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

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SUMMARY

The discussion focuses on finding the maximum natural number \( n \) such that \( n^2 - 2112 \) is a multiple of \( n + 10 \). The problem is framed within the context of number theory, specifically dealing with divisibility and quadratic equations. Participants analyze the equation and derive that the maximum value of \( n \) satisfying these conditions is 38, as substituting this value yields \( 38^2 - 2112 = 0 \), confirming it is a multiple of \( 48 \).

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

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

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

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

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

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