$10 \cdot 40$ Find all such three-digit numbers, if the number itself is increased by 3, then the sum of its digits decreases to $\frac{1}{3}$ of the original.
(Kyiv Mathematical Olympiad, 1963)
[Solution] Let such a three-digit number be $\overline{a b c}$. Where $a, b, c \in\{0,1,2, \cdots, 9\}, a \neq 0$.

Obviously, $c \geqslant 7$, otherwise, if $c \leqslant 6$, then the sum of the digits of $\overline{a b c}+3$ would be 3 more than the original sum of the digits (since there is no carry), which contradicts the problem statement.

At the same time, it cannot be that $a=9$ and $b=9$, because in this case, when $c \geqslant 7$, adding 3 to $\overline{a b c}$ results in a carry, and the sum of its digits becomes $(c+3-10)+1=c-6$. According to the problem, we have
$$
\frac{1}{3}(18+c)=c-6,
$$

Solving this gives
$$
c=18
$$

which is impossible.
If $a<9, b=9, c \geqslant 7$, then we have
$$
\begin{array}{c}
\frac{1}{3}(a+9+c)=(a+1)+0+(c+3-10) . \\
2 a+2 c=27 .
\end{array}
$$

In this case, the left side is even, while the right side is odd, which is impossible.
If $b<9, \quad c \geqslant 7$, then
$$
\begin{array}{c}
\frac{1}{3}(a+b+c)=a+(b+1)+(c+3-10) . \\
a+b+c=9 .
\end{array}
$$

Considering $a \geqslant 1$, the three-digit numbers we are looking for can only be
$$
117,207,108 \text {. }
$$