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HW7sol

# HW7sol - → 2608977 Now to decrypt we take the results...

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MATH 3360 – Applicable Algebra. HW 7 Solutions (1) Problem 13 on page 230. (graded) Solution: Consider H = { 1 , 7 , 9 , 15 } . Note that this set is closed under the operation and that each element is its own inverse. Hence this is a non-cyclic subgroup of U 16 . (2) Problem 14 on page 230. Solution: Follow the hint in the book. (3) Problem 15 on page 233. Solution: (i) { 1 , 7 , 11 } , { 2 , 14 , 3 } , { 4 , 9 , 6 } , { 5 , 16 , 17 } , { 8 , 18 , 12 } , { 10 , 13 , 15 } are distinct cosets. Note that 6 · 3 = 18. (ii) { 1 , 12 , 11 , 18 , 7 , 8 } , { 2 , 5 , 3 , 17 , 14 , 16 } , { 4 , 10 , 6 , 15 , 9 , 13 } are distinct cosets. Note that 3 · 6 = 18. (iii) { 1 , 5 , 6 , 11 , 17 , 9 , 7 , 16 , 4 } , { 2 , 10 , 12 , 3 , 15 , 18 , 14 , 13 , 8 } are distinct cosets. Note that 2 · 9 = 18. (4) Problem 33 on page 241. (graded) Solution: (i) ker L [2] = { [0] 10 , [5] 10 } . (ii) ker L [3] = { [0] 11 } . (iii) ker L [4] = { [0] 12 , [3] 12 , [6] 12 , [9] 12 } . (5) Problem 49 on page 252. Solution: We know U 8 = { 1 , 3 , 5 , 7 } . So the homomorphisms are L 1 : (1 , 3 , 5 , 7) (1 , 3 , 5 , 7), L 3 : (1 , 3 , 5 , 7) (3 , 1 , 7 , 5), L 5 : (1 , 3 , 5 , 7) (5 , 7 , 1 , 3), L 7 : (1 , 3 , 5 , 7) (7 , 5 , 3 , 1). (6) Problem 6 on page 205. Solution: To encrypt, we let c = w e mod m = 5552 17 mod 3501697 = 974771. Similarly 0307 889719, 4562 3245212 and 1587
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Unformatted text preview: → 2608977. Now to decrypt, we take the results from above to power d = 1440269 modulo m = 3501697 to get back the numbers we were encoding. This problem has to be done using some programming language, for example Java, since the calculator would not be able to deal with such large numbers. (7) Problem 3 on page 261. (graded) Solution: We have x = 10 + 15 k = 17 + 28 n , for n , k ∈ Z . By Euclidean algorithm we get 15(-13 · 7) + 28(7 · 7) = 7, so x = 10 + 15(-91) =-1355 mod 420 ≡ 325 mod 420. So the smallest positive solution is x = 325. (8) Problem 7 on page 261. (graded) Solution: Solving the ﬁrst two congruences gives x ≡ 38 mod 60. Now solving this equation with x ≡ 10 mod 14, gives x ≡ 38 mod 420. 1...
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