64. If $2^{x} \times 9^{y}$ equals the four-digit number $\overline{2 x 9 y}$, then $x^{2} y^{3}=$
Answer: 200.
Solution: If $x=0, 2^{0}=1$, and $9^{3}=729$, $9^{4}=6561>2999>\overline{2 x 9 y}$, thus $2^{x} \times 9^{y}>\overline{2 x 9 y}$.

Therefore,
$$
y<4,
$$

which means $y$ can only be 0, 2.

Since the left side has no factor of 5, $\overline{2 x 9 y}$'s $y$ cannot be 0, so $y$ can only be 2.
At this point, $\overline{2 x 92}$ on the right must be a multiple of 9, so
$$
2+x+9+2=13+x \text { must be a multiple of } 9,
$$

thus
$$
x=5 \text {. }
$$

Upon verification, $2^{5} \times 9^{2}=2592$ satisfies the given conditions,
therefore
$$
x=5, \quad y=2 .
$$

So
$$
x^{2} y^{3}=5^{2} \times 2^{3}=200 \text {. }
$$