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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?? : $C_{2^{m1}}^{k}\times 2^k$ is divisible by $2^{m}$. with for any $k$ such that $2^{m1}\geq k \geq 1$. 
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 $${k \choose m1}$$ looks strange. $k$ is normally no smaller than $(m1}$. Do you have an error? 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: $C_k^{2^{m1}} \times 2^k$ is divisible by $2^m$. Let $\nu$ denote the 2adic valuation. The above statement is then equivalent to $\nu(C_k^{2^{m1}}) \geq mk$. See if you can prove $\nu(C_k^{2^{m1}}) = m  1  \nu(k)$, 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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binomial, coefficient, divisiblity 
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