## Task A-2.4.

Determine the largest natural number $n$ such that $n+10$ divides $n^{3}+100$.
## Solution.

Notice that $n^{3}+1000$ is a sum of cubes, so it holds that $n^{3}+1000=(n+10)\left(n^{2}-10 n+100\right)$.

Now we have

$$
\begin{aligned}
\frac{n^{3}+100}{n+10} & =\frac{n^{3}+1000-900}{n+10}=\frac{(n+10)\left(n^{2}-10 n+100\right)-900}{n+10} \\
& =n^{2}-10 n+100-\frac{900}{n+10}
\end{aligned}
$$

Therefore, if $n+10$ is a divisor of $n^{3}+100$, it must also be a divisor of 900, and vice versa.

Since the greatest divisor of 900 is 900 itself, the largest natural number that satisfies the statement of the problem is $n=900-10=890$.

Note: The equality $\frac{n^{3}+100}{n+10}=n^{2}-10 n+100-\frac{900}{n+10}$ can also be achieved in some other way, for example, by polynomial division. Any (correct) derivation is worth 3 points.