hw2 - Solutions to Assignment 2 Section 2.1 2 A complete...

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September 16, 2009 Section 2.1 2. A complete residue system modulo 17 can be taken to be {- 8 , - 7 , ··· , - 1 , 0 , 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 , 0 , 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 , ··· , 0 , 1 , ··· , 4 , 5 } . Clearly it suffices to check the following: 0 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. 20. Note that 42 = 2 × 3 × 7. Now by Fermat’s Little Theorem, n 7 n mod 7 (1) for any integer n . On the other hand, the congruences n 7 n mod 3 (2) and n 7 n mod 2 (3) is easily checked to be valid for any n (for example, to check (2), it suffices to evaluate both sides on a c.r.s. modulo 3, say given by {- 1 , 0 , 1 } ). Combining these three congruences, we obtain n 7 n mod 42. 1
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This note was uploaded on 10/10/2009 for the course MATH 115 taught by Professor Mok during the Spring '07 term at Berkeley.

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hw2 - Solutions to Assignment 2 Section 2.1 2 A complete...

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