November 9th, 2017, 09:41 AM  #1 
Senior Member Joined: Dec 2015 From: Earth Posts: 227 Thanks: 26  Natural set
Show that $\displaystyle (5n)! \; \vdots \; 40^n n! \; \; ,\, n \in N$ or $\displaystyle \frac{(5n)!}{40^n n!} \in N$ If can't post proof, maybe we can (use induction). Last edited by skipjack; November 9th, 2017 at 03:19 PM. 
November 9th, 2017, 10:16 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
What happens when $n$ = 1?

November 9th, 2017, 10:52 AM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 1,979 Thanks: 1027  $5! = 120$ $40^1 = 40$ $1! = 1$ $\dfrac{120}{40}=3 \in \mathbb{N}$ When attempting induction you get to a 4th degree polynomial in $n$ with a bunch of integer coefficients divided by 8. I wasn't able to proceed any further. Maybe someone else can succeed. 
November 9th, 2017, 06:29 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 1,979 Thanks: 1027 
ok.. I leave you to turn this into a formal proof but here's an example of what's going on Let $n=3$ $(5n)! = (15)(14) \dots (4)(3)(2)(1)$ $\dfrac{(5n)!}{n!} = (15)(14) \dots (4)$ Now in those factors we have $(8,5),(4,10),(6,12,15)$ each of which multiplies to $40$. That gets you your $40^3$ multiplied by the unused factors. If $n=4$ $(5n)! = (20)(19) \dots (5)$ we have the same tuples as before to get $40^3$ and now we have $(20)(19)\dots (16)$ to extract another factor of $40$ out of. One such tuple is $(16,20)$ There are others. It shouldn't be to hard to frame this as induction. You just need to show there exists a combination of factors $\in [5n+1,5(n+1)]$ that multiply to a multiple of $40$ 
November 10th, 2017, 02:04 AM  #5 
Senior Member Joined: Dec 2015 From: Earth Posts: 227 Thanks: 26 
This is the way I made the Set , but I don't know how to get there mathematically.
Last edited by skipjack; November 10th, 2017 at 05:10 AM. 
November 10th, 2017, 05:04 PM  #6 
Member Joined: Jan 2016 From: Athens, OH Posts: 89 Thanks: 47 
Legendre's formula provides an easy proof without induction: 

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