, 1994

Let $b_{k}$ and $c_{k}$ be the number of digits in the $k$th term in lists $B$ and $C$, respectively. Then
$$2^{b_{k}-1} \leq 10^{k} < 2^{b_{k}} \Longleftrightarrow \log_{2} 10^{k} < b_{k} \leq \log_{2} 10^{k} + 1 \Longleftrightarrow b_{k} = \left\lfloor k \cdot \log_{2} 10 \right\rfloor + 1$$
and, similarly
$$c_{k} = \left\lfloor k \cdot \log_{5} 10 \right\rfloor + 1.$$
Beatty's theorem states that if $\alpha$ and $\beta$ are irrational positive numbers such that
$$\frac{1}{\alpha} + \frac{1}{\beta} = 1,$$
then the sequences $\lfloor k \alpha \rfloor$ and $\lfloor k \beta \rfloor$, $k = 1, 2, \ldots$, partition the positive integers.
Then, since
$$\frac{1}{\log_{2} 10} + \frac{1}{\log_{5} 10} = \log_{10} 2 + \log_{10} 5 = \log_{10}(2 \cdot 5) = 1,$$
the sequences $b_{k} - 1$ and $c_{k} - 1$ partition the positive integers, and therefore each integer greater than $1$ appears in $b_{k}$ or $c_{k}$ exactly once. We are done.