Problem: If the decimal expansion of $a$ contains all $10$ digits ($0,...,9$), then the number of length $n$ strings (shorted as $n$-strings) is greater than $n+8$.
If you've established the simple lemma, the solution is instant. Otherwise very impossible.
Lemma: The number $C_n$ of $n$-strings is strictly greater than $C_{n-1}$, that of $n-1$-strings.
Actually, we always have $C_n \ge C_{n-1}$: Every $n-1$-string corresponds to an $n$-string by continuing 1 more digit. The map is clearly injective. If $C_n=C_{n-1}$, it is bijective, meaning we have a way to continue uniquely, which means rationality. Rigidly, at least one of the $n-1$-strings occurs infinitely, but all digits after some $n-1$-string is totally determined by it. So if an $n-1$-string appears twice, it must appear every such distances, and so do the digits between.
(Further discussion: For a rational number, split it into a finite part, and a recurring part. If the minimal length of recurring string is $n$, then any $m$-string starting in the recurring part has exactly $n$ variations, if $m \ge n$. Additional variations brought by the finite irregular part is finite (regardless of $m$), as the starting point is finite. So $C_n$ in this case is not decreasing but bounded. So it reaches some certain number and stays stable. In a purely recurring case, the number is exactly $n$ (meaning afore-defined).
Now $C_n \ge C_{n-1}+1 \ge ... \ge 9+1$, as $10 > 9=1+8$ holds, the problem is solved.
This may not be really hard. But the most important thing is to see the principle behind it.
(I WILL FURTHER COMPLETE THIS POST.)
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