16. (1988 AIME Problem 6) Find the smallest positive integer $n$ such that the last three digits of its cube are 888.
16. Solution 1 If the cube of a positive integer ends with 8, then this number itself must end with 2, i.e., it can be written in the form $n=10k+2$ (where $k$ is a non-negative integer).
Thus,
$$
n^{3}=(10k+2)^{3}=1000k^{3}+600k^{2}+120k+8.
$$

Here, $120k$ 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 $12k$ should be 8, i.e., the units digit of $k$ is 4 or 9, so we can set $k=5m+4$ (where $m$ is a non-negative integer).
At this time, $n^{3}=[10(5m+4)+2]^{3}=125000m^{3}+315000m^{3}+264600m+74088$.
To make the hundreds digit of $n^{3}$ 8, it is necessary that the units digit of $2646m$ is 8, the smallest $m=3$.
At this time, $k=5m+4=19, n=10k+2=192$.
We can find that $n^{3}=7077888$, its last three digits are 888.
Therefore, the smallest $n$ is 192.
Nie
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Solution 2 From the problem, we have $1000 \mid n^{3}-888$, so
$1000 \mid n^{3}+112-1000$, thus
$1000 \mid n^{3}+112$.
Therefore, $n$ is even, let $n=2k$, then
$1000 \mid 8k^{3}+112=8(k^{3}+14)$.
This requires $125 \mid k^{3}+14$.
First, it must be that $5 \mid k^{3}+14$, so the units digit of $k$ is 6 or 1.
Thus, we can set $k=5m+1$ (where $m$ is a non-negative integer).
$k^{3}+14=(5m+1)^{3}+14=125m^{3}+75m^{2}+15m+15=5(25m^{3}+15m^{2}+3m+3)$.
To make $125 \mid k^{3}+14$, it is only necessary that
$25 \mid 15m^{2}+3m+3$,
thus the units digit of $15m^{2}+3m+3$ is 0 or 5. Since
$15m^{2}+3m+3=3m(5m+1)+3$,
$m(5m+1)$ is certainly even, so the units digit of $3m(5m+1)$ is certainly 2. Thus, the units digit of $m(5m+1)$ must be 4, at this time, the units digit of $m$ must be 4 or 9. For this, set $m=5t+4$ (where $t$ is a non-negative integer).
$$
\begin{aligned}
15m^{2}+3m+3 & =15(5t+4)^{2}+3(5t+4)+3=15 \cdot 25t^{2}+40 \cdot 15t+16 \cdot 15+15t+15 \\
& =15 \cdot 25t^{2}+24 \cdot 25t+15t+15 \cdot 17.
\end{aligned}
$$

To make $25 \mid 15m^{2}+3m+3$, it is only necessary that
$25 \mid 15t+15 \cdot 17$, i.e., $5 \mid 3t+51$.
Take the smallest $t=3$.
Thus, $m=5t+4=19, k=5m+1=96$.
Therefore, $n=2k=192$.