Find the Maximum of n: $n^2 - 2112$

[Albert1]

$n\in N$
$n^2 - 2112 $ is a multipler of $n+10$
please find :$max(n)$

 

[kaliprasad]

Re: Find max (n)

$n\in N$
$n^2 - 2112 $ is a multipler of $n+10$
please find :$max(n)$

Spoiler

n^2 - 2112 = n^2 - 100 - 2012

= (n+10)(n-10) - 2012

2012 is divisible by n+ 10 so max n = 2002