XVI OBM

The answer is yes. In fact, there exist two sequences $x=\dots a_n\dots$ and $y=\dots b_n\dots$ such that, for every $k\ge 1$, $2^k$ divides $a_k\dots a_2{a}_1$ and $5^k$ divides $b_k\dots b_2{b}_1$; then $10^k$ divides $a_k\dots a_2{a}_1\times b_k\dots b_2{b}_1$ and $xy=0$.

Consider first $x$. We proceed by induction. For $k=1$ choose $a_1=2$. Suppose $a_m\dots a_2{a}_1$ is a multiple of $2^m$ and let $c_m$ be $a_m\dots a_2{a}_1$ divided by $2^m$. Then $a_{m+1}{a}_m\dots a_2{a}_1=10^m\cdot a_{m+1}+a_m\dots a_2{a}_1=2^m(5^m\cdot a_{m+1}+c_m)$ and we may choose $a_{m+1}=0$ or $1$ according to the parity of $c_m$, so $5^m\cdot a_{m+1}+c_m$ is even and, consequently, $a_{m+1}{a}_m\dots a_2{a}_1$ is a multiple of $2^{m+1}$. This completes the induction.

The induction for $y$ is not that different. Indeed, choose $b_1=5$ and suppose $b_m\dots b_2{b}_1=5^m\cdot d_m$. Then $b_{m+1}{b}_m\dots b_2{b}_1=10^m\cdot b_{m+1}+b_m\dots b_2{b}_1=5^m(2^m\cdot b_{m+1}+d_m)$ and it's always possible to choose $b_{m+1}\in \{0,1,2,3,4\}$ such that $2^m\cdot b_{m+1}\equiv -d_m(\mathrm{mod}5)$, since $2^m$ admits inverse modulo $5$.