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    Please can someone help me with this?

    I'm getting nowhere with this question.


    

If 12345654321_7 \equiv x (mod 21)  with -21 < x \leq 0,  what   is   x?

    At the moment I've converted 123456543217 to denary (it's 384473664) and I've been hopelessly dividing by 21... but I feel like the question has something to do with it being in base 7.

    ... I also checked wolfram alpha and got the answer x = 15, but this isn't in the correct range. Any help there?

    Any help would be greatly appreciated!!
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    (Original post by SummerPi)
    Please can someone help me with this?

    I'm getting nowhere with this question.


    

If 12345654321_7 \equiv x (mod 21)  with -21 < x \leq 0,  what   is   x?

    At the moment I've converted 123456543217 to denary (it's 384473664) and I've been hopelessly dividing by 21... but I feel like the question has something to do with it being in base 7.

    ... I also checked wolfram alpha and got the answer x = 15, but this isn't in the correct range. Any help there?

    Any help would be greatly appreciated!!
    Converting to base 10 seems a little messy here.
    What I'd recommend doing is considering what: 7^n \pmod{21} and seeing what you get.
    Then using this in terms of the base 7 representation, the result follows reasonably easily.
    Spoiler:
    Show
    You should get 7^n \equiv 7 \pmod{21} \ \forall n \in \mathbb{N}

    For your last issue note that: 15 \equiv -6 \pmod{21}
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    (Original post by joostan)
    Converting to base 10 seems a little messy here.
    What I'd recommend doing is considering what: 7^n \pmod{21} and seeing what you get.
    Then using this in terms of the base 7 representation, the result follows reasonably easily.
    Spoiler:
    Show
    You should get 7^n \equiv 7 \pmod{21} \ \forall n \in \mathbb{N}

    For your last issue note that: 15 \equiv -6 \pmod{21}
    Thank you very much for this! A great help

    Just a bit confused about the case with 7^n \pmod{21} where n=0 i.e. where 7^n=1 ... as 1 \not \equiv 7 \pmod{21} ?
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    (Original post by SummerPi)
    Thank you very much for this! A great help

    Just a bit confused about the case with 7^n \pmod{21} where n=0 i.e. where 7^n=1 ... as 1 \not \equiv 7 \pmod{21} ?
    A true statement, which is why I used the caveat \forall n \in \mathbb{N} where \mathbb{N}, the set of natural numbers is conventionally defined to not include 0.
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    (Original post by joostan)
    A true statement, which is why I used the caveat \forall n \in \mathbb{N} where \mathbb{N}, the set of natural numbers is conventionally defined to not include 0.

    Ahh brilliant! Sorry I missed that Thank you so much for your help - I fully understand this now
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    (Original post by SummerPi)
    Ahh brilliant! Sorry I missed that Thank you so much for your help - I fully understand this now
    Haha, no worries
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    (Original post by SummerPi)
    Please can someone help me with this?

    I'm getting nowhere with this question.


    

If 12345654321_7 \equiv x (mod 21)  with -21 < x \leq 0,  what   is   x?

    At the moment I've converted 123456543217 to denary (it's 384473664) and I've been hopelessly dividing by 21... but I feel like the question has something to do with it being in base 7.

    ... I also checked wolfram alpha and got the answer x = 15, but this isn't in the correct range. Any help there?

    Any help would be greatly appreciated!!
    joostan's answer is great. I think there's a slightly more intuitive way if you know the Chinese Remainder Theorem:

    12345654321_7 is 1 mod 7 (trivially), so it is 1 or 8 or 15 mod 21. Which? CRT tells us we can reduce mod 3 to distinguish between these.
    7^n is 1 mod 3, for all n (because 7 is 1 mod 3), so the sum-the-digits-to-get-parity-mod-3 trick works in base 7 for exactly the same reason as it works in base 10. The digit sum is clearly a multiple of 3, so 12345654321_7 is 0 mod 3. Hence being 0 mod 3 and 1 mod 7, it is 15 mod 21 by Chinese Remainder Theorem.

    ETA: by the way, a \pmod{b} can be quickly written in LaTeX as a \pmod{b}. "pmod" as in "parentheses-mod".
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    (Original post by Smaug123)
    joostan's answer is great. I think there's a slightly more intuitive way if you know the Chinese Remainder Theorem:

    12345654321_7 is 1 mod 7 (trivially), so it is 1 or 8 or 15 mod 21. Which? CRT tells us we can reduce mod 3 to distinguish between these.
    7^n is 1 mod 3, for all n (because 7 is 1 mod 3), so the sum-the-digits-to-get-parity-mod-3 trick works in base 7 for exactly the same reason as it works in base 10. The digit sum is clearly a multiple of 3, so 12345654321_7 is 0 mod 3. Hence being 0 mod 3 and 1 mod 7, it is 15 mod 21 by Chinese Remainder Theorem.

    ETA: by the way, a \pmod{b} can be quickly written in LaTeX as a \pmod{b}. "pmod" as in "parentheses-mod".
    Many many thanks for this. I definitely find this method more intuitive.

    Also thanks for the LaTeX tip!
 
 
 
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