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Thread: some help with number theory

  1. #1
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    some help with number theory

    Show that if a positive integer $\displaystyle n=6a+5,a \geq 0$, then $\displaystyle n=3a'-1,a'\geq 1$.
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  2. #2
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    Hello,
    $\displaystyle n=6a+5=6(a+1)-6+5=6(a+1)-1=3\cdot2(a+1)-1$
    $\displaystyle a' = 2(a+1)\ge 1$ because $\displaystyle a\ge 0$.
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  3. #3
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    I do not know if this helps but:

    I think that the key here is variable substitution.

    $\displaystyle
    n=6a+5\Rightarrow
    $
    $\displaystyle
    n=2\cdot 3\cdot a+2+3\Rightarrow
    $
    $\displaystyle
    n=3(2a+1)+2
    $

    By setting $\displaystyle b=2a+1$ we have:

    $\displaystyle
    n=3b+2
    $

    Next:
    $\displaystyle
    n=3b+3-1\Rightarrow
    $

    $\displaystyle
    n=3(b+1)-1\
    $

    By setting $\displaystyle a'=b+1$ we have:
    $\displaystyle
    n=3a'-1
    $

    Think that this is that you want.
    But you also check the proccess.
    Last edited by gdmath; Sep 11th 2009 at 08:11 AM. Reason: Bingk correction, see bellow
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  4. #4
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    Hi

    gdmath, you might want to check/edit your work. I think there's an error.

    You got that $\displaystyle n=3b+2$, but the next line doesn't work out.

    It says $\displaystyle n=3(b+1)+3-1$, but if you look at that, what we have is $\displaystyle n=3(b+1)+2=3b+5 \neq 3b+2$

    I think what you should have is $\displaystyle n=3b+3-1=3(b+1)-1$

    Then we set $\displaystyle a'=b+1$, leaving us with $\displaystyle n=3a'-1$

    Also, $\displaystyle a'=b+1=2a+1+1=2a+2=2(a+1)$, which agrees with Taluivren's solution

    Just one thing though ... shouldn't $\displaystyle a' \geq 2$?
    Since when $\displaystyle a=0$, $\displaystyle a'=2(0+1)=2$
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