Sum of digits of sum of digits of sum of digits

To calculate the answer, all we need are some bounds on the order of magnitude of $x$ and $y$. (In particular, we won't be looking at the answer options, nor will we need to resort to any kind of reasoning "from the assumed uniqueness of the answer", which is questionable at best.)

Here's all it takes:

The starting number has 1998 digits. Therefore $x$ is at most

Therefore, $y$ is at most

Therefore, $z$ is at most

Finally, $y$ cannot actually be 99 (which is larger than 45), so $z$ cannot be 18. The only possible value for $z$ is then

My answer is

Well, given that

and we only need to find

The format of question seems to imply the other answers must be wrong.

Consider the number: 9 followed by 1997 zeros ($9*10^{1997}$). This is trivially divisible by 9 (result is 1 followed by 1997 zeros or ($1*10^{1997}$).

The sum of digits is 9 ($9 + 0 + 0 + ... + 0$). Therefore, a is a possible answer.

Does this exclude the possiblity of other choices being correct?

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Because the setting is a competitive exam, we can assume there is only one correct choice (or if they made a mistake, this answer to this question wouldn't matter). Either way, speed and correctness is important in this setting, but not for coming up with a proof the other choices are wrong.