Let $N = a_1a_2\ldots a_c$. Let us define a sequence of pair of integers by the following way. $a_0 = N$ and $b_0 = 0$.
$$\begin{cases} a_{n+1} \equiv a_n \cdot N + b_n \pmod{10^c} \\ b_{n+1} = \frac{a_n \cdot N + b_n - a_{n+1}}{10^c} \end{cases}$$
Let us show that $b_n < N$, for $n \in \mathbb{N}$. Use induction on $n$. For $n = 0$, it is trivial.
$$b_{n+1} < \frac{(a_n + 1)N - a_{n+1}}{10^c} \le \frac{(a_n + 1)N}{10^c} \le N.$$
Thus,
$$\begin{cases} b_n \equiv 10^c \cdot b_{n+1} + a_{n+1} \pmod N \\ a_n = \frac{10^c + a_{n+1} - b_n}{N} \end{cases}$$
Hence $(a_n, b_n)$ and $(a_{n+1}, b_{n+1})$ define each other unambiguously. Since $0 \le a_n, b_n \le 10^c$, the number of different pairs of $(a_n, b_n)$ is finite. Thus, $(a_n, b_n)$ is periodic sequence from the first term. Let $(a_t, b_t) = (a_0, b_0)$, then
$$M = a_t \cdot 10^{(t-1)c} + a_{t-1} \cdot 10^{(t-2)c} + \dots + a_1$$
has a desired property. This shows that for each positive integer $N$ there exists $M$ that satisfies the conditions in the problem statement.