My Math Forum divisiblity of a binomial coefficient

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 October 28th, 2018, 10:58 PM #1 Newbie   Joined: Jul 2018 From: morocco Posts: 19 Thanks: 0 Math Focus: algebraic number theory 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 $2^{m-1}\geq k \geq 1$.
 October 29th, 2018, 04:34 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra 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? Thanks from Chems Last edited by v8archie; October 29th, 2018 at 04:41 AM.
 October 30th, 2018, 11:14 AM #3 Newbie   Joined: Jul 2018 From: morocco Posts: 19 Thanks: 0 Math Focus: algebraic number theory I really dont know ^^ i tryed so many methodes
 October 30th, 2018, 04:45 PM #4 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 282 Thanks: 85 Math Focus: Algebraic Number Theory, Arithmetic Geometry 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. Thanks from topsquark and Chems

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