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 $1000$.
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 $1000$.
By division we find that $n^3 + 100 = (n + 10)(n^2 − 10n + 100)−900$.
Therefore, if $n +10$ divides $n^3 +100$, then it must also divide $900$. Since we are looking for largest $n$, $n$ is maximized whenever $n + 10$ is, and since the largest divisor of $900$ is $900$, we must have $n + 10 = 900 \Rightarrow n = 890$
The largest $n$ is therefore $890$
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. $\rm\ mod\ n+10\!:\ n\equiv -10\ \Rightarrow\ f(n)\equiv f(-10)$
$\begin{array}{r r c} \rm m=n+10: & \rm (m-10)^3+100 & \rm \equiv0 \;\bmod{m} \\ & \rm (-10)^3+100 & \rm \equiv0\; \bmod m \\ \times (-1) & 900 & \rm \equiv 0 \;\bmod m \\ \\ \hline \end{array}$
$\rm \max_m \{m:m|900\,\}=900 \implies n=890. $