# Multiple of $n$ that does not contain the digit $1$ in its decimal representation

Prove that for every positive integer $n$ coprime to $10$ there exists a multiple of $n$ that does not contain the digit $1$ in its decimal representation.

I thought about using Euler's Totient Theorem. This says that $a^{\phi(10)} \equiv 1 \pmod{10}$. How could we continue from this?

It is more fruitful considering that $10^{\phi(a)}\equiv 1\pmod a$.
Then $10^{\phi(a)}-1$ is a multiple of $a$ and it has only nines in its decimal representation.
Consider numbers $3,33,333,3333,...,\underbrace{333\ldots 3}_\text{n 3's}$
Then say $3\equiv r_1\pmod n$,$33\equiv r_2\pmod n$,$\underbrace{333\ldots 3}_\text{n 3's}\equiv r_n\pmod n$.
If any of $r_i$ is $0$ then we're done if not then there exist $r_i,r_j$ and say i>j that are equal from pignhole principal say
$$r_i\equiv r_j\pmod n$$ or $$\underbrace{333\ldots 3}_\text{i 3's}\equiv \underbrace{333\ldots 3}_\text{j 3's} \pmod n$$ then $$\underbrace{333\ldots 3}_\text{i-j 3's}\underbrace{00\ldots 0}_\text{j 3's}\equiv 0\pmod n$$Since $(n,10)=1$ then $$\underbrace{333\ldots 3}_\text{i-j 3's}\equiv 0 \pmod n$$ And we are done
The set of multiples of $n$ is a subset of $\mathbb{Z}$ with positive density, while the set $E$ of numbers without $1$s in their decimal representation is rather sparse (it is an integer Cantor set with zero density), so there is for sure a multiple of $n$ in $\mathbb{N}\setminus E$. For that to work, $\gcd(n,10)=1$ is quite irrelevant.
In particular, as soon as $m$ is an integer such that $$1-\left(\frac{9}{10}\right)^m > 1-\frac{1}{n},$$ there is a multiple of $n$ without digits $1$ in the interval $[1,10^m]$, so the least multiple of $n$ without digits $1$ in its decimal representation is always smaller than $n^{22}$. And in general $n^{22}$ is much smaller than $10^{\varphi(n)}$.