Prove that for positive number, some multiple only has 0 and d as it's digits Let $ n$ be a positive integer, and let $1<=d<=9$. Show that some multiple
of $n$ has $0$ and $d$ as its only digits.
I don't know how to even start this question. It's under the pigeonhole section of the text book so I am guessing that we have to use that somehow. Any help?

HINT: Consider the numbers written $d$, $dd$, $ddd$, and so on. E.g., if $d=3$ these are $3$, $33$, $333$, etc. Note that if you subtract one of these numbers from a larger one, you get a number whose digits are all either $d$ or $0$.
Let $r_k$ be the remainder when you divide $$\underbrace{dd\ldots dd}_{k\text{ digits}}$$ by $n$. There are only $n$ possibilities for $r_k$: it must be in the set $\{0,1,\ldots,n-1\}$. These possibilities are your pigeonholes. 
