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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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