The largest value of $n$ such that $n+10| n^3+100$

Question

If $n\in \mathbb{Z}$ and follow the property

$$n+10| n^3+100$$

then what is the largest value of $n$

Please help me to solve this!!!

Answer 1

Let $m=n+10$ then $n=m-10$ $$n^3+100=(m-10)^3+100$$ $$=m^3-30m^2+300m-900$$ condition $$(n+10)|(n^3+100)$$ now takes the form $$m|(m^3-30m^2-300m-900)$$ Since $m$ is divisor of each of the quantities $m^3,30m^2,300m$

It follows that $$m|900$$ The largest $m$ for which this is true is $m=900$, so it follows that the largest $n$ with the given property is $$n=890$$

Answer 2

Note that $$ n+10|n^3+10n^2\implies n+10|(n^3+10n^2)-(n^3+100)=10n^2-100. $$ Similarly, \begin{aligned} &\quad n+10|(10n^2+100n)-(10n^2-100)=100n+100\\ &\implies n+10|(100n+1000)-(100n+100)=900. \end{aligned} So the largest possible $n$ is $900-10=\boxed{890}$. Verify that this works: $$ \frac{890^3+100}{890+10}=783299\in\mathbb{Z}. $$

Answer 3

Notice that: $n+10| n^3+100$ is same as $$n+10| n^3+1000-900$$ $$\implies n+10| (n^3+1000)-900$$

$$\implies n+10| (n+10)(n^2-10n+100)-900\implies n+10|900$$.

The greatest integer dividing $900$ is $900$ itself. So, for maximum $n$, $$n+10=900\implies n=900-10=890$$