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# help counting

0
122
4

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.)

Mar 13, 2022

#1
+2

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

Mar 13, 2022
#2
+1

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? ElectricPavlov  Mar 13, 2022
edited by ElectricPavlov  Mar 13, 2022
edited by ElectricPavlov  Mar 13, 2022
#3
+1

I agree with EP!  If repetitions are allowed, I calculate: 3,150 such 4-digit integers.

Guest Mar 14, 2022
#4
+1

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

Mar 14, 2022