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Math Help - Help Find f(2007)

  1. #1
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    Help Find f(2007)

    I need Your Help !

    Ehiter f : \mathbb{N} \longrightarrow \mathbb{N} , Such That :

    f(n+1) > f(n) And f(f(n))=3n.

    Find f(2007) .
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  2. #2
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    Quote Originally Posted by Oricalcos View Post
    I need Your Help !

    Ehiter f : \mathbb{N} \longrightarrow \mathbb{N} , Such That :

    f(n+1) > f(n) And f(f(n))=3n.

    Find f(2007) .
    I don't understand what the inequality has to do with the problem. If f(f(n))=3n, let n=2007 so that f(f(2007))=3(2007). I could be wrong, but this makes sense to me.
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    I don't understand what the inequality has to do with the problem. If f(f(n))=3n, let n=2007 so that f(f(2007))=3(2007). I could be wrong, but this makes sense to me.
    Read more precisely the question... It's f(2007), not f(f(2007)) that we're looking for.
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  4. #4
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    Quote Originally Posted by VonNemo19 View Post
    I don't understand what the inequality has to do with the problem. If f(f(n))=3n, let n=2007 so that f(f(2007))=3(2007). I could be wrong, but this makes sense to me.
    Quote Originally Posted by Moo View Post
    Read more precisely the question... It's f(2007), not f(f(2007)) that we're looking for.
    Von Nemo wasn't the only one who misread the question. I assumed that the answer must be f(n) = \sqrt3\,n. But of course that doesn't map \mathbb{N} to \mathbb{N}.

    The answer is actually a good deal more subtle than that. Start by thinking about f(n) when n is small. We know that f(f(1)) = 3, and that f is an increasing function. That means that f(1) must be 2 (since that is the only available number greater than 1 and less than 3). Then f(2) = f(f(1)) = 3. Next, f(f(2)) = 6, in other words f(3) = 6, and therefore f(6) = f(f(3)) = 9. But f(4) and f(5) must be sandwiched between f(3) and f(6), which implies that f(4) = 7 and f(5) = 8.

    Continuing in that way, you see that f(n) is uniquely determined for each n, and that f(n) seems to increase in steps of either 1 or 3. That suggests that it might be helpful to express n in base 3.

    Spoiler:
    Let n = (a_1a_2\ldots a_k)_3 be the ternary expansion of n. Define f(n) as follows. If a_1=1 then f(n) = (2a_2\ldots a_k)_3. If a_1=2 then f(n) = (1a_2\ldots a_k0)_3.

    Here's my answer:
    Spoiler:
    2007 = 2202100_3, and so f(2007) = 12021000_3, which is 3834 in decimal notation.
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