M135F08A5S

M135F08A5S - MATH 135 Assignment#5 Solutions Hand-In...

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MATH 135 Fall 2008 Assignment #5 Solutions Hand-In Problems 1. (a) NO By inspection, gcd(15 , 39) = 3. Since 3 6 | 28, there are no solutions. (b) NO Since 148 = 4(37) and 370 = 10(37), then gcd(148 , 370) = 74. Since 74 6 | 111, there are no solutions. (c) Set z = - y . We solve 1980 x + 1923 y = 9 first. We find gcd(1980 , 1923) using the Extended Euclidean Algorithm: 1 0 1980 0 1 1923 1 - 1 57 1 - 33 34 42 33 34 - 35 15 1 - 101 104 12 2 135 - 139 3 1 - 641 660 0 4 So gcd(1980 , 1923) = 3 | 9 so there are solutions. Since 1980(135) + 1923( - 139) = 3, then, multiplying both sides by 3, we get 1980(405) + 1923( - 417) = 9. Therefore, (405 , - 417) is a particular solution to 1980 x + 1923 z = 9. Therefore, the complete solution to 1980 x + 1923 z = 9 is x = 405 + 1923 3 n = 405 + 641 n z = - 417 - 1980 3 n = - 417 - 660 n ± n Z and so the complete solution to 1980 x - 1923 y = 9 is x = 405 + 641 n y = - z = 417 + 660 n ± n Z (d) 22680 = 2 × 11340 = 2 2 × 5670 = 2 3 × 5 × 657 = 2 3 × 5 × 3 2 × 63 = 2 3 × 3 4 × 5 × 7 (e) 66150 = 2 × 5 × 6615 = 2 × 5 2 × 1323 = 2 × 5 2 × 3 2 × 147 = 2 × 5 2 × 3 2 × 7 × 21 = 2 × 3 3 × 5 2 × 7 2 (f) Since 22680 = 2 3 3 4 5 1 7 1 and 66150 = 2 1 3 3 5 2 7 2 , then gcd(22680 , 66150) = 2 1 3 3 5 1 7 1 = 1890 (g) Since 22680 = 2 3 3 4 5 1 7 1 , then its divisors are of the form 2 a 3 b 5 c 7 d with a , b , c , d Z and 0 a 3, 0 b 4, 0 c 1, 0 d 1. Thus, there are 4 × 5 × 2 × 2 = 80 positive divisors. (h) For 2 a 3 b 5 c 7 d with a , b , c , d Z to be an even positive divisor of 22680, we need a 1, so 1 a 3, 0 b 4, 0 c 1, 0 d 1, so there are 3 × 5 × 2 × 2 = 60 even positive divisors. Since there are 80 positive divisors in total, then the probability of choosing an even positive divisor at random is 60 80 = 3 4 .
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2. Using the Extended Euclidean Algorithm, 0 1 256 1 0 176 - 1 1 80 1 3 - 2 16 2 - 16 11 0 5 so gcd(176 , 256) = 16. Since 16 | 5200 (with quotient 325), there are solutions. Since 176(3) + 256( - 2) = 16, then, multiplying both sides by 325, we get 176(975) + 256( - 650) = 5200. Therefore, (975 , - 650) is a particular solution to 176 x + 256 y = 5200. The complete solution is thus x = 975 + 256 16 n = 975 + 16 n y = - 650 - 176 11 n = - 650 - 11 n ± n Z For non-negative solutions, x 0 975 + 16 n 0 n ≥ - 975 16 = - 60 15 16 n ≥ - 60 since n Z y 0 ⇒ - 650 - 11 n 0 n ≤ - 650 11 = - 59 1 11 n ≤ - 60 since n Z Combining these inequalities, the only non-negative solution is when n = - 60, giving ( x, y ) = (15 , 10). 3. Since the complete solution to 1911 x + 741 y = 273 is x = 49 + 19 n y = - 126 - 49 n ± n Z then the complete solution to 1911 u - 741 v = 273 is u = x = 49 + 19 n v = - y = 126 + 49 n ± n Z For non-negative solutions, u 0 49 + 19 n 0 n ≥ - 49 19 = - 2 11 19 n ≥ - 2 as n Z v 0 126 + 49 n 0 n ≥ - 126 49 = - 2 28 49 n ≥ - 2 as n Z For v 2 u + 50,
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This note was uploaded on 11/01/2008 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

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M135F08A5S - MATH 135 Assignment#5 Solutions Hand-In...

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