Brazilian Mathematical Olympiad

The answer is yes. In fact, there exist two sequences $x = \ldots a_n \ldots$ and $y = \ldots b_n \ldots$ such that, for every $k \ge 1$, $2^k$ divides $a_k \ldots a_2 a_1$ and $5^k$ divides $b_k \ldots b_2 b_1$; then $10^k$ divides $a_k \ldots a_2 a_1 \times b_k \ldots b_2 b_1$ and $xy = 0$.

Consider first $x$. We proceed by induction. For $k=1$ choose $a_1=2$. Suppose $a_m \ldots a_2 a_1$ is a multiple of $2^m$ and let $c_m$ be $a_m \ldots a_2 a_1$ divided by $2^m$. Then $a_{m+1} a_m \ldots a_2 a_1 = 10^m \cdot a_{m+1} + a_m \ldots a_2 a_1 = 2^m (5^m \cdot a_{m+1} + c_m)$ 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, $a_{m+1} a_m \ldots a_2 a_1$ 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 $b_m \ldots b_2 b_1 = 5^m \cdot d_m$. Then $b_{m+1} b_m \ldots b_2 b_1 = 10^m \cdot b_{m+1} + b_m \ldots b_2 b_1 = 5^m (2^m \cdot b_{m+1} + d_m)$ and it's always possible to choose $b_{m+1} \in \{0, 1, 2, 3, 4\}$ such that $2^m \cdot b_{m+1} \equiv -d_m \pmod{5}$, since $2^m$ admits inverse modulo $5$.