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# hmk5sol - COT 3100 Homework 5 Solutions Fall 2000 1 Use...

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COT 3100 Homework # 5 Solutions Fall 2000 1) Use Euclid’s Algorithm to find the greatest common divisor(GCD) of 245 and 455. 455 = 1x245 + 210 245 = 1x210 + 35 210 = 6x35 Thus, the gcd(455, 245) = 35. 2) Show that if 11 | (6x + 5y), then there are no integer solutions to the equation 9x + 13y = 20000 Since 11 | (6x + 5y), we must have that 6x+5y = 11A, for some integer A. 9x + 13y = 33(x+y) – 4(6x+5y) (You can verify this by multiplying out the right hand side.) = 11*3(x+y) – 4*11A = 11(3(x+y) – 4A) Thus, we can conclude that 11 | (9x + 13y). However, we can do long division to verify that 11 | 20000 is false. Thus, since one side of the equation is divisible by 11 and the other is not, the equation can never be true, thus having no integer solutions in x and y. 3) Prove that if 3 | 2x + 7y, then 60 | (20x – 80y). Since 3 | 2x + 7y, 2x+7y = 3A for some integer A. We must prove that 20x – 80y = 60c, for some integer c. 20x – 80y = 20(x – 4y) = 20(3(x+y) – (x – 4y)), once again you can verify this by multiplying out what is inside the parentheses. = 20(3(x+y) – 3A) =(20)(3)(x+y – A) =60(x + y – A) Thus we can conclude that 60 | (20x – 80y).

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4) Using mod rules, find the remainder when you divide 5 192 by 7. 5 192 (5 2 ) 96 (mod 7) (25) 96 (mod 7) (4) 96 (mod 7) (4 2 ) 48 (mod 7) (16) 48 (mod 7) (2) 48 (mod 7) (2 3 ) 16 (mod 7) (8) 16 (mod 7) (1) 16 (mod 7) 1 (mod 7) Thus, the desired remainder is 1. 5) Let a and b be two positive integer. If a is odd, show that 9ab + 5b 2 is even. First we will show that the product of two odd numbers is odd. Let x and y be two odd numbers. Then we can express x=2x’ + 1 and y=2y’ + 1.
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