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#1 2007-03-19 04:37:29
- luca-deltodesco
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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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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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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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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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Re: apparently simple proof
Either n is divisible by 3 or its 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
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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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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
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Re: apparently simple proof
I even find it odd.
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#9 2007-03-19 15:55:06
- Stanley_Marsh
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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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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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