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Math Help - find the 1991st term

  1. #1
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    find the 1991st term

    Let 2,3,5,6,7,10,11,... be an increasing sequence of positive integers that are neither the perfect squares nor the cubes. Find the 1991st term.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chris86 View Post
    Let 2,3,5,6,7,10,11,... be an increasing sequence of positive integers that are neither the perfect squares nor the cubes. Find the 1991st term.
    Let's define  f(n) to be the number of numbers less than or equal to  n such that they are neither the perfect squares or cubes.

    We then have  \displaystyle f(n) = n - \lfloor \sqrt{n}\rfloor - \lfloor \sqrt[3]{n}\rfloor + \lfloor \sqrt[6]{n}\rfloor .

    The intuition behind  f(n) is we take every number up to  n and then subtract all of the perfect squares and cubes. We must then account for any duplicates.

    Now solve  f(n) = 1991 .
    Last edited by chiph588@; January 13th 2011 at 09:01 PM.
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by chiph588@ View Post
    Let's define  f(n) to be the number of numbers less than or equal to  n such that they are neither the perfect squares or cubes.

    We have  \displaystyle f(n) = n - \lfloor \sqrt{n}\rfloor - \lfloor \sqrt[3]{n}\rfloor + \lfloor \sqrt[6]{n}\rfloor .

    Intuition behind  f(n) is we take every number up to  n and then subtract all of the perfect squares and cubes. We then must account for duplicate counting.

    Now solve  f(n) = 1991 .
    Don't know a great way to solve that though. Perhaps binary search in the interval  [0,4000] or something like that. Or you could just plot it and solve via inspection (that's how I did it).
    Last edited by chiph588@; January 13th 2011 at 09:00 PM.
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