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

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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**JaneFairfax****Member**- Registered: 2007-02-23
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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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**luca-deltodesco****Member**- Registered: 2006-05-05
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proof its divisable by 3?

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**luca-deltodesco****Member**- Registered: 2006-05-05
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oh right, thats common sense, nevermind

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**JaneFairfax****Member**- Registered: 2007-02-23
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Either *n* is divisible by 3 or its not.

If it is, fine. If not, then *n* = 3*k*+1 or *n* = 3*k*−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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**MathsIsFun****Administrator**- Registered: 2005-01-21
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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"?

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

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**luca-deltodesco****Member**- Registered: 2006-05-05
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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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**MathsIsFun****Administrator**- Registered: 2005-01-21
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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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**Stanley_Marsh****Member**- Registered: 2006-12-13
- Posts: 345

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

Numbers are the essence of the Universe

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**George,Y****Member**- Registered: 2006-03-12
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If you divide a natural by 3, the remainder is 0, 1, or 2. No other possibilities.

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**George,Y****Member**- Registered: 2006-03-12
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