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Show n^3 + 11n Divisible by 6

Date: 12/11/2002 at 00:41:56
From: Jasmine
Subject: Algebra

If n is a natural number, show that for all values of n, (n^3+11n) is 
divisible by 6.


Date: 12/11/2002 at 10:33:52
From: Doctor Ian
Subject: Re: Algebra

Hi Jasmine,

I would probably try to use induction. That is, I know that the 
formula works for small values of n: n=1, n=2, and so on. 

So what I want to show is that if it works for any particular n, it
must also work for the next value of n, i.e., (n+1).  That is, 

  If n^3 + 11n is divisible by 6, 
  then (n+1)^3 + 11(n+1) is also divisible by 6. 

Now, 

    (n+1)^3 + 11(n+1) 

  = (n^3 + 3n^2 + 3n + 1) + (11n + 11)

  = (n^3 + 11n) + (3n^2 + 3n + 12)

We're already assuming that the first part,

  (n^3 + 11n) 

is divisible by 6; so now we have a smaller problem, which is to show
that the second part, 

  3n^2 + 3n + 12

is always divisible by 6. (Hint: Show that it's divisible by both 2
and 3.)

Can you take it from here? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 12/11/2002 at 19:47:05
From: Jasmine
Subject: Thank you (Algebra)

Got the hint. Thank you so much. You have been a great help.

Jasmine
Associated Topics:
High School Number Theory

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