Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

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

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Remains? Remainder?

- anonymous

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

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?

- anonymous

yes you got it by the way (m+2)^2=m^2+4m+4

- anonymous

Great! Oh, of course it's, my bad ;)

- anonymous

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?

- anonymous

We did answer the remain will be all Interger #

- anonymous

Then you divide n^2/3, which remainders can be left over like the remainder in the euclidean algorithm

- anonymous

Well you can only get remainders of 0, 1, and 2... that's obvious.

- anonymous

What about the 4 I got above when n= (m+2) -> m^2+4m+4 ?

- anonymous

@wio could you please explain why the remainders only can be 0,1,2 and why it's so obvious, would really appreciate it :)

- anonymous

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

- anonymous

Okey I get it, thanks a lot! :)

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.