I will try to solve this problem in least number of complex mathematical steps possible. I will not use any maths greater than grade 8.

ABCD*4 = DCBA implies A must be less than 3 since A is at 1000s place and 3*4 = 12, which makes the given number a five digit number.

Thus A can be 1 or 2.

Since DCBA is a multiple of 4 it implies that A must be even and hence A is 2.

A is 2 again implies D must be 8 or 9. (D cannot be less than 8 since it is at the 1000s place of DCBA, which is multiple of a four digit number of the form 2@#$%. Hence DCBA is greater than 8000).

Thus two possible values of D ar 8 or 9.

This is because D*4 = A and only for 8*4 = 32 so we get units place as 2.

Now A = 2 and D = 8.

Hence we have ABCD = 2BC8, where B and C have to be determined.

Since at the 1000s place we have 2 and 2*4 = 8, we can now ignore 1000s place of both number ABCD and DCBA.

Hence now the problem reduces to BC8*4 = CB2.

Going by similar logic for four digits we can assume that B is less than 3 since for B = 3, we will end up with a four digit number.

Hence B = 0, 1 or 2.

Since CB2 is a multiple of four which in turn implies that B2 must be divisible by 4 (Divisibility test of 4).

Thus possible combinations are 12, 32, 52, 72, 92. Of all these we have only 12 satisfying both conditions of divisibility of 4 and being less than 3.

Hence now our number is 1C8*4 = C12

Just using simple decimal expansion we have

400 + 40C + 32 = 100C + 12

Implies 60C = 420

Implies C = 7.

Thus the original number is 2871.