10. If $n$ is a natural number, and the value of $n^{3}$ ends exactly with “2016”, find the minimum value of $n$.
Solve:
$n^{3} \equiv 6(\bmod 10) \Rightarrow n \equiv 6(\bmod 10)$, i.e., the unit digit of $n$ is 6.
If $n$ is a two-digit number, then $n=10 a+6 \Rightarrow n^{3}=(10 a+6)^{3}=10^{3} a^{3}+216+180 a(10 a+6)$
$=10^{3} a^{3}+1800 a^{2}+1080 a+216 \equiv 16(\bmod 100) \Rightarrow a=5$.
After calculation, $56^{3}=175616$, contradiction;
If $n$ is a three-digit number, then $n=10 \overline{b c}+6 \Rightarrow n^{3}=(10 \overline{b c}+6)^{3}$
$$
=10^{3} \overline{b c}^{3}+1800 \overline{b c}^{2}+1080 \overline{b c}+216 \equiv 16(\bmod 100) \Rightarrow c=0,5.
$$

If $c=0$, then $n^{3}=(100 b+6)^{3}=100^{3} b^{3}+216+1800 b(100 b+6)$
$$
=100^{3} b^{3}+180000 b^{2}+10800 b+216 \equiv 016(\bmod 1000) \Rightarrow b=1,6.
$$

And $106^{3}=1191016$, contradiction; $606^{3}=222545016$, contradiction;
If $c=5$, then $n^{3}=(100 b+56)^{3}=100^{3} b^{3}+175616+16800 b(100 b+56)$
$$
=100^{3} b^{3}+1680000 b^{2}+940800 b+175616 \equiv 016(\bmod 1000) \Rightarrow b=3,8.
$$

And $356^{3}=45118016$, contradiction; $856^{3}=627222016$, satisfies the condition.
Therefore, the minimum value of $n$ is 856.