TITLE: What is the largest positive $n$ for which $n^3+100$ is divisible by $n+10$
QUESTION [11 upvotes]: What is the largest positive $n$ for which $n^3+100$ is divisible by $n+10$
I tried to factorize $n^3+100$, but $100$ is not a perfect cube. I wish it were $1000$.

REPLY [12 votes]: Hint $\rm\quad\ \ n\!+\!10\ \,|\,\ f(n) \:\iff\: n\!+\!10\ \,|\,\ f(-10),\, $ for any $\rm\:f(x)\in \mathbb Z[x],\, $ by the Factor Theorem 
i.e. $\rm\ mod\,\ n\!+\!10\!:\,\ \color{#c00}{n\equiv -10}\ \Rightarrow\ f(\color{#c00}n)\equiv f(\color{#c00}{-10})\ $ by the Polynomial Congruence Rule.
Thus the largest such $\rm\,n\,$ occcurs when $\,\rm n\!+\!10\,$ is the largest divisor of $\rm\,f(-10)$.