A periodic decimal expansion Let us suppose that $\{\alpha_{n}\}_{n \in \mathbb{N}}$ is a strictly increasing sequence of natural numbers and that the number obtained by concatenating the decimal representations of the elements of $\{\alpha_{n}\}_{n \in \mathbb{N}}$ after the decimal point, i.e.,
$0.\alpha_{1}\alpha_{2}\alpha_{3}\ldots$
has period $s$ (e.g., $0.12 \mathbf{12} \mathrm{121212}...$ has period 2).
If $a_{k}$ denotes the number of elements in $\{\alpha_{n}\}_{n \in \mathbb{N}}$ with exactly $k$ digits in their decimal representation, does the inequality
$a_{k} \leq s$
always hold?
What would be, in your opinion, the right way to approach this question? I've tried a proof by exhaustion without much success. I'd really appreciate any (self-contained) hints you can provide me with.

If the period is $s$ then there are essentially $s$ starting places in the recurring decimal for a $k$-digit integer - begin at the first digit of the decimal, the second etc - beyond $s$ you get the same numbers coming round again. If you had $a_k > s$ then two of your $\alpha_n$ with $k$ digits would be the same by the pigeonhole principle.
