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## #1 2007-03-19 04:37:29

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

### apparently simple proof

apparently there is some simple proof for this, but i cant find it.

i have to proove that n(n+1)(n+2) where n is a natural number, is always a multiple of 6

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## #2 2007-03-19 04:58:27

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: apparently simple proof

The expression (call it A) is a product of three consecutive integers, and any three consecutive integers always contain a multiple of 3. Therefore A is divisible by 3.

It is also divisible by 2 because any three consecutive integers must also contain at least one even number.

Since A is divisible by both 2 and 3, and 2 and 3 are coprime, it follows that A is divisible by 6.

Last edited by JaneFairfax (2007-03-19 04:59:43)

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## #3 2007-03-19 05:03:36

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

### Re: apparently simple proof

proof its divisable by 3?

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## #4 2007-03-19 05:03:59

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

### Re: apparently simple proof

oh right, thats common sense, nevermind

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## #5 2007-03-19 06:09:28

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

### Re: apparently simple proof

Either n is divisible by 3 or its not.

If it is, fine. If not, then n = 3k+1 or n = 3k−1 for some integer k. It then follows that (n+2) or (n+1) respectively would be divisible by 3.

So one of n, (n+1), (n+2) is always divisible by 3. Hence the product n(n+1)(n+2) is always divisible by 3.

Last edited by JaneFairfax (2007-03-19 12:00:01)

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## #6 2007-03-19 09:16:41

MathsIsFun
Registered: 2005-01-21
Posts: 7,685

### Re: apparently simple proof

Or to put it another way, the next "divisible by 3" number is only 3 numbers away, so if you have 3 consecutive numbers, one will be.

Isn't it odd that we can say even, but there is no easy word for "divisible by 3" ... maybe "threven" ?  And the two in-between would be "throdds"?

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## #7 2007-03-19 10:53:38

luca-deltodesco
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Registered: 2006-05-05
Posts: 1,470

### Re: apparently simple proof

and isn't it odd that you happened to use odd when talking about that, referring to the common pair of odd and even

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## #8 2007-03-19 14:02:33

MathsIsFun
Registered: 2005-01-21
Posts: 7,685

### Re: apparently simple proof

I even find it odd.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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## #9 2007-03-19 15:55:06

Stanley_Marsh
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Registered: 2006-12-13
Posts: 345

### Re: apparently simple proof

6=2x3,
There are two kinds of integers 2k and 2k+1 , thus it can be divided by 2
There are three kinds of integers 3k , 3k+1, 3k+2 , thus it can be divided by 3

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## #10 2007-03-20 02:32:49

George,Y
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Registered: 2006-03-12
Posts: 1,332

### Re: apparently simple proof

If you divide a natural by 3, the remainder is 0, 1, or 2. No other possibilities.

X'(y-Xβ)=0

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## #11 2007-03-20 02:34:09

George,Y
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Registered: 2006-03-12
Posts: 1,332

### Re: apparently simple proof

This post has inspired me in my question!

X'(y-Xβ)=0

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