RULE. Divide the given number by any prime factor; divide the quotient in the same manner, and so continue the division until the quotient is a prime number. The several divisors and the last quotient will be the prime factors required. PROOF. The product of all the prime factors will be the given number. 3. What are the prime factors of 29925? 6. Find the prime factors of 2310. 7. Find the prime factors of 2205. 36= 8. What are the prime factors of 13981 ? 9. What are the prime factors of 532? 10. What are the prime factors of 2500? Ans. 2, 2, 3, 5, 19. 97. To resolve a number into all the different sets of factors possible. 1. In 36 how many sets of factors are there, and what are they? OPERATION. 2 x 18 3 x 12 4 × 9 6 × 6 2 x X 9 6 3 × 3 × 4 × 3 × 3 SOLUTION. Writing the 36 at the left of the sign =, we arrange all the different sets of factors into which it can be resolved, under each other, as shown in the operation, and we find that 36 can be resolved into 8 sets of factors. 2. How many sets of factors are there in the number 24? What are they? Ans. 6 sets. 3. In 125 how many sets of factors are there? What are they? Ans. 2 sets. CANCELLATION. 98. Cancellation is the process of rejecting equal factors from numbers sustaining to each other the relation of dividend and divisor. It has been shown (77) that the dividend is equal to the product of the divisor multiplied by the quotient. Hence, if the dividend can be resolved into two factors, one of which is the divisor, the other factor will be the quotient. 1. Divide 63 by 7. OPERATION. Divisor, 779 Dividend. 9 Quotient. EXAMPLES. 24 x 56 7 x 48 the factor 7, and the remaining factor, 9, is the quotient. = SOLUTION.- We see in this example that 63 is composed of the factors 7 and 9, and that the factor 7 is equal to the divisor. Therefore we reject 99. Whenever the dividend and divisor are each composite numbers, the factors common to both may first be rejected without altering the final result. (86, Prin. III.) 2. What is the quotient of 24 times 56 divided by 7 times 48? OPERATION. 4× 6 × 7×8 7 × 6 × 8 a line, and those which constitute the divisor below it. Instead of multiplying 24 by 56, in the dividend, we resolve 24 into the factors 4 and 6, and 56 into the factors 7 and 8; and 48 in the divisor into the factors 6 and 8. We next cancel the factors 6, 7, and 8, which are common to the dividend and divisor, and we have left the factor 4 in the dividend, which is the quotient. When all the factors or numbers in the dividend are canceled, 1 should be retained. 100. If any two numbers, one in the dividend and one in the divisor, contain a common factor, we may reject that factor. 3. In 54 times 77, how many times 63 are there? SOLUTION.In this example we see that 9 will divide 54 and 63; so we reject 9 as a factor of 54, and retain the factor 6, and we also reject 9 as a factor of 63, and retain the factor 7. Again, 7 will divide 7 in the divisor, and 77 in the dividend. Dividing both numbers by 7, 1 will be retained in the divisor, and 11 in the dividend. Finally, the product of 6 × 11=66, the quotient. OPERATION. 6 11 54 × 77 63 7 . 4. Divide 25 x 16 x 12 by 10 x 4 x 6 x 7. OPERATION. 5 4 2 25 × 16 × 12 10×4×6×7 SOLUTION. In this, as in the preceding example, we reject all the factors that are common to both dividend and divisor, and we have remaining the 2 factor 7 in the divisor, and the factors 5 and 4 in the dividend. Completing the work, we have 20 = 20 ÷ 7 or 2, Ans. 5 × 4 = 30=24. 5. Divide the product of 120 × 44 × 6 × 7 by 72 × 33 × 14. OPERATION. SOLUTION.-Rejecting all the factors common to both dividend and divisor, we have remaining the factor 3 in the divisor, and the factors 10 and 2 in the dividend. Completing the work, we have 20=20÷3 or 63. Ans. 72 × 33 × 14 6 3 2 2 10 4 120 × 44 × 6 × 7_10 × 2 = = 3 From the preceding examples and illustrations we derive the following rule: RULE.-I. Write the numbers composing the dividend above a horizontal line, and the numbers composing the divisor below it. II. Cancel all the factors common to both dividend and divisor. III. Divide the product of the remaining factors of the dividend by the product of the remaining factors of the divisor, and the result will be the quotient. 1. Rejecting a factor from any number is dividing the number by that factor. 2. When a factor is canceled, the unit, 1, is supposed to take its place. 3. One factor in the dividend will cancel only one equal factor in the divisor. 4. If all the factors or numbers of the divisor are canceled, the product of the remaining factors of the dividend will be the quotient. 30. 6. What is the quotient of 16 × 5 × 4 divided by 20 X 8? Ans. 2. 7. Divide the product of 90 x 66 x 8 by 4 x 11 x Ans. 36. 8. Divide the product of 33 × 35 × 28 by 11 × 15 X 14. Ans. 14. 9. What is the quotient of 21 × 11 × 26 divided by 14 x 13? Ans. 33. 10. Divide the product of the numbers 48, 72, 28, and 5, by the product of the numbers 84, 15, 7, and 6, and give the result. Ans. 94. 11. Divide 140 × 39 × 13 × 7 by 30 × 7 × 26 × 21. Ans. 4. 12. What is the quotient of 66 × 9 × 18 × 5 divided by 22 × 6 × 40 ? Ans. 101. 13. Divide the product of 200 x 36 x 30 x 21 by 270 x 40 x 15 x 14. Ans. 2. 14. Multiply 240 by 56, and divide the product by 60 multiplied by 28. Ans. 8. 15. The product of the numbers 18, 6, 4, and 42 is to be divided by the product of the numbers 4, 9, 3, 7, and 6. What is the result? Ans. 4. 16. How many tons of hay, at $12 a ton, must be given for 30 cords of wood, at $4 a cord? Ans. 10 tons. GREATEST COMMON DIVISOR. 101. A Common Divisor of two or more numbers is a number that will exactly divide each of them. 102. The Greatest Common Divisor of two or more numbers is the greatest number that will exactly divide each of them. Numbers prime to each other have no common divisor. A common divisor is sometimes called a Common Measure; and the greatest common divisor, the Greatest Common Measure. OPERATION. 103. When the numbers are readily factored. 1. What is the greatest common divisor of 6 and 10? We readily find by inspection that 2 will divide both the given numbers; hence, 2 is a common divisor; and since the quotients 3 and 5 have no common factor, but are prime to each other, the common divisor, 2, must be the greatest common divisor. 26 10 3 5 2, Ans. EXAMPLES. SOLUTION. 2. What is the greatest common divisor of 42, 63, and 105? OPERATION. 3 42 63 105 7 14 21 35 2 3 SOLUTION. We observe that 3 will exactly divide each of the given numbers, and that 7 will exactly divide each of the resulting quotients. Hence, each of the given numbers can be exactly divided by 3 times 7; and these numbers must be component factors of the greatest common divisor. Now, if there were any other component factor of the greatest common divisor, the quotients, 2, 3, 5, would be exactly divisible by it. But these quotients are prime to each other. Hence, 3 and 7 are all the component factors of the greatest common divisor sought. 3 x 721, Ans. |