that 2 is a factor of each of them, it is therefore a factor of the L. C. M. (Prin. 3); dividing the quotients that will contain it by 3, we see that s is a factor of some of the numbers, it is therefore a factor of the L. . M. Dividing the next quotients by 5, we see that 5 is a factor of some of them, hence 5 is a factor of the L. C. M.; and the quotients having no other common factor, we see that the factors of the given numbers are 2, 3, 5, 2, and 7, hence their product, which is 420, is the L. C. M. Hence the following 5 2 2 5 35 7 2×3×5x2x7=420 Rule.-I. Write the numbers one beside another, divide by any prime number that will xactly divide two or more, and write the quotients and undivided numbers beneath. II. Divide the quotients in the same manner, and thus continue until no two numbers in the lowest line have a common factor. III. Take the product of the divisors and final quotients; the result will be the least common multiple required. Find the least common multiple of 2. 16, 20, and 30. 3. 28, 56, and 84. 4. 48, 60, and 30. 6. 40, 96, 100, and 120. 7. 120, 180, 200, and 240 j8. 140, 280, 160, and 320. Ans. 240. Ans. 168. Ans. 24 Ans. 3000. Ans. 2400. Ans. 3600. Ans. 2240. CASE II. 130. When the numbers are large and cannot be eadily factored. OPERATION. 28=4x7; 63=9x7 L C. M. 1. Find the least common multiple of 28 and 63. SOLUTION.-The greatest common divisor of these numbers is 7; 28 equals 4 times 7, and 63 equals 9 times 7; hence the L. C. M., as found in the first method, is 4 x 7 x 9, which equals 28 multiplied by 63 divided by 7: or the first number multiplied by the second divided by their greatest common divism. = 4X7X9 63 =28 X Rule.-I. Find the greatest common divisor of two num bers, divide one number by it, and multiply the other number by the quotient. II. When there are more than two numbers, find the least common multiple of two of the numbers, and then of this number and the third number, etc. WRITTEN EXERCISES. Find the least common multiple of 2. 671 and 793. 3. 3503 and 4859. 4. 6527 and 7597. Ans. 8723. Ans. 150629. Ans 463417. 5. What is the smallest sum of money for which I could hire workmen for one month, paying either $16, $20, $28, or $35 a month? Ans. $560. 6. A can dig in rods of ditch in a week, B can dig 18 rods, C 22 rods, and D 24 rods; what is the least number of rods that would afford an exact number of weeks' work for each one of them? Ans. 5544 rods. 7. What is the smallest number of bushels of corn that will fill a number of barrels containing 3 bushels each, a number of sacks containing 5 bushels each, a number of casks containing 14 bushels each, or a number of bins containing 48 bushels each? Ans. 1680 bu. 12. Four men start at the same place to walk around a garden; A can go around in nine minutes, B in 10 minutes, C in 12 minutes, and D in 15 minutes; in what time will they all meet at the starting point? Ans. 180 minutes. 13. A, B, C, and D start from the same point, A traveling a mile in 18 minutes, B in 24 minutes, C in 30 minutes, and D in 35 minutes; what is the least whole number of miles each may travel that they may return to the starting point at the same moment? Ans. A, 140; B, 105; C, 84; D, 72. INTRODUCTION TO CANCELLATION. MENTAL EXERCISES. 1. If we omit the factor 2 from 12 and 6, what factors will remain? 2. Divide 24 by 6. Divide 24 by of 6. Divide § of 24 by 6. 3. Divide of 24 by of 6. Divide 36 by 18, first taking out the common factor 6. 4. Is there any difference in the quotient of 48 divided by 12, and of 48 divided by { of 12? 5. Divide 72 by 48, first omitting common factors. Divide 90 by 60 in the same way; 144 by 96. 6. Divide 2×2×2 by 2×2; 3×3×4 by 2×3; 3×4×5 by 3×5. 7. Divide 2×3×7 by 2×7; 2×3×4 by 2×3; 3×5×8 by 3×8; 6x 6x8 by 3×6; 2×7×9×10 by 9×2. CANCELLATION. 131. Cancellation is a process of abbreviating arithmetical operations by rejecting common factors in both dividend and divisor. PRINCIPLES. 1. The cancelling of a factor from any number divides the number by that factor. DEM. Thus if we take the factor 3 out of 24 we shall divide 24 by 3. 2. The cancelling of a factor in both dividend and divisor will not change the quotient. DEM.-Cancelling a factor in both dividend and divisor is the same as dividing them both by the same number, which, by the principles of division, does not change the quotient. 1. Divide 84 × 60 by 24 × 63. 2 OPERATION. 4 5 24×833 SOLUTION.-We cancel the common factor 12 from 60 and 24, writing 5, the other factor of 60, above 60, and 2, the other factor of 24, below 24; we then cancel the common factor 21 from 84 and 63, writing 4, the other factor of 84, above 84, and 3, the other factor of 63, below 63; we then cancel 2 from 2 and 4, writing 2 above the 4; the product of the remaining factors of the dividend is 10, the product of the remaining factors of the divisor is 3; hence the quotient 10 divided by 3, or 3. Rule.-I. Cancel the common factors from the dividend and divisor. II. Then divide the product of the remaining factors of the dividend by the product of the remaining factors of the divisor. NOTES.-1. The unit 1 takes the place of a cancelled factor, but need not be written, except in the dividend of the quotient, when there are no other factors of the dividend remaining. 2. A factor in one term will cancel two or more factors in the other term, when their product is equal to the former. 3. Some prefer to place the dividend upon the right and the divisor upon the left, of a vertical line. 2. Divide 12× 14 × 16 by 6×7x8. 3. Divide 20 × 32 × 35 by 4 × 5 × 16. WRITTEN EXERCISES. Ans. 8. Ans. 10 Ans. Ans. 42. Ans. 1. 1. How many yards of muslin, worth 12 cents a yard, may be bought for 16 pounds of butter, worth 15 cents a pound? SOLUTION. If one pound of butter is worth 15 cents, 16 pounds are worth 16x15 cents; for 16x15 cents, át 12 cents a yard, we can get as many yards of muslin as 12 is contained times in 15×16, which we find, by cancellation, to be 20. OPERATION. 20 12 4 2. How many bushels of corn, worth 45 cents a bushel, must be exchanged for 125 pounds of butter, at 18 cents a pound? Ans. 50. 3. A exchanged rye, worth 84 cents per bushel, for 78 bushels of wheat, worth 98 cents per bushel; required the number of bushels of rye. Ans. 91. 4. How many bushels of corn, at 42 cents a bushel, must be given in exchange for 7 pieces of cloth, each containing 40 yards, at 36 cents a yard? Ans. 240. 5. How many boxes of tea, each containing 24 pounds at 90 cents a pound, must be given for 27 firkins of butter of 56 pounds each, at 20 cents a pound? Ans. 14. 6. A farmer sold a grocer 9 loads of apples, each load containing 18 bags, and each bag 2 bushels, at 35 cents a bushel, and received in payment 12 boxes of sugar, each con. taining 135 pounds; what was the sugar worth a pound? Ans. 7 cents PRIME NUMBERS. 132. No general method of determining prime numbers, beyond a certain limit, has yet been discovered, although much time has been spent in the investigation. 133. We give the following practical method, which consists in writing a series of numbers, and sifting out those which are composite. METHOD. Since the even numbers after 2 are composite, we write the series of odd numbers; thus, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41. 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, Now, commencing at 3, since every third term is divisible by 3, every third number is composite, which we indicate by putting the figure 3 over it. Commencing at 5, every fifth number is divisible by 5, and is therefore composite, hence we place a figure 5 over every fifth number. Proceed in the same manner with 7, and the numbers unmarked will be the prime numbers up to 100. This method was discovered by Eratosthenes, a Greek mathematician. He inscribed the series of odd numbers on parchment, and then cutting out the composite numbers, his parchment with its holes resembled a sieve; hence the method has been called Eratosthenes' Sieve. TABLE OF PRIME NUMBERS. 1 389 47 113 197 281 379 463 571 659 761 863 877 479 587 673 773 881 61 137 223 307 593 677 787 883 NOTE.-This page will be of interest to the student to read, but is not to be recited. |