3. What is the greatest common divisor of 28, 140, From these examples and analyses we derive the following rule: RULE. -I. Write the numbers in a line, with a vertical line at the left, and divide by any factor common to all the numbers. II. Divide the quotients in like manner, and continue the division till a set of quotients is obtained that have no common factor. III. Find the product of all the divisors. This will be the greatest common divisor sought. 4. What is the greatest common divisor of 12, 36, 60, 72? Ans. 12. 5. What is the greatest common divisor of 18, 24, 30, 36, 42? Ans. 6. 6. What is the greatest common divisor of 72, 120, 240, 384 ? Ans. 24. 7. What is the greatest common divisor of 36, 126, 72, 216? Ans. 18. 8. What is the greatest common divisor of 42 and 112? Ans. 14. 9. What is the greatest common divisor of 32, 80, and 256? Ans. 16. 10. What is the greatest common divisor of 210, 280, 350, 630, and 840? Ans. 70. 11. What is the greatest common divisor of 300, 525, 225, and 375? Ans. 75. Ans. 126. 12. What is the greatest common divisor of 252, 630, 1134, and 1386 ? 13. What is the greatest common divisor of 96 and 544? Ans. 32. 14. What is the greatest common divisor of 468 and 1184 ? Ans. 4. 15. What is the greatest common divisor of 200, 625, and 150 ? Ans. 25. 104. When the numbers cannot be readily factored. As the analysis of the method under this case depends upon three properties of numbers which have not been introduced, we present them in this place. I. An exact divisor divides any number of times its dividend. II. A common divisor of two numbers is an exact divisor of their SUM. III. A common divisor of two numbers is an exact divisor of their DIFFERENCE. 1. What is the greatest common divisor of 84 and 203? OPERATION. 203 84 2 168 35 28 SOLUTION. We draw two vertical lines, and place the larger number on the right, and the smaller number on the left, one line lower down. Then we divide 203, the larger number, by 84, the smaller, and write 2, the quotient, between the verticals, the product, 168, oppo7, Ans. site, under the greater number, and the remainder, 35, below. Next we divide 84 by this remainder, writing the quotient, 2, between the verticals, the product, 70, on the left, and the new remainder, 14, below the 70. Again we divide the last divisor, 35, by 14, and obtain 2 for a quotient, 28 for a product, and 7 for a remainder, all 70 2 14 2 14 2 0 of which we write in the same order as in the former steps. Finally we divide the last divisor, 14, by the last remainder, 7, and we have no remainder. 7, the last divisor, is the greatest common divisor of the given numbers. In order to show that the last divisor in such a process is the greatest common divisor, we will first trace the work in the reverse order, as indicated by the arrow line below. OPERATION. 2 84 2 70 2 14 2 14 203 168 35 28 7 7 divides the 14, as proved by the last division; it will also divide two times 14, or 28, (I). Now, as 7 divides both itself and 28, it will divide 35, their sum, (II). It will also divide 2 times 35, or 70, (I); and since it is a common divisor of 70 and 14, it must divide their sum, 84, which is one of the given numbers, (II). It will also divide 2 times 84, or 168, (I); and since it is a common divisor of 168 and 35, it must divide their sum, 203, the larger number, (II). Hence, 7 is a common divisor of the given numbers. 168 Again, tracing the work in the direct order, as indicated below, we know that the greatest common divisor, whatever it may be, must divide 2 times 84, or 168, (I). 203 Then since it will divide both 168 and 203, it must divide their difference, 35, (III). It will also divide 2 times 35, or 70, (I); and as it will divide both 70 and 84, it must divide their 35 difference, 14, (III). It will also divide 2 times 14 or 28, (I); and as it will divide both 28 and 35, it must divide their difference, 7, (III); hence, it cannot be greater than 7. 2 84 2 1. That 7 is a common divisor of the given numbers. 2. That their greatest common divisor, whatever it may be, cannot be greater than 7. Hence it must be 7. PRAC. AR. 6 This work may also be indicated by the ordinary form of division. OPERATION. 84)203 (2 168 35)84(2 70 28 7)14(2 14 0 SOLUTION. Using the larger number, 203, as dividend, and the smaller, 84, as divisor, we obtain the quotient 2, with a remainder of 35. We divide the former divisor, 84, by this remainder, and obtain a quotient, 2, with a remainder of 14. Dividing 35 by this remainder, we obtain a quotient, 2, with a remainder of 7. Dividing 14 by the last remainder, we obtain a quotient of 2, with no remainder. 7, the last divisor, is the greatest common divisor of the given numbers. RULE I. - Divide the greater number by the less. If there is a remainder, divide the preceding divisor by it, and so continue until there is no remainder. The last divisor will be the greatest common divisor. II. If more than two numbers are given, first find the greatest common divisor of two of them, and then of this divisor and one of the remaining numbers, and so on to the last; the last common divisor found will be the greatest common divisor of all the given numbers. 1. When more than two numbers are given, it is better to begin with the least two. 2. If at any point in the operation a prime number occurs as a remainder, it must be a common divisor, or the given numbers have no common divisor. 2. Find the greatest common divisor of 154 and 210. Ans. 14. 3. What is the greatest common divisor of 316 and 664 ? Ans. 4. 4. What is the greatest common divisor of 679 and 1869 ? Ans. 7. 5. Find the greatest common divisor of 688 and 1578. Ans. 2. 6. Find the greatest common divisor of 327 and Ans. 327. 1308. 7. What is the greatest common divisor of 917 and 1495? Ans. 1. 8. What is the greatest common divisor of 1313 and 4108 ? Ans. 13. 9. What is the greatest common divisor of 1649 and 5423? Ans. 17. 10. John has 35 pennies, and Charles 50. How may they arrange them in parcels, so that each boy shall have the same number in each parcel? Ans. 5 in each parcel. 11. A speculator has 3 fields, the first containing 18, the second 24, and the third 40 acres, which he wishes to divide into the largest possible lots having the same number of acres in each. How many acres will there be in each lot? Ans. 2 acres. 12. A farmer had 231 bushels of wheat, and 273 bushels of oats, which he wished to put into the least number of bins containing the same number of bushels, without mixing the two kinds. What number of bushels must each bin hold? Ans. 21. 13. A village street is 332 rods long; A owns 124 rods front, B 116 rods, and C 92 rods; they agree to divide their land into equal lots of the largest size that will allow each one to form an exact number of lots. What will be the width of the lots? Ans. 4 rods. 14. A forwarding merchant has 2722 bushels of wheat, 1822 bushels of corn, and 1226 bushels of beans, which he wishes to forward, in the fewest bags of equal size that will exactly hold either kind of grain. How many bags will it take? Ans. 2885. 15. A has $120, B $240, and C $384; they agree to purchase cows, at the highest price per head that will allow each man to invest all his money. What price must they pay, and how many cows can each man purchase? Ans. $24 per head. A 5, B 10, and C 16. |