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

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To find the maximum value of n in the equation n^2 - 2112 being a multiple of n + 10, the problem requires analyzing the relationship between these expressions. By substituting n + 10 into the equation, it can be determined that n must satisfy specific divisibility conditions. The maximum value of n can be calculated by solving the quadratic equation derived from the original expression. Ultimately, the solution reveals that the maximum integer value of n is 42. This conclusion is reached through careful manipulation and evaluation of the given mathematical conditions.
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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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