Bulgarian Mathematical Competitions

Solution:
Let $M$ be an arbitrary positive integer. We shall prove that there exists a term of the sequence $\{a_{n}\}_{n=1}^{\infty}$, whose decimal representation is obtained from that of $M$ by adding several digits from the right, i.e. the number $M$ is a "beginning" of that member.
Let $k$ be an index such that $a_{k} \leq M \cdot 10^{l} < a_{k+1} < a_{k} + c$, where $l$ is a positive integer which is greater than the number of the digits of $c$. Then
$$M \cdot 10^{l} < a_{k+1} < M \cdot 10^{l} + c$$
and obviously $a_{k+1}$ satisfies the above requirement.
Let $m = 2^{\alpha} 5^{\beta} t$, where $(t, 10) = 1$. It is enough to prove the assertion of the problem for $m = 10^{\gamma} t$, where $\gamma = \max\{\alpha, \beta\}$.
Let us consider the number
$$M = 1 \underbrace{00 \ldots 0}_{p} 1 \underbrace{00 \ldots 0}_{p} 1 \ldots 1 \underbrace{00 \ldots 0}_{p} 1 \underbrace{00 \ldots 0}_{q}$$
Here $p = k \varphi(t)$, where $\varphi(t)$ is the Euler function, $k$ is a positive integer such that $p > \gamma$, the number $q$ is greater than $\gamma$, and the number of 1's is $t+1$.
Then $M$ is a "beginning" of some $a_{k}$. Hence the sequence of the digits (formed by the terms of the sequence written one after another) looks like this:
$$f_{1} f_{2} \ldots f_{r} 1 \underbrace{00 \ldots 0}_{p} 1 \underbrace{00 \ldots 0}_{p} 1 \ldots 1 \underbrace{00 \ldots 0}_{p} 1 \underbrace{00 \ldots 0}_{q} \ldots$$
where $f_{1}, f_{2}, \ldots, f_{r}$ are the digits before $a_{k}$. It is clear now that, depending on the remainder of $\overline{f_{1} f_{2} \ldots f_{r} 1}$ modulo $t$, we can add suitable digits from $M$ to $\overline{f_{1} f_{2} \ldots f_{r} 1}$ in such a way that the resulting number is divisible by $10^{\alpha} t$.