$1 \cdot 53$ A student did not notice the multiplication sign written between two 7-digit numbers, mistaking it for a 14-digit number. This 14-digit number happens to be 3 times the original product. Try to find these 3 numbers.
[Solution] The original expression is $1666667 \times 3333334$.
In fact, let these two 7-digit numbers be $x$ and $y$. According to the problem,
$$
3 x y=10^{7} \cdot x+y \text {, }
$$

thus
$$
y=\frac{10^{7} \cdot x}{3 x-1},
$$

so $10^{7} x$ should be a multiple of $3 x-1$. However,
$$
10^{7} x=3333333 \cdot(3 x-1)+x+3333333,
$$

therefore, there exists a positive integer $k$, such that
$$
x+3333333=k(3 x-1) .
$$

When $k=1$, we get $x=1666667, y=3333334$.
When $k \geqslant 2$, we get $x<10^{6}$, which does not meet the requirement.
Therefore, the three numbers sought are $1666667,3333334,16666673333334$.