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hw3 - EECS 203 Homework 3 Solution Section 1.6 1(E 6 An odd...

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EECS 203 Homework 3 Solution Sep 26, 2009 Section 1.6 1. (E) 6 An odd number has the form of , where is an integer. Let and be two integers, say and . Their product is which is also in the form of . Therefore is odd. 2. (E) 18ab a. We prove the contrapositive: if is odd, then is odd. Assume , then . Therefore is two times some integer plus 1, so it is odd. b. Suppose that is odd. Then is odd. (Because from the previous exercise, we know that the product of two odd numbers is odd). Therefore is odd. (An odd number plus 2 is odd). But the problem assumes that is even. Such contradiction indicates that our supposition was wrong, and therefore the proof by contradiction is complete. 3. (E) 24 Solution 1 : We give a proof by contradiction. If there were at most two days falling in the same month, then we could have at most days, as there are 12 months. However we have chosen 25 days, therefore at least three of them must fall in the same month. Solution 2 : think days as balls and months as bins . We put 25 balls into 12 bins. Then at least one bin should contain at least three balls. Otherwise, if every bin contains two or less balls, then the total number of balls will be no more than , yielding a contradiction since we have a total number of 25 balls.
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