Multiple of real numbers with only $0$'s and $1$'s in the integer part

Solution 1:

If we allow $n$ to be an arbitrary integer (and $x$ be positive), then the answer is positive.

Imitating the case where $x$ is a fraction we consider the numbers $a_1=1, a_2=11, a_3=111, ... $. Let $r_1,r_2,r_3, ...$ be the remainders when the $a_i$ are divided by $x$ (i.e. $a_j=n_j*x+r_j$ where $n_j$ are integers and $0 \le r_j<x$). If $i>[x]+1,$ then the distance between some remainders $r_k, r_l$ $(1 \le k<l<=i)$ is smaller than 1. As $|r_l - r_k|<1,$ then $a_k-a_l=(n_k-n_l)*x+r_k- r_l$ with $a_k-a_l$ having only 0's and 1's as digits in its decimal expansion and $-1<r_k-r_l<1.$ Hence $[(n_k-n_l)*x] = a_k-a_l$ if $r_k \le r_l$ and $[(n_k-n_l)*x] = a_k-a_l-1 = -111...10000...0001$ if $r_k > r_l.$

However, this answers your question if $x<0$ as $[(n_k-n_l)*z]=[(n_l-n_k)*(-z)]$ for $z>0$ and $n_l-n_k>0.$

My guess is that you are mainly interested in the case where $n$ is a natural number and $x>0.$