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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$n\in N$

$n^2 - 2112 $ is a multipler of $n+10$

please find :$max(n)$

$n^2 - 2112 $ is a multipler of $n+10$

please find :$max(n)$

- Thread starter Albert
- Start date

- Thread starter
- #1

- Jan 25, 2013

- 1,225

$n\in N$

$n^2 - 2112 $ is a multipler of $n+10$

please find :$max(n)$

$n^2 - 2112 $ is a multipler of $n+10$

please find :$max(n)$

- Mar 31, 2013

- 1,309

$n\in N$

$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

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

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