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 November 9th, 2017, 10:41 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 831 Thanks: 113 Math Focus: Elementary Math 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 04:19 PM. November 9th, 2017, 11:16 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,113 Thanks: 2327 What happens when $n$ = 1? November 9th, 2017, 11:52 AM   #3
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Quote:
 Originally Posted by skipjack What happens when $n$ = 1?
$5! = 120$

$40^1 = 40$

$1! = 1$

$\dfrac{120}{40}=3 \in \mathbb{N}$

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, 07:29 PM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,641 Thanks: 1475 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 $40$ Thanks from topsquark November 10th, 2017, 03:04 AM #5 Senior Member   Joined: Dec 2015 From: Earth Posts: 831 Thanks: 113 Math Focus: Elementary Math 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 06:10 AM. November 10th, 2017, 06:04 PM #6 Member   Joined: Jan 2016 From: Athens, OH Posts: 93 Thanks: 48 Legendre's formula provides an easy proof without induction:  Tags natural, set 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 lenoramarie Pre-Calculus 2 November 1st, 2014 05:09 AM mathkid Calculus 4 August 27th, 2012 10:59 AM tsl182forever8 Calculus 2 March 1st, 2012 07:16 PM jinjouk Number Theory 12 June 3rd, 2008 07:11 AM tsl182forever8 Algebra 1 December 31st, 1969 04:00 PM

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