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Moderator Message: this question is from an ongoing competition.

Define a prime $p$ as having $k$ hands in $n$'s hair if $p^k|n$ and $n|2^n+1$ . Does there exist an integer $n$ with $2012$ hands in its hair? Does there exist an integer $n$ with $2012$ distinct primes' hands in its hair?

Furthermore it would be fantastic if you could show your full logic, thank you.

For example, $3$ has 1 hand in its own hair, and I can't think of any other nice examples.

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    Interesting terminology.2012-08-02
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    When you say "with $2012$ hands in its hair", do you mean $2012$ hands of the same prime, or the sum of all hand counts of all primes? I. e., does there have to be a factor $p^{2012}$ in $n$, or does $n$ just have to have at least $2012$ prime factors, not necessarily the same?2012-08-02
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    For the first question all hands could be from the same prime, and for the second no. In general it is the sum of hands from all primes.2012-08-02
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    In general I am hypothesizing that p has to be 3 and k has to be a power of 3.2012-08-02
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    Can we prove that it works whenever that's the case?2012-08-02
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    That's quite a bit of hands in hair, someone must not be too happy about it!2012-08-02
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    Proposition: If $n$ is bald, then $n=0$.2012-08-02
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    I rolled back to the original formulation of the question because Robert's answer and most of the comments depend on it. Changing the entire terminology of a question long after it's been answered is a bad idea.2012-12-10

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