Every integer divides some number made of 1s and 0s

Prove that for each positive integer $n$ there exists a positive multiple of $n$ whose decimal digits are all $0$ or $1$.

Use the pigeonhole principle on repunits. Look at $1, 11, 111, \dots$ up to the number with $n+1$ ones. There are $n+1$ of them but only $n$ residues mod $n$, so two share a remainder. Their difference is divisible by $n$ and has the form $11\cdots1\,00\cdots0$, which uses only the digits $1$ and $0$.