It is obviosly proved by induction. 1)2|2 Let $2^{n}|A_{n}$, we must find $A_{n+1}=A_{n}(mod \ 2^{n})$. Because it had unique solution $A_{n+1}=A_{n}+c*10^{n}$, were $c=2$ or $c=5$. Obviosly $c=2$ if $2^{n+1}|A_{n}$ else $c=5$.

We use pigeonhole and induction: Let \[ S_n = \{s|s_{10}\text{ consists of only } 2 \text{ and }5\text{ and is }n\text{ digits long}\}\] Consider $ a,b\in S_{2005}$ that are equal $ \mod{2^{2005}}$. Then consider $ a - b\mod{10}$; it must be $ 0$ or $ \pm3$. Obviously the latter case fails. So, $ a - b$ can be written as $ 5$ times a multiple of $ 2^{2005}$, that is $ 10$ times a multiple of $ 2^{2004}$. Now we take $ a - b$, divide by $ 10$, and get a new number that can be written as the difference of some $ a_2,b_2\in S_{2004}$. By induction we continue in this way until we are left with two numbers in $ S_1$ that are congruent mod $ 2$. Thus the two numbers are equal, either both 2 or both 5, and the two numbers where forced to be equal at every step along the induction. Hence $ S_{n}$ consists of the entire residue class $ \mod{2^n}$. In particular, it contains the class 0, and we are done.

Well, here 2 and 5 are dummy—they can be replaced by an odd number and an even number (except 0).