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 Number Theory Number Theory Math Forum

 October 28th, 2018, 09:58 PM #1 Newbie   Joined: Jul 2018 From: morocco Posts: 26 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, 03:34 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,689 Thanks: 2669 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 03:41 AM. October 30th, 2018, 10:14 AM #3 Newbie   Joined: Jul 2018 From: morocco Posts: 26 Thanks: 0 Math Focus: algebraic number theory I really dont know ^^ i tryed so many methodes October 30th, 2018, 03:45 PM #4 Senior Member   Joined: Aug 2017 From: United Kingdom Posts: 313 Thanks: 112 Math Focus: Number Theory, Algebraic 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 Tags binomial, coefficient, divisiblity Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Casper Number Theory 4 January 5th, 2017 08:06 AM panky Algebra 1 September 30th, 2016 05:31 PM pikachu26134 Number Theory 3 July 26th, 2011 07:03 PM sunflower Number Theory 0 April 2nd, 2011 07:15 AM coolhandluke Applied Math 1 March 26th, 2010 04:55 PM

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