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# hmk4_key - COT 3100 Homework 4 Spring 2001 Assigned Due...

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COT 3100 Homework # 4 Spring 2001 Assigned: 3/26/01 Due: 4/9,10/01 (in class) 1) Use Euclid’s Algorithm to find the greatest common divisor(GCD) of 851 and 444. 851 = 1x444 + 407 444 = 1x407 + 37 407 = 11x37, so gcd(851, 444) = 37. 2) Show that if 17 | (3x + 8y), then there are no integer solutions to the equation 22x + 2y = 1602 22x + 2y = 34(x + y) – 4(3x + 8y) = 17(2x+2y) – 4(17)(c), for some integer c, using given information. = 17(2x + 2y – 4c) Thus, if x and y are integers, then 17 | (22x + 2y). However, 1602 is NOT divisible by 17, thus the two sides of the given equation can not be equal since there are not equivalent mod 17. 3) Using mod rules, find the remainder when you divide 7 214 by 13. 7 214 (7 2 ) 107 (mod 13) (49) 107 (mod 13) (-3) 107 (mod 13) ((-3) 4 ) 26 (-3) 3 (mod 13) (81) 26 (-3) 3 (mod 13) (3) 26 (-27)(mod 13) (3) 26 (3)(-9)(mod 13) (3) 27 (-9) (mod 13) (3 3 ) 9 (-9) (mod 13) (27) 9 (4) (mod 13) 1 9 (4) (mod 13) 4 (mod 13), so the remainder desired is 4. 4) Let a and b be two positive integer. Show that 3ab + 7ab 2 is even. 3ab + 7ab 2 = ab(3 + 7b)

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If either a or b is even, the quantity above is automatically even since a and b are both divisors of the quantity. Thus, if the quantity is to be odd, then both a and b must be odd. But in this case, 7b would be an odd number, making 3+7b an even number. Thus, in all possible cases, at least one of the three factors a, b, (3+7b) is even, making 3ab + 7ab 2 even for all positive integers a and b.
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