Homework 2 - Solutions to Assignment 2 September 13, 2007...

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Unformatted text preview: Solutions to Assignment 2 September 13, 2007 Section 2.1 2. A complete residue system modulo 17 can be taken to be {- 8 ,- 7 , ,- 1 , , 1 , , 7 , 8 } For each element, we can add suitable multiples of 17 to make it divisible by 3. One possible answer is: { 9 ,- 24 ,- 6 , 12 ,- 21 ,- 3 , 15 ,- 18 , , 18 ,- 15 , 3 , 21 ,- 12 , 6 , 24 ,- 9 } 8. We check the values of n 2 mod 10 for the complete residue system modulo 10 given by {- 4 ,- 3 , , , 1 , , 4 , 5 } . Clearly it suffices to check the following: 2 0 mod 10 1 2 1 mod 10 2 2 4 mod 10 3 2 9 mod 10 4 2 6 mod 10 5 2 5 mod 10 as stated. 10. For this problem, we can either use Eulers formula for ( n ), or we can list a a reduced residue system for that modulus and count the number of elements: the result is: n ( n ) 1 1 2 1 3 2 4 2 5 4 6 2 1 7 6 8 4 9 6 10 4 11 10 12 4 20. Note that 42 = 2 3 7. Now by Fermats Little Theorem, n 7 n mod 7 (1) for any integer n . On the other hand, the congruences....
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Homework 2 - Solutions to Assignment 2 September 13, 2007...

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