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.)
+2668 There are 2 cases: The second digit is a 2 or it is a 4
If the second digit is a 2: 9×1×10×5=450
If the second digit is a 4: 9×1×10×9=810
Add these together for a total of 1,260 ways
+37165 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?
+118696 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
