9*9*9*1 or 9*9*9*2 coz they said A, B, C, D are 4 positive number either D could be 5. Can anyone clear my doubt? As 0 can't be taken as positive integer

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\(A, B, C, D\) all positive integers.

\(AD\) & \(DC\) both divisible by \(5\)

So \(D\) & \(C\) both will be \(5\) as they can't be \(0\)

\(A\) & \(B\) can take \(1-9\)

Total \(4\) digit nos \(= 9 * 9 * 1 * 1 *4! = 81 * 4! = 1944\)

Carcass
the answer should be 81
how is it that it is 1944

D and C are 5 so
9*9*1*1 = 81

why 4! is multiplied on top of it ??

cuz as far as im looking at it the order is A B C and D

Case 1: $D=5$

If $D$ is 5 , then both products ( $A D$ and $D C$ ) are guaranteed to be divisible by 5 , because 5 is a factor in both.
- A can be any of the 9 digits ( $1-9$ ).
- B can be any of the 9 digits ( $1-9$ ).
- $\(\mathbf{C}\)$ can be any of the 9 digits ( $1-9$ ).
- $\(\mathbf{D}\)$ is fixed as 5 (1 choice).

Total for Case 1: $\(9 \times 9 \times 9 \times 1=\mathbf{7 2 9}\)$

Case 2: $D \neq 5$
If $D$ is not 5 , the only way for the products to be divisible by 5 is if the other factors are 5 .
- For $A D$ to be divisible by 5 , A must be 5 .
- For $D C$ to be divisible by 5, $\(\mathbf{C}\)$ must be $\(\mathbf{5}\)$.

For this case:
- $\(\mathbf{A}\)$ is fixed as 5 ( 1 choice).
- B can be any of the 9 digits ( 9 choices).
- $\(\mathbf{C}\)$ is fixed as 5 ( 1 choice).
- D can be any digit except $\(5(1,2,3,4,6,7,8,9 \Longrightarrow 8\)$ choices).

Total for Case 2: $\(1 \times 9 \times 1 \times 8=\mathbf{7 2}\)$

Final Total
Since these cases are mutually exclusive, we add them together:

$$
\(729+72=801\)
$$


There are $\(\mathbf{8 0 1}\)$ possible four-digit numbers.