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# tut01sols - 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 1 1. In each case find the greatest common divisor of a and b , and integers r, s such that gcd( a, b ) = ra + sb . ( i ) a = 14, b = 35; ( ii ) a = 168, b = 132; ( iii ) a = 847, b = 510. Solution. ( i ) A B 35 14 7 0 Q 2 2 L K 0 1 2 5 N M 1 0 1 2 . This table is computed as follows. Firstly, the first two columns of numbers are filled in, using the given values of a and b (namely 14 and 35) as the initial values of A and B , and 0, 1, 1, 0 as the initial values of L , K , N and M (respectively). Choose A to be the larger of the two given numbers and B the smaller – that is, put A = 35 and B = 14 – rather than vice versa. (Putting A = 14 and B = 35 works too, it just takes one step longer.) Now divide B into A , and put the remainder R into the top row (the new value of B ) and the quotient Q underneath it. Underneath Q put QK + L (the new K ) and under that put QM + N (the new M ). Repeat until B = 0. The second last entry in the top row will be the gcd (7 in this case) and the integers r and s that we seek are either the final values of N and - L (if completion occurs after an even number of steps) or the final values of - N and L (for an odd number of steps). In this example we finished in two steps; so r is 1 (the 2nd last number in the last row) and s is - 2 (the negative of the 2nd last number in the second last row). We readily check that 1 × 35 + ( - 2) × 14 = 7.

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tut01sols - 2 The University of Sydney MATH2068 Number...

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