hw-solutions-ch3

hw-solutions-ch3 - MATH 3012 Homework Problems and...

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MATH 3012 Homework Problems and Solutions Chapter 3, Spring 2009, WTT and MTK Group 1: Basic Problems 1. Factor m = 2205 and n = 1925 into products of primes. Then use this factorization to find the greatest common divisor and the least common multiple of m and n . Answer: 2205 = 3 2 · 5 · 7 2 and 1925 = 5 2 · 7 · 11, so gcd(2205 , 1925) = 5 · 7 = 35 and lcm(2205 , 1925) = 3 2 · 5 2 · 7 2 · 11. 2. Use the algorithm developed in our class to find the greatest common divisor d of m = 2205 and n = 1925. Also, find integers a and b such that 2205 a + 1925 b = d . Answer: First, note by long division that: 2205 = 1 · 1925 + 280 1925 = 6 · 280 + 245 280 = 1 · 245 + 35 245 = 7 · 35 + 0. From this it follows that gcd(2205 , 1925) = 35. Also, we can rewrite the first three calculations as: 280 = 1 · 2205 - 1 · 1925 245 = 1 · 1925 - 6 · 280 35 = 1 · 280 - 1 · 245. Now we start with the last of these statements and back-substitute as follows: 35 = 1 · 280 - 1 · (1 · 1925 - 6 · 280) = - 1 · 1925 + 7 · 280 35 = - 1 · 1925 + 7(1 · 2205 - 1 · 1925) = 7 · 2205 - 8 · 1925. That is, we may take a = 7 and b = - 8. Note that 7 · 2205 = 15435 and 8 · 1925 = 15400. 3. Can you find integers a and b so that 2205 a + 1925 b = 280? The answer is yes if and only if the greatest common divisor of 2205 and 1925 is also a divisor of 280. Note that 280 = 8 · 35, so the answer is yes. Then from the preceding work, we can take a = 56 = 8 · 7 and b = - 64 = 8( - 8). Can you find integers a and b so that 2205 a + 19255 b = 285? Here the answer is no since 35 is not a divisor of 285. 4. Let r n denote the number of regions into which the plane is divided by n circles that intersect in general position. Note that r 1 = 2 and r 2 = 4. Develop a recursive formula satisfied by r n and use it to find
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hw-solutions-ch3 - MATH 3012 Homework Problems and...

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