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Math Help - the House Problem

  1. #1
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    the House Problem

    Consider this series: 1, 2, 3,..., N-1, N, N+1,..., M.
    Given 1 + 2 + 3 + ... + (N-1) = (N+1) + (N+2) +...+ M. Find N.
    (hint: N is a 4 digits number)
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  2. #2
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    Quote Originally Posted by john_n82 View Post
    Consider this series: 1, 2, 3,..., N-1, N, N+1,..., M.
    Given 1 + 2 + 3 + ... + (N-1) = (N+1) + (N+2) +...+ M. Find N.
    (hint: N is a 4 digits number)

    Note, 1+2+...+(N-1) = \tfrac{1}{2}N(N-1) \text{ and }(N+1)+(N+2)+...+M = \tfrac{1}{2}(M-N)(M-N+1) + N(M-N)

    Therefore, N^2 = \tfrac{1}{2}M(M+1), therefore M^2 + M - 2N^2 = 0.
    We need the discriminant to be a square, so we need 1 + 8N^2 = x^2 .
    This is a Pellian equation.
    Last edited by ThePerfectHacker; March 7th 2009 at 02:08 PM.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Note, 1+2+...+(N-1) = \tfrac{1}{2}N(N-1) \text{ and }(N+1)+(N+2)+...+M = \tfrac{1}{2}(M-N)(M-N+1) + N(M-N)

    Therefore, N^2 = \tfrac{1}{2}M(M+1), therefore M^2 + M - 2N^2 = 0.
    We need the discriminant to be a square, so we need 1 + 8N^2 = x^2 .
    This is a Pellian equation.
    I found M = 8, so N = 6.
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  4. #4
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    Quote Originally Posted by john_n82 View Post
    Consider this series: 1, 2, 3,..., N-1, N, N+1,..., M.
    Given 1 + 2 + 3 + ... + (N-1) = (N+1) + (N+2) +...+ M. Find N.
    (hint: N is a 4 digits number)
    i think you mean N^2 is a four-digit number

    in this case M = 49, N = 35
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  5. #5
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    i found 2 more values for M

    M = 288, N = 204
    M 9800, N = 6930

    What is the exact answer for this problem?
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  6. #6
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    Quote Originally Posted by john_n82 View Post
    i found 2 more values for M

    M = 288, N = 204
    M 9800, N = 6930

    What is the exact answer for this problem?
    There are many answers to this problem.
    You are told that you need N to be a 4 digit number so N=6930
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