IMO 2006 Shortlisted Problems

Since $x$ is rational, its digits repeat periodically starting at some point. We wish to show that this is also true for the digits of $y$, implying that $y$ is rational.

Let $d$ be the length of the period of $x$ and let $d=2^u\cdot v$, where $v$ is odd. There is a positive integer $w$ such that $$2^w\equiv 1(\mathrm{mod}v).$$ (For instance, one can choose $w$ to be $\phi (v)$, the value of Euler's function at $v$.) Therefore $$2^{n+w}=2^n\cdot 2^w\equiv 2^n(\mathrm{mod}v)$$ for each $n$. Also, for $n\ge u$ we have $$2^{n+w}\equiv 2^n\equiv 0\left(\mathrm{mod}{2}^u\right)$$ It follows that, for all $n\ge u$, the relation $$2^{n+w}\equiv 2^n(\mathrm{mod}d)$$ holds. Thus, for $n$ sufficiently large, the $2^{n+w}$th digit of $x$ is in the same spot in the cycle of $x$ as its $2^n$th digit, and so these digits are equal. Hence the $(n+w)$th digit of $y$ is equal to its $n$th digit. This means that the digits of $y$ repeat periodically with period $w$ from some point on, as required.