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Unformatted text preview: Homework #3 Solutions p 23, #4. s = 3 and t = 2 work since 7 s + 11 t = 21 + 22 = 1 . These choices are not unique since s = 8, t = 5 also work: 7 s + 11 t = 56 55 = 1 . p 24, #30. Given any integer n , at least one of the three consecutive numbers n 1 ,n,n +1 must be divisible by 2 and at least one of them must be divisible by 3. Therefore the product ( n 1) n ( n + 1) = n ( n 2 1) = n 3 n must be divisible by 2 3 = 6. It follows that n 3 and n must have the same remainder when divided by 6, i.e. n 3 mod 6 = n mod 6. p 54, #8 We must verify that the 4 group axioms hold for the set S = { 5 , 15 , 25 , 35 } together with the operation of multiplication modulo 40. Since we know that this operation is associative on all of Z n , it will be associative on S as well. We need only verify closure, the existence of an identity, and the existence of inverses. We can do this by building a Cayley table for S : 5 15 25 35 5 25 35 5 15 15 35 25 15 5 25 5 15 25 35 35 15 5 35 25 The table shows that S is closed under multiplication mod 40, that 25 is the identity of S , and, since 25 appears in each row, every element has an inverse. In fact, each element is its own inverse!...
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This note was uploaded on 02/29/2012 for the course MATH 3362 taught by Professor Ryandaileda during the Fall '10 term at Trinity University.
 Fall '10
 RyanDaileda

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