How many 4-digit numbers have the second digit even and the fourth digit at least twice the second digit? (Here, digits are read from the left, so the first digit is the leftmost digit, and so on.)
The second digit must be even, so it must be one of 0, 2, 4, 6, or 8 However, it cannot be 6 or 8 since then the fourth digit could not be twice the second digit. Thus, there are different possible combinations of second and fourth digits, as shown in the following table:\[
\begin{array}{|c|c|}\hline
\text{Second digit} & \text{Fourth digit} \\ \hline
0 & 0,1,2,3,4,5,6,7,8,9 \\ \hline
2 & 4,5,6,7,8,9 \\ \hline
4 & 8,9 \\ \hline
\end{array}
\]
The first digit can be any of the nonzero digits, and the third digit can be any of the 10 digits. The answer is $18\times 9\times 10=\boxed{1620}.$
