anonymous
  • anonymous
Find all the whole number values of n, that would make the following statement true. (3n+9)/(n+1)
Mathematics
schrodinger
  • schrodinger
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mayankdevnani
  • mayankdevnani
means answer should be 0
mayankdevnani
  • mayankdevnani
@Mello
anonymous
  • anonymous
Thanks, but there was a set of numbers on the example question. 0 works, what else?

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

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