28. If $n$ is divisible by 20, then every multiple of $n$ has two last digits even and hence it is not alternate. We shall show that any other $n$ has an alternate multiple. (i) Let $n$ be coprime to 10. For each $k$ there exists a number $A_{k}(n)=$ $\overline{10 \ldots 010 \ldots 01 \ldots 0 \ldots 01}=\frac{10^{m k}-1}{10^{k}-1}(m \in \mathbb{N})$ that is divisible by $n$ (by Euler's theorem, choose $\left.m=\varphi\left[n\left(10^{k}-1\right)\right]\right)$. In particular, $A_{2}(n)$ is alternate. (ii) Let $n=2 \cdot 5^{r} \cdot n_{1}$, where $r \geq 1$ and $\left(n_{1}, 10\right)=1$. We shall show by induction that, for each $k$, there exists an alternative $k$-digit odd number $M_{k}$ that is divisible by $5^{k}$. Choosing the number $10 A_{2 r}\left(n_{1}\right) M_{2 r}$ will then solve this case, since it is clearly alternate and divisible by $n$. We can trivially choose $M_{1}=5$. Let there be given an alternate $r$-digit multiple $M_{r}$ of $5^{r}$, and let $c \in\{0,1,2,3,4\}$ be such that $M_{r} / 5^{r} \equiv$ $-c \cdot 2^{r}(\bmod 5)$. Then the $(r+1)$ digit numbers $M_{r}+c \cdot 10^{r}$ and $M_{r}+(5+c) \cdot 10^{r}$ are respectively equal to $5^{r}\left(M_{r} / 5^{r}+2^{r} \cdot c\right)$ and $5^{r}\left(M_{r} / 5^{r}+2^{r} \cdot c+5 \cdot 2^{r}\right)$, and hence they are divisible by $5^{r+1}$ and exactly one of them is alternate: we set it to be $M_{r+1}$. (iii) Let $n=2^{r} \cdot n_{1}$, where $r \geq 1$ and $\left(n_{1}, 10\right)=1$. We show that there exists an alternate $2 r$-digit number $N_{r}$ that is divisible by $2^{2 r+1}$. Choosing the number $A_{2 r}\left(n_{1}\right) N_{r}$ will then solve this case. We choose $N_{1}=16$, and given $N_{r}$, we can prove that one of $N_{r}+$ $m \cdot 10^{2 r}$, for $m \in\{10,12,14,16\}$, is divisible by $2^{2 r+3}$ and therefore suitable for $N_{r+1}$. Indeed, for $N_{r}=2^{2 r+1} d$ we have $N_{r}+m \cdot 10^{2 r}=$ $2^{2 r+1}\left(d+5^{r} m / 2\right)$ and $d+5^{r} m / 2 \equiv 0(\bmod 4)$ has a solution $m / 2 \in$ $\{5,6,7,8\}$ for each $d$ and $r$. Remark. The idea is essentially the same as in (SL94-24).