1.It`s a famous problem that there exist a $n-digit$ number with only odd digits such that divisible by $5^n$ . 2.It`s a more famous !! problem that there exist a $n-digit$ number with only $1,2$ as digits such that divisible by $2^n$ . 3.If $(n,10)=1$ then $\frac{10^{\phi(n)} -1}{9}$ fulfill the conditions .

Let $n=2^k3^lm$ or $n=5^k3^lm$, were $(m,30)=1$. exist $a=per(10)\mod m$ suth that $m|10^a-1\to m|\frac{10^m-1}{9}$. Therefore $A=111...111$ (all digits 1) with $3^la$ digits satysfyed $3^lm|A$. It give solution for case $(n,10)=1$. By induction we can built numbers $B_k$ and $C_k$ with k digits, suth that $2^k|B_k$ and all digits of $B_k$ are $1$ or $2$, $5^k|C_k$ and all digits of $C_k$ are $1,2,3,4,5$. Let $3^lm|10^a-1,a>k$, then $2^k3^lm|10^aB_k-B_k$ and $5^k3^lm|10^aC_k-C_k$. These numbers have not digits 0, because last digit of $B_k$ is 2 and last digit of $C_k$ is 5.

We start with the following Lemma. For any positive integer $k$ there exist a multiple of $n=5^k$ containing exactly $k$ digits in its decimal representation. Proof. The statement obviously holds for $k=1$. Assume it holds for $k=m$ and let us denote the multiple of $5^m$ by $x_m$. If $a$ is a nonzero digit such that $2^{m}a+\frac{x_m}{5^m} \equiv 0(\text{mod}\ 5)$ (such a digit obviously exists) then it can be seen that the number $x_{m+1}=x_m+10^{m}a$ satisfies the requirements for $k=m+1$. Remark. Notice that a similar proof is valid for $n=2^k$. There are two cases. Case 1. $n=5^{k}m$, where $m$ is coprime to $10$. Case 2. $n=2^{k}m$, where $m$ is coprime to $10$. Let's prove the statement for the first case, since the proof in the second case goes in the same way. By the Lemma one can find an integer $A$, which does not contain 0 in its decimal representation and is divisible by $5^k$. If $s$ is the number of digits of $A$ then then number $$N=A+10^{s}A+...+10^{(l-1)s}A=A\frac{10^{ls}-1}{10^{s}-1}$$can be shown to be divisible by $n$ for $l=m\varphi(m)$. Furthermore, $N$ does not contain $0$ in its decimal representation, and therefore satisfies the requirements.