Zeros and ones only

Being a multiple of $63$ means being a multiple of both $9$ and $7$. What does divisibility by $9$ say about how many ones are needed?

1. A multiple of $9$ has digit sum divisible by $9$, so the number needs at least $9$ ones, and the smallest candidate with exactly $9$ ones and no zeros is $111111111$.
2. That number leaves remainder $6$ modulo $7$, so it fails, and any other number with fewer than $10$ digits needs $9$ ones and is the same number.
3. With $10$ digits and exactly $9$ ones, the single zero can sit in positions $2$, $3$, $4$, $5$ and so on, and in increasing order the candidates $1011111111$, $1101111111$, $1110111111$, $1111011111$ leave remainders $3$, $2$, $4$, $0$ modulo $7$.
4. The first candidate divisible by $7$ is therefore the smallest multiple of $63$ of this kind, and it is $1111011111$.

Answer: $1111011111$