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Tags: number theory, divisible, divides
Find the largest positive integer $n$ such that $n^2 + 10$ is divisible by $n-5$.
$\frac{n^2+10}{n-5}=n+5+\frac{35}{n-5}.$ Thus, our answer is $n-5=35$ which is $\boxed{40}$
Write $\frac{n^2+10}{n-5}=n+5+\frac{35}{n-5}$. The maximum is hence $n=35+5=\boxed{40}$.
$\frac{n^2+10}{n-5}=n+5+\frac{35}{n-5} \implies n=40$
$n^2+10=(n-5)^2+10(n-5)+35$