$2 \cdot 64$ Find the smallest natural number $x$, such that in decimal notation, the number $x^{3}$ has the first three digits and the last four digits all as 1.
［Solution］ Clearly, the last four digits of the number $x^{3}$ are only related to the last four digits of $x$.
Since the last digit of the number $x^{3}$ is 1, the last digit of the number $x$ can only be 1.
Cubing the numbers $01,11,21, \cdots, 91$ reveals that only $71^{3}=357911$. Therefore, the last two digits of the number $x$ are 71.

Cubing the numbers $071,171,271, \cdots, 971$ reveals that only $471^{3}=104487111$. Therefore, the last three digits of $x$ are 471.

Cubing the numbers $0471,1471, \cdots, 9471$ reveals that only $8471^{3}$ ends with 1111, so the last four digits of the number $x$ are 8471.
Let $n$ be the number of digits in $x^{3}$ between 111 and 1111, there are three possible cases.
(1) $n=3 k, k \geqslant 0$, then
$$
111 \times 10^{3 k+4}<x^{3}=11 \underset{3 k \text { digits }}{1 \cdots 11}<112 \times 10^{3 k+4}
$$

This gives
$$
\begin{aligned}
10.353988 \times 10^{k+1} & \approx \sqrt[3]{1110} \cdot 10^{k+1}<x<\sqrt[3]{1120} \cdot 10^{k+1} \\
& \approx 10.384988 \times 10^{k+1} .
\end{aligned}
$$

When $k=0$, it is easy to see that the required number $x$ does not exist. If $k \geqslant 1$, then the required number $x$ starts with 103, i.e., $x=103 \cdots 8471$. The smallest number is 1038471. It is not difficult to verify that this number meets the required conditions.
(2) $n=3 k+1, k \geqslant 0$, then
$$
111 \times 10^{3 k+5}<x^{3}=11 \underbrace{1 \cdots 1}_{3 k+1 \text { digits }} 111<112 \times 10^{3 k+5} .
$$

Therefore
$$
\begin{aligned}
22.306991 \times 10^{k+1} \approx & \sqrt[3]{11100} \times 10^{k+1}<x<\sqrt[3]{11200} \times \\
& 10^{k+1} \\
\approx & 22.373779 \times 10^{k+1} .
\end{aligned}
$$

When $k=0$, the required number $x$ does not exist. When $k \geqslant 1$, the required number has the form $x=$ $223 \cdots 8471$, and there is at least one digit in the deleted node position, but any number of this form is clearly greater than the already found (seven-digit) 1038471. Thus, none of these numbers are what we are looking for.
(3) $n=3 k+2, k \geqslant 0$, this case is discussed similarly to case (2), and similarly, numbers greater than 1038471 are obtained. Therefore, the required number is 1038471.