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

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frxBest ResponseYou've already chosen the best response.0
Sorry some difficulities with the language. What i mean is if you divide for example 4 with 3 you get the remainder 1..
 one year ago

ktnguyen1Best ResponseYou've already chosen the best response.1
all I number it present as this a/3 and a can be (0,1,2,3...n)
 one year ago

frxBest ResponseYou've already chosen the best response.0
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 3m then n = m > m^2 n= (m+1) > m^2+2m+1 n= (m+2) > m^2+4m+2 since 3m then 3m^2+2m+1 and 3m^2+4m+2 so every integer should be devisible by 3, am I on the right track?
 one year ago

ktnguyen1Best ResponseYou've already chosen the best response.1
yes you got it by the way (m+2)^2=m^2+4m+4
 one year ago

frxBest ResponseYou've already chosen the best response.0
Great! Oh, of course it's, my bad ;)
 one year ago

wioBest ResponseYou've already chosen the best response.0
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?
 one year ago

ktnguyen1Best ResponseYou've already chosen the best response.1
We did answer the remain will be all Interger #
 one year ago

frxBest ResponseYou've already chosen the best response.0
Then you divide n^2/3, which remainders can be left over like the remainder in the euclidean algorithm
 one year ago

wioBest ResponseYou've already chosen the best response.0
Well you can only get remainders of 0, 1, and 2... that's obvious.
 one year ago

frxBest ResponseYou've already chosen the best response.0
What about the 4 I got above when n= (m+2) > m^2+4m+4 ?
 one year ago

frxBest ResponseYou've already chosen the best response.0
@wio could you please explain why the remainders only can be 0,1,2 and why it's so obvious, would really appreciate it :)
 one year ago

ktnguyen1Best ResponseYou've already chosen the best response.1
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
 one year ago

frxBest ResponseYou've already chosen the best response.0
Okey I get it, thanks a lot! :)
 one year ago

wioBest ResponseYou've already chosen the best response.0
@frx If you have some number, 3m+4, then you have 3m+3+1, which is 3(m+1) + 1... remainder 1!
 one year ago
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