Find all the whole number values of n, that would make the following statement true. (3n+9)/(n+1)

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Find all the whole number values of n, that would make the following statement true. (3n+9)/(n+1)

Mathematics
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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.

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means answer should be 0
Thanks, but there was a set of numbers on the example question. 0 works, what else?

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(3n+9)/(n+1) it is wholly divided....
What do you mean by, wholly divided?
means the remainder is zero(0)
Right. But I'm pretty sure there are other values of n, that also leaves a remainder of 0
n = 2 ?
no one whole no. can be fully satisfied with n
there would be some integers..
I think \[n = 3k-1, k \in \mathbb{N}\]
Oh I got it, 0; 1; 2; 5; -2; -3; -4; -7 Heh, thanks for your help :)
thanx for who??
are you there??
if n is whole number, so negative integers can't be the solution, since whole number starts at 0,1,2,3,....
\[\frac{3n+9}{n+1}=\frac{3n+3+6}{n+1}=3+\frac{6}{n+1}\]so \(n+1|6\) and we have\[n+1=\pm1,\pm2,\pm3,\pm6\]
@mukushla are negative numbers also the solutions? since the question ask for n to be whole number
u r right negatives are not solution
only solutions are \(n=0,1,2,5\)
@mayankdevnani everyone who helped. @chihiroasleaf @mukushla Sorry, I translated this from another language. I think the correct term was integer. Thanks again to everyone for your input!
np :)

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