Let's assume that $\pi = \sum_{i=0}^{\infty}\frac{a_i}{10^i}.$ Also, let $n$ be an arbitrary natural number.

Is there $j,k,m \in \mathbb{N}, (j < k < m)$ such that $|b_1 - b_{2}| = n$ when $b_1 = \sum_{i=j}^{k}a_i10^{k-i}, b_2 = \sum_{i=k+1}^{m}a_i10^{m-i}$?

**for example**: 0.14(15)(92)... so for |92-15|=77 there are j,k,m as wanted.

the **original question** was:

can a number $x = \sum_{i=0}^{\infty}\frac{x_i}{10^i}$ such that $\forall n \in \Bbb N\ \exists j,k,m \in \mathbb{N}, (j < k < m)$ such that $|b_1 - b_{2}| = n$ when $b_1 = \sum_{i=j}^{k}a_i10^{k-i}, b_2 = \sum_{i=k+1}^{m}a_i10^{m-i}$ is a rational number?

i would like for now a solution only for the first part but any solution will be welcomed.