Integers whose multiples contain a specific digit

Question

In base 10 (or any other base, for that matter) is there an integer all whose multiples contain a specific digit, say 1?

Answer 1

In base $b$, we can represent a number $n$ as $xy$ with $\gcd(x,b)=1$ and $y|b^k$ for some $k$. Let $z=b^k/y$, so $xb$ is a multiple of $n$. Then, if we consider

$$b^k\left(b^{\varphi(x)}-1\right),$$

it is a multiple of $x$ by Euler's theorem, and only contains the "digits" $0$ and $b-1$. Similarly,

$$b^k\left(\frac{b^{(b-1)\varphi(x)}-1}{b-1}\right)$$

is a multiple of $n$ and has only the digits $1$ and $0$. So, the only digit that could necessarily exist in every multiple of $n$ is $0$, and if $b|n$ then every multiple of $n$ has a $0$.

Note: this only works in bases $b>2$. If $b=2$, then every nonzero number has a one.