It's ABCD

We have 4* abcd = dcba . Since 4 times abcd is again a 4 digit number, so abcd is less than 2500. So a is either 2 or 1. But a cannot be 1 since then dcba will not be divisible by 4. So, a = 2. Now d can be 8 or 9. But if d is 9 then a must be 6. So d = 8. Now trivially, 4c + 3 has last digit b. Let 4c + 3 = b + 10k, where k is the carry-over value to the 3rd place of abcd. A little logic shows that 4b + k = c. Hence, 16b + 4k + 3 = b + 10k or, 15b + 3 = 6k or, 5b + 1 = 2k ------------------------(1) k can either be 0,1,2 or 3. Since b is an integer, k = 3. Hence b = 1 and c = 7. So the number is 2178.

First observation: $a$ must be small, otherwise the product would have five digits. Particularly, $a = 1, 2$ . But since $a$ as the last digit of the product must be even, we are left with $a=2:\ \ \ 4\times2bcd = dcb2.$ Next, the product should be at least 8000: $d = 8, 9$ . But since the last digit of the product is 2, we must have $d = 8:\ \ \ 4\times 2bc 8 = 8cb 2.$ The carry over from the last digit is 3, which will be added to $4c$ . Therefore the second-last digit in the product must be odd. On the other hand, if (\c > 2), the first digit will become greater than 8. Thus $b = 1:\ \ \ 4\times 21c8 = 8c12.$ Finally, $4\cdot 18 = 72$ , so that $c = 7:\ \ \ 4\times 2718 = 8172.$