Let $S=\{ x\in(0,1)\setminus\mathbb{Q} \,:\, \lfloor 100x \rfloor\in \{ 0,11,22,33,44,55,66,77,88,99\}\} $. Find the smallest $k$ such that any $X\in(0,1)$ can be written as $X=x_1+x_2+\dotsb+x_k$ where $x_t\in S$ for $1\le t\le k$ and no two of the $x_i$ are equal.
So if we take $X=0.10\overline{9}$ then we must have $k\ge 12$. But i'm having difficulty showing that $k=12$ suffices. Any ideas?
To show $k=12$ suffices, let $y=\frac {0.11}{12}$ and let $z$ be an irrational that is less than $\frac 1{1000}$. Then $y-6z,y-5z,y-4z,y-3z,y-2z,y-z,y+z,y+2z,y+3z,y+4z,y+5z,y+6z$ add to $0.11$