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April 14th, 2017, 04:22 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 134 Thanks: 20  Divisible for natural numbers
Show that :$\displaystyle 5^n$mod $\displaystyle (40^n n!)=0$ $\displaystyle n\in N$ 
April 14th, 2017, 05:09 AM  #2  
Senior Member Joined: Feb 2010 Posts: 600 Thanks: 87  Quote:
A) Show that $\displaystyle 5^n \equiv 0$ mod $\displaystyle 40^nn!$ or B) Show that $\displaystyle 40^nn! \equiv 0$ mod $\displaystyle 5^n$ A is not true and B is trivial.  
April 14th, 2017, 08:23 AM  #3 
Senior Member Joined: Dec 2015 From: Earth Posts: 134 Thanks: 20 
Show that $\displaystyle \frac{(5n)!}{40^n n!}\in N$
Last edited by idontknow; April 14th, 2017 at 08:28 AM. 
April 15th, 2017, 07:21 AM  #4 
Senior Member Joined: Feb 2010 Posts: 600 Thanks: 87  
April 15th, 2017, 08:56 AM  #5 
Member Joined: Jan 2016 From: Athens, OH Posts: 32 Thanks: 19 
Here's a solution for your corrected problem: I just noticed the last post and realized a much easier induction proof is the way to go. 

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