Find all integer $n$ such that $n-5$ divide $n^2+n-27$.
Problem
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Tags: algebra, polynomial, calculus, integration
arberiii
04.03.2013 21:18
my soution was $\frac{n^2+n-27}{n-5}=\frac{n^2-10n+25+11n-55+3}{n-5}=n-5+11+\frac{3}{n-5}$ $n-5\in{-3,3,-1,1}$
nikoma
04.03.2013 21:21
Using polynomial division \[\frac{n^2 + n - 27}{n - 5} = n + 6 + \frac{3}{n - 5}\] which will be integral for $n - 5 = \pm 3$ and $n - 5 = \pm 1$, from this we get solution $\boxed{n \in \{2,4,6,8\}}$.
Aiscrim
18.03.2013 11:09
Taking $n-5=x$, we obtain: $x|(x+5)^2+(x+5)-27\Leftrightarrow x|3 \Leftrightarrow n-5|3$
IDMasterz
18.03.2013 11:44
LOL! This is simple polynomial division theorem: $n^2 + n - 27 = d(n)(n-5) + r(n) \implies r(n)$ is constant $\implies $ sub in $n = 5 \implies r(5) = 3 \implies n-5 \mid 3$.
dip1729
27.10.2018 18:29
IDMasterz wrote: LOL! This is simple polynomial division theorem: $n^2 + n - 27 = d(n)(n-5) + r(n) \implies r(n)$ is constant $\implies $ sub in $n = 5 \implies r(5) = 3 \implies n-5 \mid 3$. So, 2,4,6,8.