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frx

  • 3 years ago

Which remains can occur when dividing \[n^2\] with 3, let \[n \in \mathbb{N} \]

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  1. wio
    • 3 years ago
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    Remains? Remainder?

  2. frx
    • 3 years ago
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    Sorry some difficulities with the language. What i mean is if you divide for example 4 with 3 you get the remainder 1..

  3. ktnguyen1
    • 3 years ago
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    all I number it present as this a/3 and a can be (0,1,2,3...n)

  4. frx
    • 3 years ago
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    Ok I think i get what you say but what if I want to be more general. Let's say that n is an interger which can be written on the form m, m+1,m+2..., where 3|m then n = m -> m^2 n= (m+1) -> m^2+2m+1 n= (m+2) -> m^2+4m+2 since 3|m then 3|m^2+2m+1 and 3|m^2+4m+2 so every integer should be devisible by 3, am I on the right track?

  5. ktnguyen1
    • 3 years ago
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    yes you got it by the way (m+2)^2=m^2+4m+4

  6. frx
    • 3 years ago
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    Great! Oh, of course it's, my bad ;)

  7. wio
    • 3 years ago
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    Sorry, but it's still not a clear question. You want the remainder of n^2 / 3 in terms of the remainder of n/3?

  8. ktnguyen1
    • 3 years ago
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    We did answer the remain will be all Interger #

  9. frx
    • 3 years ago
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    Then you divide n^2/3, which remainders can be left over like the remainder in the euclidean algorithm

  10. wio
    • 3 years ago
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    Well you can only get remainders of 0, 1, and 2... that's obvious.

  11. frx
    • 3 years ago
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    What about the 4 I got above when n= (m+2) -> m^2+4m+4 ?

  12. frx
    • 3 years ago
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    @wio could you please explain why the remainders only can be 0,1,2 and why it's so obvious, would really appreciate it :)

  13. ktnguyen1
    • 3 years ago
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    the 4 because I mean n=(m+2) but we have to find n^2 mean (m+2)^2= m^2+4m+4 I just want to corrct the formula you typed previous

  14. frx
    • 3 years ago
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    Okey I get it, thanks a lot! :)

  15. wio
    • 3 years ago
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    @frx If you have some number, 3m+4, then you have 3m+3+1, which is 3(m+1) + 1... remainder 1!

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