I am assuming ABCD is a 4-digit number and DCBA is the number reversed and ABCD does not stand for A \times B \times C \times D .

What we have essentially is to find 0 \leq A, B, C, D \leq 9 such that

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

This essentially boils down to

A \times 3999 + B \times 390 = C \times 60 + D \times 996 .

If we concentrate on the unit's place, the values of B and C become irrelevant, we have to find A and D such that the number in units place of A \times 9 is the same as the number in the units place of D \times 6 . It is also clear from the first equation that D \geq A \times 4 . This gives us A = 2 and D = 8 .

Substituting these values in the equation above, we get

30 + B \times 390 = C \times 60 \implies 1 + B \times 13 = C \times 2 \implies C = 7, B = 1 .

Therefore, A = 2, B = 1, C = 7, D = 8 and 2178 \times 4 = 8712 .