# Division in $1$ variable

Let the sum of all integers $n$ such that $(2n^2+9/n+3)$ is an integer be $A$, what is $|A|$?

I tried expressing it to:

$k$ is an integer, $(n+3)k + 0 = 2n^2 + 9$, $n(2n-k) + 3(3-k) = 0$, from which I couldn't move on,

I also tried $(n+3)k + 0 = 2n^2 + 9$, $k = 2n$, $2n^2 - 2n^2 + 9 - 6n$. $9-6n/n+3$ is an integer, $9 - 6n = 3(3-n)$ though now I also do not know how to move on.

• Should there be parentheses around $2n^2+9$ and $n+3$ in the first line? As written the expression is $2n^2+\frac 9x+3$, so just add up the values for $n=1,3,9$ and declare victory. – Ross Millikan Nov 7 '17 at 15:45

Let $n+3=m\ne0$

$2n^2+9=2(m-3)^2+9=m(2m-12)+27$

So. $m|27$

I found quotient and remainder so the fraction became $$\frac{2 n^2 + 9}{ n + 3}=2 n-6+\frac{27}{n+3}$$ and can easily seen that it is integer only for $n=6;\;n=24$

Therefore $A=30$

Hope this helps

• The answer is said to be 24 though, I used the fact that 27/n+3 can be negative, so n can be n= -2, -6, -12, -30; (30+6+12+2-30) = 20? The answer is 24 though I cant pinpoint what i did wrong – SuperMage1 Nov 8 '17 at 12:42
• oh I forgot -4 mb the answer is 24 – SuperMage1 Nov 8 '17 at 13:03