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tut05sols - 2 The University of Sydney MATH2068 Number...

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The University of Sydney MATH2068 Number Theory and Cryptography (http://www.maths.usyd.edu.au/u/UG/IM/MATH2068/) Semester 2, 2008 Lecturer: R. Howlett Tutorial 5 1. Let m be a positive integer. In our terminology, a residue mod m is a natural number less than m , and the residue of k mod m is the remainder when k is divided by m . If a and b are residues mod m then we say that a is the inverse of b (mod m ) if ab 1 (mod m ). (Then, equally, b is the inverse of a (mod m ).) ( i ) Let m = 21. Find, if possible, the inverses of 8, 2, 5, 17, 15, 7. ( ii ) Let m = 84. Find, if possible, the inverses of 17, 83, 33, 23. Solution. ( i ) 8 2 = 64 1 (mod 21); so 8 is the inverse of 8. Obviously 11 is the inverse of 2, since 22 1 (mod 21). Since 5 × 4 ≡ - 1 (mod 21) the inverse of 5 is the residue of - 4, namely 17. The inverse of 17 is therefore 5. Since gcd(15 , 21) = 1 there is no k such that 15 k 1 (mod 21); so 15 has no inverse. Similarly, gcd(7 , 21) = 1, and so 7 has no inverse mod 21. ( ii ) We can immediately spot that 17 × 5 1 (mod 84); so the inverse of 17 is 5. Since 83 ≡ - 1 (mod 84) and ( - 1) 2 = 1, it follows that 83 must be its own inverse. Since gcd(84 , 33) = 3 = 1, the residue 33 has no inverse mod 84. Finally to find the inverse of 23 we probably need to use the Euclidean Algorithm. We find 84 = 3 × 23 + 15, then 23 = 15 + 8, then 15 = 8 + 7, then 8 = 7 + 1 and finally 7 = 7 × 1. We construct the Extended Euclidean Algorithm table 84 23 15 8 7 1 0 3 1 1 1 7 0 1 3 4 7 11 84 1 0 1 1 2 3 23 and deduce that 11 × 23 ≡ ± 1 (mod 84). In fact 11 × 23 = 253 = 3 × 84 + 1. So the inverse of 23 is 11. 2. Solve the following systems of simultaneous congruences. ( i ) 23 x 5 (mod 84) 4 x 15 (mod 37) ( ii ) 3 x 1 (mod 5) 2 x 10 (mod 12) 7 x 2 (mod 17) 2 Solution.
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