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Winter2005 Final with solutions

# Winter2005 Final with solutions - MATH 135 Final Exam...

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Unformatted text preview: MATH 135 Final Exam Solutions Page 1 of 11 Winter 2005 1. (a) Determine the complete solution to the linear Diophantine equation [6] 391 x + 253 y = 2760 . Solution: The Extended Euclidean Algorithm gives 1 391 1 253 1 1-1 138 1-1 2 115 1 2-3 23 5-11 17 Thus, 391(2) + 253( − 3) = 23 and gcd(391 , 253) = 23. Further, 391(240) + 253( − 360) = 2760 . The complete solution is x = 240 + 11 n y = − 360 − 17 n ∀ n ∈ Z . (b) Suppose that the complete solution to a linear Diophantine equation is given by [4] j = − 23 − 3 n k = 521 + 50 n ∀ n ∈ Z . List all positive solutions. Solution: j ≥ k ≥ − 23 − 3 n ≥ 521 + 50 n ≥ 3 n ≤ − 23 50 n ≥ − 521 n ≤ − 7 . ¯ 6 n ≥ − 10 . 42 Thus − 7 . ¯ 6 ≥ n ≥ − 10 . 42. Valid values of n are -10, 9, -8. The positive solutions are: n j k-10 7 21-9 4 71-8 1 121 MATH 135 Final Exam Solutions Page 2 of 11 Winter 2005 2. (a) Prove that if a, b, c ∈ Z with c | ab and gcd( a, c ) = 1 then c | b . [4] Solution: Since gcd( a, c ) = 1, ∃ x, y ∈ Z with ax + cy = 1. Multiplying by b gives axb + cby = b . Since c | ab and | cby , thus c | b . (b) If z = x + iy is a complex number in standard form and z negationslash = 0, give an expression for z- 1 [3] and prove that z · z- 1 = 1 . Solution: z- 1 = x- iy x 2 + y 2 . z · z- 1 = ( x + iy )( x − iy x 2 + y 2 ) = x 2 − i 2 y 2 x 2 + y 2 = 1 MATH 135 Final Exam Solutions Page 3 of 11 Winter 2005 3. Prove that 1 − 3 + 5 − 7 + ... + ( − 1) n- 1 (2 n − 1) = ( − 1) n- 1 n ∀ n ∈ P . [6] Solution: Proof by mathematical induction. When n = 1, 1 = ( − 1) 1- 1 · 1. Suppose that 1 − 3 + 5 − 7 + ... + ( − 1) k- 1 (2 k − 1) = ( − 1) k- 1 ( k ). Consider 1 − 3 + 5 − 7 + ... + ( − 1) k- 1 (2 k − 1) + ( − 1) k (2 k + 1) = ( − 1) k- 1 ( k ) + ( − 1) k (2 k + 1) by the induction hypothesis = ( − 1) k- 1 ( k − 2 k − 1) = ( − 1) k- 1 ( − k − 1) = ( − 1) k ( k + 1) By mathematical induction, 1 − 3 + 5 − 7 + ... + ( − 1) n- 1 (2 n − 1) =) − 1) n- 1 n ∀ n ∈ P . MATH 135 Final Exam Solutions Page 4 of 11 Winter 2005 4. (a) State Fermat’s Little Theorem. [3] Solution: If p is a prime and p negationslash | a then a p- 1 ≡ 1 (mod p ). (b) Find all solutions to 14 x 11 + 33 x 8 + 32 ≡ 0 (mod 77). [6] Solution: By the Chinese Remainder Theorem and Fermat’s Little Theorem, this is equivalent to the following congruences: braceleftBigg 14 x 11 + 33 x 8 + 32 ≡ 0 (mod 7) 14 x 11 + 33 x 8 + 32 ≡ 0 (mod 11) OR braceleftBigg 5 x 8 + 4 ≡ 0 (mod 7) 3 x 11 ≡ 1 (mod 11) OR braceleftBigg 5 x 2 ≡ 3 (mod 7) (since x 7 ≡ x (mod 7)) 3 x ≡ 1 (mod 11) (since x 11 ≡ x (mod 11))...
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Winter2005 Final with solutions - MATH 135 Final Exam...

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