How many 4-digit numbers have the second digit even and the fourth digit is at most twice the second digit? (Here, digits are read from the left, so the first digit is the leftmost digit, and so on.)
There are 2 cases: The second digit is a 2 or it is a 4
If the second digit is a 2: \(9 \times 1 \times 10 \times 5 = 450\)
If the second digit is a 4: \(9 \times 1 \times 10 \times 9 = 810\)
Add these together for a total of \(\color{brown}\boxed{1,260}\) ways
the fourth digit is at most twice the second digit .... I think you may need to recalculate .....
second digit = 0 fourth = 0
2 fourth 0 1 2 3 4
4 fourth 0 - 8
6 fourth 0-9
8 fourth 0-9 Right?
How many 4-digit numbers have the second digit even and the fourth digit is at most twice the second digit? (Here, digits are read from the left, so the first digit is the leftmost digit, and so on.)
0 0 9*10 = 90
2, (0,1,2) 9*10*3 = 270
4, (0-8) 9*10*9 = 810
6 (0-9) 9*10*10=900
8 (0-9) 9*10*10=900
90+270+810+1800 = 2970