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.)
When the first digit is 1, there are only 3 choices for the 2nd digit: 0, 2, 4
When the 2nd digit is 0, there are 10 choices for the 3rd digit and only 8 choices for the 4th digit:2, 3, 4, 5, 6, 7, 8, 9. So: 10 x 8 ==80
When the 2nd digit is 2, there are 10 choices for the 3rd digit and only 6 choices for the 4th digit: 4, 5, 6, 7, 8, 9. So:10 x 6 ==60
When the 2nd digit is 4, there are 10 choices for the 3rd digit and only 2 choices for the 4th digit: 8 and 9. So: 10 x 2 ==20
Therefore beginning with 1 as the 1st digit from the left, we have a total of: 80 + 60 + 20 =160 such integers.
This pattern is repeated for every 1st digit from 1 to 9. Therefore, there are a total of: 9 x 160 =1,440 such 4-digit integers.
Note: I wrote a short computer code in C++ and confirms the above total of 1440 integers. If needed, it can be printed.
