A good problem on number theory

We know that $N + 10 | N^3 + 1000$ , since ( $a+b|a^3 + b^3$ ).Also, $N + 10 | N^3 +100$ .

Therefore,

$N + 10 |( N^3 + 1000) -( N^3 + 100) \Rightarrow N + 10 | 900$

Since we want the maximum $N$ , we have $N + 10 = 900 \Rightarrow N = \boxed{890}$ .

let $m=n+10$

then $n=m-10$

$n^3+100$ = $(m-10)^3+100$ = $(m^3$ - 30 $m^2$ $+300m - 900)$

condition $(n+10)$ divides $(n^3+100)$ now translates into $m$ divides $(m^3$ - 30 $m^2$ $+300m - 900$ )

since $m$ is a divisor of each of the quantity $m^3$ , $30m^2$ , $300m$ it follows that $m$ divides 900,the largest $m$ it is true for is 900 so it follows that $n=890$