Problem 539

Given that the unit digit of $n^{3}$ is 8, we know $n=10 k+2$. Thus, $n^{3}=(10 k+2)^{3}=$ $1000 k^{3}+600 k^{2}+120 k+8 \equiv 88(\bmod 100)$, therefore $k=4(\bmod 5)$. Let $k=5 m+4(m$ be a non-negative integer), then $n^{3} \equiv 600 \times 4^{2}+120 \times(5 m+4)+8 \equiv 88+600 m \equiv 888(\bmod 1000)$, hence $m=3, n=192$.