ABCD ×4 = DCBA

The long multiplication can be written as:

$\begin{aligned} \overline{ABCD} \times 4 & = \overline{DCBA} \\ 4000A + 400B + 40C + 4D & = 1000D+100C+10B+A \end{aligned}$

Since the RHS or $\overline{DCBA} < 9999$ , $A$ can only be 1 or 2. And since $\overline{DCBA}$ is even, $A=2$ . Then

$\begin{aligned} 8000 + 400B + 40C + 4D & = 1000D+100C+10B+2 \end{aligned}$

Since the LHS $\ge 8000$ , this means that $1000D \ge 8000$ or $D$ is either 8 or 9. Since $4\blue D=4\times \blue 8 = 3\red 2 = \overline{3\red A}$ , $\blue{D=8}$ . Then

$\begin{aligned} 8000 + 400B + 40C + 32 & = 8000+100C+10B+2 \\ 390B+30 & = 60C \\ 13B + 1 & = 2C \end{aligned}$

For $B,C < 10$ , there is only one solution, $B=1$ and $C=7$ . Therefore, $\overline{ABCD} = \boxed{2178}$ .

As ABCD and DCBA are 4 digit numbers.

ABCD < 2500

So A lies b/w 1 to 2 . Also there is not such number that is divisable by 4 and last digit is 1. So A = 2.

We know,

4 × 3 = 12 & 4 × 8 = 32

So D is either 3 or 8 Same case here, no such number is there where last digit is 3 and divisable by 4. So D = 8

By algebra we can write

4 (1000A + 100B + 10C +D) = 1000D +100C + 10B + A

Putting values of A and D we get

8000 + 400B + 40C + 32 = 8000 + 100C + 10B + 2

390B + 30 = 60C

13B + 1 = 2C

By solving the equation we get

B = 1 & C = 7

So the Answer is 2178