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    (Original post by marioman)
    The way you worded your post suggested that you didn't understand how it could be divisible by 2 whilst not having a 2 that can be taken out as a factor.
    I was providing the example to show the OP that what he said wasn't necessarily true.
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    (Original post by marioman)
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    lol, you totally missed his point...
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    (Original post by marioman)
    The way you worded your post suggested that you didn't understand how it could be divisible by 2 whilst not having a 2 that can be taken out as a factor.
    It's ok, you misinterpreted what is said. Easy mistake to make.
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    (Original post by B_9710)
    I was providing the example to show the OP that what he said wasn't necessarily true.
    i not an intelligent man

    which example of mine were you proving not always true?
    (Original post by Zacken)
    lol.Essentially you have 4 * (even number), so that's always divisible by 8 since it will always have a factor of 8 in it, to make that clear to the reader, you put it in the form 8g(k) so that the factor of 8 is apparent.
    Ah ok thanks!
    (Original post by B_9710)
     k^2+k \ (k\in \mathbb{Z}) is always divisible by 2 but doesn't have a 2 out that can be taken out as a factor seemingly.
    So you have to make it obvious?
 
 
 
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