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April 14th, 2017, 04:22 AM   #1
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Divisible for natural numbers

Show that :$\displaystyle 5^n$mod $\displaystyle (40^n n!)=0$
$\displaystyle n\in N$
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April 14th, 2017, 05:09 AM   #2
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Quote:
Originally Posted by idontknow View Post
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.
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April 14th, 2017, 08:23 AM   #3
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Show that $\displaystyle \frac{(5n)!}{40^n n!}\in N$

Last edited by idontknow; April 14th, 2017 at 08:28 AM.
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April 15th, 2017, 07:21 AM   #4
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Quote:
Originally Posted by idontknow View Post
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.
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April 15th, 2017, 08:56 AM   #5
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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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