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Math Help - some help with number theory

  1. #1
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    some help with number theory

    Show that if a positive integer n=6a+5,a \geq 0, then n=3a'-1,a'\geq 1.
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  2. #2
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    Hello,
    n=6a+5=6(a+1)-6+5=6(a+1)-1=3\cdot2(a+1)-1
    a' = 2(a+1)\ge 1 because 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.

    <br />
n=6a+5\Rightarrow<br />
    <br />
n=2\cdot 3\cdot a+2+3\Rightarrow<br />
    <br />
n=3(2a+1)+2<br />

    By setting b=2a+1 we have:

    <br />
n=3b+2<br />

    Next:
    <br />
n=3b+3-1\Rightarrow<br />

    <br />
n=3(b+1)-1\<br />

    By setting a'=b+1 we have:
    <br />
n=3a'-1<br />

    Think that this is that you want.
    But you also check the proccess.
    Last edited by gdmath; September 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 n=3b+2, but the next line doesn't work out.

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

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

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

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

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