10-112 Find all $n \in N$, such that if the numbers $n^{3}$ and $n^{4}$ are written side by side in decimal notation, then each of the ten digits appears exactly once.
(Former Yugoslavia Mathematical Olympiad, 1983)
[Solution] Let $f(m)$ denote the number of digits in the decimal representation of the natural number $m$.
Then for the required $n$, we have

Obviously,
$$
\begin{array}{c}
f\left(n^{3}\right)+f\left(n^{4}\right)=10 . \\
f\left(n^{3}\right) \geqslant 4 .
\end{array}
$$

Otherwise, if $f\left(n^{3}\right) < 4$, then $n^3 < 10000$, so
$$
n < 17 \text{. }
$$

Thus,
$$
18 \leqslant n \leqslant 21 .
$$

Since any natural number is congruent to the sum of its digits modulo 9, we have
$$
\begin{array}{l}
n^{3}+n^{4} \equiv (0+1+2+\cdots+9) \quad(\bmod 9), \\
n^{3}(n+1) \equiv 0 \quad(\bmod 9) .
\end{array}
$$

Clearly, $n=19$ and $n=20$ do not satisfy the above equation.
When $n=21$, since the last digit of both $21^{3}$ and $21^{4}$ is 1, it does not meet the problem's requirement.
Therefore, we only consider $n=18$. Since
$$
18^{3}=5832, \quad 18^{4}=104976,
$$

$n=18$ satisfies the problem's requirement.