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October 28th, 2018, 10:58 PM   #1
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divisiblity of a binomial coefficient

Hello, how to show the following?? :

$C_{2^{m-1}}^{k}\times 2^k$ is divisible by $2^{m}$.
with for any $k$ such that $2^{m-1}\geq k \geq 1$.
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October 29th, 2018, 04:34 AM   #2
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I would suggest induction.

I realise now that your $${k \choose m-1}$$ looks strange. $k$ is normally no smaller than $(m-1}$. Do you have an error?
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Last edited by v8archie; October 29th, 2018 at 04:41 AM.
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October 30th, 2018, 11:14 AM   #3
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I really dont know ^^
i tryed so many methodes
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October 30th, 2018, 04:45 PM   #4
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I assume you mean: $C_k^{2^{m-1}} \times 2^k$ is divisible by $2^m$.

Let $\nu$ denote the 2-adic valuation. The above statement is then equivalent to $\nu(C_k^{2^{m-1}}) \geq m-k$. See if you can prove $\nu(C_k^{2^{m-1}}) = m - 1 - \nu(k)$, or more generally that $\nu_p(C_s^{p^r}) = r - \nu_p(s)$, where $p$ is a prime and $\nu_p$ is the $p$-adic valuation. The result is immediate from this.
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