Q:

Getting from $N$ to $N-8$ with finitely-many steps of adding $3$ or removing an instance of the digit "$1$"

Here is a strange question that a friend gave to me:

Suppose we have an $n$-digit number $N$ written on the board. We are given the following challenge:
At every step, we can either add $3$ to the number $N$ and write the new number, or delete one of the instances of the digit "$1$" within the number and write the new number on the board.
Is it possible that after a finite number of steps we have the number $N-8$?

I really don't know where to start from!
I did a little experimenting with some examples but didn't make it :)
Any help is highly appreciated!

A:

An easy to describe approach is to start with $N$ which has $n$ digits.  Add $3\ \frac 13(10^n-1)$ times.  This will add $10^n-1$ to the number, which will put a $1$ in front of it and decrease the rest by $1$.  Now cancel the $1$ at the front.  Repeat $8$ times.  This takes more operations than other approaches, but it will work.

