5. (19th Nordic Mathematical Contest) Find all positive integers $k$, such that in decimal notation, the product of the digits of $k$ equals $\frac{25}{8} k - 211$.
5. Let $k$ be a decimal number, and $s$ be the product of the digits of $k$.

It is easy to see that $s \in \mathbf{N}$, hence $8 \mid k$ and $\frac{25}{8} k-211 \geqslant 0$, i.e.,
$$
k \geqslant \frac{1688}{25} \text {. }
$$

Since $k \in \mathbf{N}_{+}$, we have $k \geqslant 68$.
Also, $8 \mid k$, so the last digit of $k$ is even, thus, $s$ is even.
Since 211 is odd, $\frac{25}{8} k$ must be odd, so, $16 \nmid k$.
Let $k=\overline{a_{1} a_{2} \cdots a_{t}} ， 0 \leqslant a_{i} \leqslant 9(i=2,3, \cdots, t), 1 \leqslant a_{1} \leqslant 9$. By definition,
$$
S=\prod_{i=1}^{t} a_{i} \leqslant a_{1} \times 9^{t-1}s=\frac{25}{8} k-211$, so, $k \leqslant 99$.
Given $8 \mid k, 16 \times k$, we get $k=72$ or 88.
Upon verification, $k$ is 72 or 88.