# Repeatedly multiplying a number by $10^k+1$, where the $k$ rightmost digits are nonzero

Let $n$ be an integer not ending with zero in its decimal representation. Then there are $k \geq 1$ nonzero digits in $n$ (starting from the right side until we hit a zero). Let's multiply $n$ by $10^k+1$ and do the same procedure for the resulting number.

For example, given $n=102035$, we see $k=2$ and multiply by $101$ to get $10305535$, Then we multiply by $10^4+1$ and so on.

This procedure "kills" the rightmost zero of the number: the first index from the right in which there is a zero strictly increases after the multiplication. However, the number of digits increases by the multiplication.

I want to prove that there is an $n$ such that this iterated procedure will keep returning numbers containing a zero digit in them. In other words, no matter how many times we do the multiplication, we never reach a number without zeros.

It seems that this is the behaviour for the "typical" input, like $n=102035$ mentioned above. However, It is not clear to me how to prove that there is such a starting number $n$.

$n = 1011$ appears to work. Iterating your procedure, one gets $$\begin{cases} 1011\\ 102111\\ 1021212111\\ 102121212121212111\\ 1021212121212121212121212121212111 \end{cases}$$ The pattern is clear, and the proof straightforward.