Adding sum of leftmost digits

Question

Starting with a number of at least $9$ digits, every minute we add the sum of the leftmost $9$ digits of the current number to the number itself. Will we always, at some point, see $9$ consecutive numbers that are all not divisible by $9$?

The sum of the leftmost $9$ digits is at most $9\times 9=81$, so we never add more than $81$ each time. Consequently, when the number is large enough, the leftmost $9$ digits will stay the same for more than $9$ minutes. We can wait until the leftmost $9$ digits is divisible by $9$, to make sure that we have $9$ consecutive numbers that leave the same remainder when divided by $9$. If we can ensure that at some point this remainder is not $0$, we would be done.

Answer 1

Let $a_n$ be the $n$th number. Clearly, the sequence $\{a_n\}_n$ is strictly increasing and hence unbounded. Let $M=10^m$ for some $m\ge 11$ and such that $M>a_1$. Then there is a minimal $n_0$ with $a_{n_0}\ge M$. As noted by the OP, $a_{n_0}\le a_{n_0-1}+ 81$, hence certainly $a_{n_0}$ has leading digits $100000000$. This means that the sequence will grow in steps of $1$ for a while, namely until we hit $10^m+10^{m-8}$ exactly. From here on, we will advance in steps of $2$ until we hit $10^m+2\cdot 10^{m-8}$ exactly. Now follow steps of $3$ until we hit $10^m+3\cdot 10^{m-8}+2$, then steps of size $4$ until we hit $10^m+4\cdot 10^{m-8}+2$, then steps of size $5$ until we hit $10^m+5\cdot 10^{m-8}+2$, then steps of size $6$ until $10^m+6\cdot 10^{m-8}+4$. Note that $10^m+5\cdot 10^{m-8}+2+k\cdot 6$ is not a multiple of $9$ (or even a multiple of $3$). Thus we have found $\approx \frac{10^{m-8}}6\gg 9$ consecutive terms that are not multiples of $9$