10. (1991 6th Latin American Mathematical Olympiad) $N$ is a 5-digit number composed of 5 different non-zero digits, and $N$ is equal to the sum of all three-digit numbers formed by taking 3 different digits from these 5 digits. Find all such 5-digit numbers $N$.


10. Let the 5-digit number $N=\overline{a_{1} a_{2} a_{3} a_{4} a_{5}}$.
When $a_{1}$ is in the hundreds place of a three-digit number, there are $P_{4}^{2}=12$ such numbers,
When $a_{1}$ is in the tens place of a three-digit number, there are $P_{4}^{2}=12$ such numbers,
When $a_{1}$ is in the units place of a three-digit number, there are $P_{4}^{2}=12$ such numbers.
Thus, according to the problem, we have
$$
\begin{aligned}
N & =\overline{a_{1} a_{2} a_{3} a_{4} a_{5}}=\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)(100 \cdot 12+10 \cdot 12+12) \\
& =1332\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right) .
\end{aligned}
$$

Since $9 \mid 1332$, then
$$
9 \mid N=\overline{a_{1} a_{2} a_{3} a_{4} a_{5}} \text {. }
$$

Thus, $9 \mid\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}\right)$.
Therefore, $N$ must be a multiple of $1332 \cdot 9=11988$.
Since $15=1+2+3+4+5 \leqslant a_{1}+a_{2}+a_{3}+a_{4}+a_{5} \leqslant 9+8+7+6+5=35$, $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}$ can only be 18 or 27.
(1) When $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=18$,
$$
\overline{a_{1} a_{2} a_{3} a_{4} a_{5}}=1332 \cdot 18=23976 \text {, }
$$

but $2+3+9+7+6=27 \neq 18$.
(2) When $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=27$,
$$
\overline{a_{1} a_{2} a_{3} a_{4} a_{5}}=1332 \cdot 27=35964 \text {, }
$$

at this point $3+5+9+6+4=27$.
Therefore, the only 5-digit number that satisfies the condition is 35964.