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Unformatted text preview: 64 , 125 agrees with 0 , 1 , 2 , 3 , 4 , 5.) Putting this all together we have n 3 mod 6 = r 3 mod 6 = r mod 6 = n mod 6. Another proof could be based on the fact that n 3n = ( n1) n ( n +1). The right hand side is even since either n or n +1 is even. Since n1 , n and n +1 are three consecutive numbers one of them is divisible by three. This implies that the right hand side is also divisible by three. Since ( n1) n ( n + 1) is divisible by 2 and by 3 it is divisible by 6. By Chap. 0, Number 9 (see back of book), since 6 divides n 3n that means that n 3 mod 6 = n mod 6. 1...
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This note was uploaded on 10/27/2009 for the course MATH 330 taught by Professor Staff during the Spring '08 term at Ill. Chicago.
 Spring '08
 STAFF
 Algebra, Remainder

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