Number on a blackboard

Lemma: If the last digit of $n$ is greater than 5 then the last digit of $9n$ is less than or equal to 5.
Proof: Let $d$ be the last digit of $n$ . Then the last digit of $9n$ is the same as the last digit of $9d$ . But $9d \in \{54,63,72,81\}$ . $\square$

Clearly, all $a<5$ will become zero. Assume that, for all $a<N$ , $a$ eventually leaves the blackboard empty. If the last digit of $N$ is less than or equal to 5, it will be erased and on day two we will have a number smaller than $N$ which, by assumption, will eventually give us an empty blackboard. If the last digit of $N$ is greater than 5 then on day two we will have $9N$ which, by our lemma, will have a last digit of at most 5. Thus, on day 3, we will erase the last digit of $9N$ , yielding $\left\lfloor \frac{9N}{10} \right\rfloor$ . But $\left\lfloor \frac{9N}{10} \right\rfloor < \frac{9N}{10} < N$ and therefore, by assumption, will eventually give us an empty blackboard. By strong induction, all $N$ will eventually yield an empty blackboard.

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Alternatively, by contradiction: assume there are integers for which we never get an empty blackboard. Then there must be a smallest such integer. But, as show above, after at most two days we will have a smaller integer on the blackboard. Contradiction.

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Solved by meta-reasoning: If $a$ never leaves the blackboard empty, then neither will $10^na$ for any $n$ . Thus, if one such $a$ exists, there must be infinite such $a$ . Therefore the sum of all such $a$ would diverge, and it would be impossible to input an answer. So no such $a$ can exist.