A multiple of n written with only 0s and 1s

1. The $n+1$ numbers $R_0,R_1,\dots,R_n$ have only $n$ possible remainders modulo $n$, so $R_i\equiv R_j\pmod n$ for some $0\le i<j\le n$. Then $n\mid R_j-R_i$. But $R_j-R_i$ is the number made of $j-i\ge1$ ones followed by $i$ zeros, which is positive and uses only the digits $0$ and $1$.

2. Assume $\gcd(n,10)=1$. Then $\gcd(9n,10)=1$ as well, so by Euler $10^{k}\equiv1\pmod{9n}$ for $k=\varphi(9n)$. Hence $9n\mid10^k-1=9R_k$, so $n\mid R_k$.

Check. $n=7$: $7\mid111111$. $n=3$: $3\mid111$.

Remark. If $n$ has factors $2$ or $5$, part 2 fails, but part 1 still holds: $2\mid10$, $5\mid10$.

Lesson. Pigeonhole on a sequence of residues gives a nonzero difference divisible by $n$; the structure of the sequence ($R_j-R_i$) tells you what that difference looks like.