Increasing the number of nonzero digits (from AOPS) I found this currently unsolved problem in the AOPS site (link):

Let $n$ be a positive integer with exactly $d$ non-zero digits. Show that there is a multiple of $n$ which has exactly $d+1$ non-zero digits.

It seems to be quite challenging and I have no idea how to solve it. Any hints how to proceed?
 A: You can combine three simple tricks to solve the problem:


*

*Take a digit greater than $1$, and split it into two nonzero digits by adding $10^t-10^s$.

*If each digit in your number $n$ is $0$ or $1$, replace $n$ by $2n$.

*In order to get rid of the prime factors $2$ and $5$, multiply $n$ by a power of $10$.

If $n$ has a digit greater than $1$ then let $m=n$. Otherwise, if each digit of $n$ are $0$ or $1$ then let $m=2n$. Hence $m$ is a multiple of $n$, it has the same nonzero digits as $n$, and $m$ has at least one digit grater than $1$.
Let $m=\sum_{i=0}^{k-1} a_i\cdot 10^i$ where $a_0,\ldots,a_{k-1}$ are the decimal digits of $m$ and let $g$ be an index with $a_g\ge2$. 
Take two integers $t,s$ such that $s>g$, $t>s+k$ and $10^t\equiv 10^s\pmod{n}$, and choose
$$
M = 10^{s-g}m + (10^t-10^s).
$$
The factor $10^{s-g}$ shifts the digit $a_g$ to the $s$th position.
The term $10^t-10^s$ replaces the digit $a_g$ by $a_g-1$ (which is nonzero), and adds a new leading digit $1$. Due to $t-s>k$, this leading $1$ is not added to the previous nonzero digits.
