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Unformatted text preview: 2 p , p is not divisible by 2. So p cannot be equal to 6 k , 6 k +2, 6 k +4 (because these are divisible by 2). Similarly, p is a prime and 3 p , p is not divisible by 3, and hence 6k+3 is not possible. Thus, we have only two cases; namely, 1 6 + = k p or 5 6 + = k p . 45. (4pts) They are equal. Proof. Suppose m, n , and d are integers and ) ( n m d . By definition of divisibility, dk n m = for some integer k . Hence dk n m + = . Also denote d n mod by r . Then by definition of mod, we have r dq n + = where q and r are integers and d r < . By substitution, r k q d r dk dq dk r dq dk n m + + = + + = + + = + = ) ( ) ( ) ( . Since k q + is an integer and d r < , the above equation means the quotient of the division of n by d is ( q+k ) and the remainder is r . That is r d m = mod . This shows that d m mod and d n mod are equal....
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 Spring '10
 seckin
 Integers

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