indicates the degree of the root. Thus 4, or 4, indicates the second root; 27, the third root, etc. 165. Cases.-The subject embraces seven cases. The development of these cases is based upon the following principles: PRINCIPLES. 1. A divisor of a number (excepting unity and the number itself) is a factor of the number. 2. A divisor of a factor of a number (excepting unity) is a factor of the number. 3. A number is divisible by its prime factors or by any 1roduct of them. 4. A number is divisible only by its prime factors, or some product of them, or by unity. CASE I. 166. To resolve a number into its prime factors. 1. Find the prime factors of 165. SOLUTION.-Dividing by 3, we find that 3 is a factor of 165 (Prin. 1). Dividing the quotient by 5, we find that 5 and 11 are also factors of 165 (Prin. 2); and since these numbers 3, 5, and 11, are prime, they are the prime factors of 165. OPERATION. 3)165 11. Rule.-I. Divide the given number by any prime number, greater than 1, that will exactly divide it. II. Divide the quotient, if composite, in the same manner, and thus continue until the quotient is prime. III. The divisors and last quotient will be the prime factors required. Find the prime factors 2. Of 385. 3. Of 1365. 4. Of 1260. 5. Of 3465. 6. Of 39270. 7. Of 7780500. Ans. 5, 7, 11. Ans. 3, 5, 7, 13. Ans. 22, 32, 5, 7. Ans. 32, 5, 7, 11. Ans. 2, 3, 5, 7, 11, 17. Ans. 22, 32, 53, 7, 13, 19. CASE II. OPERATION. 167. To resolve a number into equal factors. 1. Find the two equal factors of 225. SOLUTION. We first resolve the number into its prime factors. Now, since there are two 3's, we take one 3 for each factor; and since there are two 5's, we take one 5 for each factor; hence each of the two equal factors is 3x5 or 15; therefore 15 is one of the two equal factors of 225. 225=32X52 3x5=15, Ans. Rule.-I. Resolve the number into its prime factors. II. Take the continued product of one of each of the two equal factors, when we wish the two equal factors; one of each of the three, for the three equal factors; etc. 2. Find one of the two equal factors of 256, 576, 5184, 9216, and 20736. 3. Find one of the three 1728, 2744, 3375, 5832. Find the value of Ans. 16, 24, 72, 96, 144. equal factors of 216, 512, 1000, Ans. 6, 8, 10, 12, 14, 15, 18. 168. To resolve a number into factors bearing certain relations to each other. 1. Resolve 192 into two factors, one of which shall be 3 times the other. OPERATION. 3)192 SOLUTION.-By the condition of the problem, 3 times the second factor equals the first. Now, the second factor multiplied by 3 times the second factor equals 3 times the square of the second factor, which equals 192; hence the square of the second factor equals of 192, or 64; if 64 is the square of the second factor, 64, or 8, is the second factor, and 8×3, or 24, is the first factor. 648, 2d factor. 3x8=24, 1st factor. Rule.-I. Divide the given number by the product of the numbers indicating the relation of the factors to the smallest factor, and extract that root of the quotient indicated by the number of factors; the result will be the smallest factor. II. Multiply the smallest factor by the numbers indicating the relation of the other factors to it, and the result will be the other factors. 2. Resolve 4096 into three factors, such that the second shall equal twice the first, and the third twice the second. Ans. 8, 16, 32. 3. Resolve 2592 into three factors, of which the second is twice the first and the third three times the second. Ans. 6, 12, 36. 4. Resolve 82944 into four factors, so that the second shall be twice the first, the third twice the second, and the fourth twice the third. Ans. 6, 12, 24, 48. 5. Resolve 373248 into four factors, the second being twice the first, the third 3 times the second, and the fourth 4 times the third. Ans. 6, 12, 36, 144. 6. The contents of a cistern are 1728 cu. ft., the length being 2 times the breadth, and the breadth 2 times the depth; what are the dimensions? Ans. 24 ft.; 12 ft.; 6 ft. 7. A farmer had a bin containing 324 cu. ft., whose length was 3 times the breadth, and breadth twice the depth; required the dimensions. Ans. 18 ft.; 6 ft.; 3 ft. 8. A person wished to dig a tank to hold 1024 cu. ft., but having but little ground he was obliged to make the breadth half the length and the length of the depth; what were the dimensions? Ans. 4 ft.; 8 ft.; 32 ft. CASE IV. 169. To find all the divisors of a number whose factors are all unequal. 1. Find all the divisors of 66. SOLUTION. The prime factors of 66 are 2, 3, and 11. If we multiply 1-2 by 1-3 we obtain 1, 2, 3, and the product of 2 and 3, and these multiplied by 1-11 will give 1, 2, 3, 11, and all the products that can be formed out of 2, 3, and 11. Hence, according to Prin. 4, these are all the divisors of 66. OPERATION. 1-2 1-2-36 1-2-3-6-11-22-33-66 Rule.-Resolve the number into its prime factors, multiply 1 and the first factor by 1 and the second factor, the products by 1 and the third factor, etc., until all the factors have been used; the result will be the divisors required. Find all the divisors 2. Of 105. 3. Of 385 4. Of 570. 5. Of 1001. Ans. 1, 3, 5, 7, 15, 21, 35, 105. Ans. 1, 5, 7, 11, 35, 55, 77, 385. Ans. 1, 2, 3, 5, 6, 10, 15, 19, 38, etc. Ans. 1, 7, 11, 13, 77, 91, 143, 1001. 6. Of 19019. Ans. 1, 7, 11, 13, 19, 77, 91, 133, 143, etc. CASE V. 170. To find all the divisors of a number, some of whose factors are equal. 1. To find all the different divisors of 108. SOLUTION. We find the factors of 108 are two 2's and three 3's. Since 3 is a factor three times, 1, 3, 32, 33 is the first series of divisors; and since 2 is a factor twice, 1, 2, 4, is the second series of divisors; and OPERATION. 108 2×2×3×3×3 1-3-9-27-2-6-18-54-4-12-36-108 the products of the terms of these two series will give the prime factors and all possible products of them, and therefore all the divisors of the given numbers. Rule.-I. Resolve the number into its prime factors, form a series, consisting of 1 and the successive powers of one factor, under this write 1 and the successive powers of another factor, and take the products of the terms of the two series. II. Proceed in a similar manner with these products and the remaining factors, if any, and the terms of the last product will be all the divisors of the given number. Find all the divisors 2. Of 48. 3. Of 72. 4. Of 100. 5. Of 360. 6. Of 810. 7. Of 840. 8. Of 960. Ans. 1, 2, 4, 8, 16, 3, 6, 12, 24, 48. Ans. 1, 2, 4, 8, 3, 6, 9, 12, 24, 18, 36, 72. Ans. 1, 2, 4, 5, 10, 20, 25, 50, 100. Ans. 1, 2, 4, 8, 9, 3, 6, 12, 24, 10, etc. Ans. 1, 2, 3, 5, 9, 27, 10, 45, 90, etc. Ans. 1, 2, 3, 5, 7, 12, 4, 8, 24, etc. Ans. 1, 2, 3, 4, 5, 6, 8, 10, 12, 16, etc. CASE VI. 171. To find the number of divisors of a number. 1. How many divisors has 108? 108=22X33 SOLUTION. It is evident that 1 with the 1st, OPERATION. 2d, and 3d powers of 3 will give a series of four divisors, and 1 with the 1st and 2d powers of 2 will give a series of three divisors; hence, their product will give a series of 4×3 or 12 divisors. Hence the product of the exponents of the factors increased by 1, will give the number of divisors. = 12, Ans. = Rule. Resolve the number into its prime factors, increase the exponent of each factor by 1, and take the product of the results. 172. To find all the divisors common to two or more numbers. 1. Find the divisors common to 108 and 144. Rule.-I. Resolve the numbers into their prime factors. II. Take 1 and all the common prime factors, and all the numbers arising from their combination. Find the divisors common to 2. 36 and 48. 3. 48, 96, and 120. 4. 480, 720, and 840. 5. 576, 864, 1152, and 1728. Ans. 1, 2, 3, 4, 6, 12. Ans. 1, 2, 3, 4, 6, 8, 12, 24. Ans. 1, 2, 3, 4, 5, 6, 8, etc. Ans. 1, 2, 3, 4, 6, 8, 9, 12, etc. |