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# Modular Arithmetic Question watch

1. Please can someone help me with this?

I'm getting nowhere with this question.

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!!
2. (Original post by SummerPi)
Please can someone help me with this?

I'm getting nowhere with this question.

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: 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

For your last issue note that:
3. (Original post by joostan)
Converting to base 10 seems a little messy here.
What I'd recommend doing is considering what: 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

For your last issue note that:
Thank you very much for this! A great help

Just a bit confused about the case with where i.e. where ... as ?
4. (Original post by SummerPi)
Thank you very much for this! A great help

Just a bit confused about the case with where i.e. where ... as ?
A true statement, which is why I used the caveat where , the set of natural numbers is conventionally defined to not include 0.
5. (Original post by joostan)
A true statement, which is why I used the caveat where , 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
6. (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
7. (Original post by SummerPi)
Please can someone help me with this?

I'm getting nowhere with this question.

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, can be quickly written in LaTeX as a \pmod{b}. "pmod" as in "parentheses-mod".
8. (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, 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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