Ashutosh's Number Decipherence Streak

It can easily be observed that a and b can only be 1 or 2 (Or else the number becomes 5 digit).solving some equations we get 3999a+390b=60c+996d. Now for a=2 and b=1 we have 8388=60c+996d which has a solution for c=7 and d=8 (as a=2 ,d can only be 3 or 8) which gives us the number 2178.

since ABCDx4 = DCBA, the number ABCD <2500.. Now A=1 or A=2, but since DCBA is even A has to be even so A=2.. Therefor D=8.. So till now the number is 2XX8.. Now B cannot be greater than 2 since there wont be any carry to A.. So B=0,1,2.. So the BA in DCBA will be 02 or 12 or 22, but since DCBA is a multiple of 4 so 02 and 22 is not possible.. So B=1.. So till now the number is 21X8.. Now 8x4 will generate a carry of 3.. Let Xx4+3 = 10Y+1 since B=1 and Y is the carry.. Similarly, (1*4)+Y=X since C=X.. Solving, we get Y=3 and X=7.. So the required number is 2178.