$1 \cdot 32$ Find the smallest positive integer $n$, such that the last three digits of its cube are 888.
(6th American Invitational Mathematics Examination, 1988)
[Solution 1] If the cube of a positive integer ends in 8, then this number itself must end in 2, i.e., it can be written as
$$
n=10 k+2 \text { ( } k \text { is a non-negative integer) }
$$

Thus,
$$
\begin{aligned}
n^{3} & =(10 k+2)^{3} \\
& =1000 k^{3}+600 k^{2}+120 k+8 .
\end{aligned}
$$

Here, $120 k$ determines the tens digit of $n^{3}$. Since it is required that the tens digit of $n^{3}$ is 8, then the units digit of $12 k$ should be 8, i.e., the units digit of $k$ is 4 or 9, so we can set
$$
k=5 m+4 \text { ( } m \text { is a non-negative integer). }
$$

At this time, $n^{3}=[10(5 m+4)+2]^{3}$
$$
=125000 m^{3}+315000 m^{2}+264600 m+74088 \text {. }
$$

To make the hundreds digit of $n^{3}$ be 8, it is necessary that the units digit of $2646 m$ is 8, the smallest $m=3$.
At this time
$$
\begin{array}{c}
k=5 m+4=19, \\
n=10 k+2=192 .
\end{array}
$$

We can find that $n^{3}=7077888$, its last three digits are 888.
Therefore, the smallest $n$ is 192.
[Solution 2] By the problem, we have
$$
1000 \mid n^{3}-888
$$

So
$$
\begin{array}{c}
1000 \mid n^{3}+112-1000, \\
1000 \mid n^{3}+112 .
\end{array}
$$

So $n$ is even, let $n=2 k$, then
$$
1000 \mid 8 k^{3}+112=8\left(k^{3}+14\right) \text {. }
$$

This requires
$$
125 \mid k^{3}+14 \text {. }
$$

First, it must be $5 \mid k^{3}+14$, so the units digit of $k$ is 6 or 1.
Thus, we can set $k=5 m+1$ ( $m$ is a non-negative integer)
$$
\begin{aligned}
k^{3}+14 & =(5 m+1)^{3}+14 \\
& =125 m^{3}+75 m^{2}+15 m+15 \\
& =5\left(25 m^{3}+15 m^{2}+3 m+3\right) .
\end{aligned}
$$

To make $125 \mid k^{3}+14$, it is only necessary that
$$
25 \mid 15 m^{2}+3 m+3 \text {. }
$$

Thus, the units digit of $15 m^{2}+3 m+3$ is 0 or 5, since
$$
15 m^{2}+3 m+3=3 m(5 m+1)+3 \text {. }
$$
$m(5 m+1)$ is definitely even, so the units digit of $3 m(5 m+1)$ is definitely 2. Thus, the units digit of $m(5 m+1)$ must be 4, at this time the units digit of $m$ must be 4 or 9. For this, set $m=5 t+4,(t$ is a non-negative integer $)$.
$$
\begin{aligned}
& 15 m^{2}+3 m+3 \\
= & 15(5 t+4)^{2}+3(5 t+4)+3 \\
= & 15 \cdot 25 t^{2}+40 \cdot 15 t+16 \cdot 15+15 t+15 \\
= & 15 \cdot 25 t^{2}+24 \cdot 25 t+15 t+15 \cdot 17
\end{aligned}
$$

To make $25 \mid 15 m^{2}+3 m+3$, it is only necessary that
i.e.,
$$
\begin{array}{c}
25 \mid 15 t+15 \cdot 17, \\
5 \mid 3 t+51 .
\end{array}
$$

Take the smallest $t=3$.
Then $\quad m=5 t+4=19, k=5 m+1=96$.
Thus $\quad n=2 k=192$.