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Math Help - Numbers

  1. #1
    Senior Member tukeywilliams's Avatar
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    Numbers

    Suppose the positive integer  n is odd. First Al writes the numbers  1,2, \ldots, 2n on the blackboard. Then he picks any two numbers  a,b erases them and writes instead  |a-b| . Prove that an odd number will remain at the end.
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  2. #2
    Senior Member tukeywilliams's Avatar
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    Think I got it. I had to look for the invariant.
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  3. #3
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    In the beginning there are n even integers there are n odd integers.

    When we replace two the parity of the odd integers stays odd at each step. So we have 2 integers left 1 is odd and 1 is even. And when we subtract them we have an odd integer.
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  4. #4
    Senior Member tukeywilliams's Avatar
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    I said the sum initially is  n(2n+1) which is odd. The sum is reduced by  2 \text{min}(a,b) each time which is even. So the parity stays constant, and the sum is  1\mod 2 . The parity is odd at the beginning. So it is odd at the end.
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  5. #5
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    Quote Originally Posted by tukeywilliams View Post
    I said the sum initially is  n(2n+1) which is odd. The sum is reduced by  2 \text{min}(a,b) each time which is even. So the parity stays constant, and the sum is  1\mod 2 . The parity is odd at the beginning. So it is odd at the end.
    It is similar to mine. But I find your approach more elegant.
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