28 Find the smallest positive integer $n$, such that in decimal notation $n^{3}$ ends with the digits 888.
28. From the condition, we know $n^{3} \equiv 888(\bmod 1000)$, hence
$$n^{3} \equiv 888(\bmod 8), n^{3} \equiv 888(\bmod 125)$$

From the former, we know $n$ is even, let $n=2 m$, then
$$m^{3} \equiv 111(\bmod 125)$$

Therefore
$$m^{3} \equiv 111 \equiv 1(\bmod 5)$$

Noting that when $m=0,1,2,3,4(\bmod 5)$, correspondingly
$$m^{3} \equiv 0,1,3,2,4(\bmod 5)$$

So, from $m^{3} \equiv 1(\bmod 5)$ we know $m \equiv 1(\bmod 5)$, we can set $m=5 k+1$, at this time
$$m^{3}=(5 k+1)^{3}=125 k^{3}+75 k^{2}+15 k+1 \equiv 111(\bmod 125)$$

Hence

Thus
That is, $\square$
Therefore
This requires
Hence
$$\begin{array}{c}
75 k^{2}+15 k \equiv 110(\bmod 125) \\
15 k^{2}+3 k \equiv 22(\bmod 25) \\
15 k^{2}+3 k+3 \equiv 0(\bmod 25) \\
5 k^{2}+k+1 \equiv 0(\bmod 25) \\
5 k^{2}+k+1 \equiv 0(\bmod 5) \\
5 \mid k+1
\end{array}$$

We can set $k+1=5 l$, then
$$\begin{aligned}
5 k^{2}+k+1 & =5 \times(5 l-1)^{2}+5 l \\
& =125 l^{2}-50 l+5(l+1) \\
& \equiv 0(\bmod 25) \\
& 5 \mid l+1
\end{aligned}$$

We can set $l+1=5 r$, therefore
$$\begin{aligned}
n & =2 m=10 k+2=10(5 l-1)+2 \\
& =50 l-8=50(5 r-1)-8 \\
& =250 r-58
\end{aligned}$$

Combining that $n$ is a positive integer, we know
$$n \geqslant 250-58=192$$

Also, $192^{3}=7077888$ meets the requirement, hence the smallest positive integer satisfying the condition is 192.