# Adding sum of leftmost digits

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.

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 $$9$$