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 April 14th, 2017, 05:22 AM #1 Senior Member   Joined: Dec 2015 From: Earth Posts: 192 Thanks: 23 Divisible for natural numbers Show that :$\displaystyle 5^n$mod $\displaystyle (40^n n!)=0$ $\displaystyle n\in N$
April 14th, 2017, 06:09 AM   #2
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Quote:
 Originally Posted by idontknow Show that :$\displaystyle 5^n$mod $\displaystyle (40^n n!)=0$ $\displaystyle n\in N$
I'm not certain which of these you mean:

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, 09:23 AM #3 Senior Member   Joined: Dec 2015 From: Earth Posts: 192 Thanks: 23 Show that $\displaystyle \frac{(5n)!}{40^n n!}\in N$ Last edited by idontknow; April 14th, 2017 at 09:28 AM.
April 15th, 2017, 08:21 AM   #4
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Quote:
 Originally Posted by idontknow Show that $\displaystyle \frac{(5n)!}{40^n n!}\in N$
Well this is a totally different question from what you first asked. Thus it should be in a brand new post.

It is a routine induction.

 April 15th, 2017, 09:56 AM #5 Member   Joined: Jan 2016 From: Athens, OH Posts: 69 Thanks: 37 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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