2006 IMO Shortlist Problems/N2

Problem

(Canada) For $x \in (0,1)$ let $y \in (0,1)$ be the number whose $\displaystyle n$ th digit after the decimal point is the $\displaystyle (2^n)$ th digit after the decimal point of $\displaystyle x$ . Show that if $\displaystyle x$ is rational then so is $\displaystyle y$ .

Solution

For any real ${ a \in (0,1) }$ and any natural number $\displaystyle n$ , let $\displaystyle f_a(n)$ the $\displaystyle n$ th digit after the decimal point of $\displaystyle a$ . We note that $\displaystyle a$ is rational if and only if $\displaystyle f_a(n)$ is periodic for sufficiently large $\displaystyle n$ , i.e., if $\displaystyle f_a(n)$ is determined by the residue of $\displaystyle n$ mod $\displaystyle m$ , for some integer $\displaystyle m$ .

Suppose $\displaystyle x$ is rational, and let $\displaystyle m$ be an integer such that for sufficiently large $\displaystyle n$ , $\displaystyle f_x(n)$ is determined by the residue of $\displaystyle n$ mod $\displaystyle m$ . Let $\displaystyle m = 2^r \cdot k$ , for some odd integer $\displaystyle k$ and some nonnegative integer $\displaystyle r$ . We note that the residue of $\displaystyle n$ mod $\displaystyle m$ is uniquely determined by the residues of $\displaystyle n$ mod $\displaystyle 2^r$ and mod $\displaystyle k$ . Then for sufficiently large $\displaystyle n$ ,