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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.)

 Feb 17, 2021
 #1
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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 $18\times 9\times 10=\boxed{1620}.$

 Feb 17, 2021
 #2
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Correct! Nice job! Good explanation! I like your method!

 Feb 17, 2021

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