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Question: only ic and ii please...

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Example 5. Find gcd(41,12) and express it as a linear combination of 41 and 12. Solution. The algorithm is not needed to find gcd(41,12). In fact, 1 and 41 are the only positive divisors of 41, so gcd(41,12) 1 because 41 does not divide 12. However, guessing a linear combination1-x .41 +y .12 is not easy. The euclidean algorithm gives 41 3.12 + 5 12-2 5+ 2 5 -2.2+1 2 -2.10 Hence, gcd(41, 12)-1 as expected. Elimination of remainders gives 1-5-2.2 =5-2(12-2-5) 55- 2 12 -5(41- 3-12) -2 12 5.41-17.12 which is the required linear combination. Problem 1. (i) Do the following by imitating the procedure in Example 5 in page 35 of the textbook. (3 points) (a) Find g -gcd(37, 13) (b) Express g = 37a +13b with a, b E Z. (c) Find the unique integer n f0,...,36} such that 13n1 mod 37. (ii) Let u and v be positive integers. Prove that gcd(u, v)- 1 if and only if there exists an integer w such that uw 1 mod v (3pts). Hints: use the fact that gcd(u, v) can be expressed as a linear combination of u and v.only i)c) and ii) please

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