Verified answer

refer to

http://math-kali.blogspot.com/search?updated-min=2...

for soultion

ABCD x 4 = DCBA

When A is 2 then D must be 8 because 2 x 4 = 8

then

2BC8 x 4 = 8CB2

8 x 4 = 32 last digit already is 2 so it complies

then as we know how manual multiplication is normally done next digit which is B will be last digit of summ of 3 and 4C

so B = (3 + 4C) % 10 // 3 as first digit of 32

C must be less than 10 not to change first digit D=8

C = whole part of ((3 + 4C) / 10) + 4B

from this B must be less than 3. It can not be 2 because A alreadu is 2 the

the only option is that B is 1

then

C = whole part of ((3 + 4C) / 10) + 4

now lets try possible values 3, 5, 6, 7, 9 // because 1,2,4,8 are already taken

the equision complies only for value of 7

so C = 7

so

2178 x 4 = 8712

2178x4=8712

No formula, you have to do it logically.

D has to be bigger than A. A is even.

Dx4 ends in A, so D can't be 1 (then A=4),

D can't be 2, then A=8, D can't be 3...

Also if ABCDE x 4 = EDCBA

then 21978 x 4 =87912

2178*4 = 8712

You have that A has to either be 1 or 2.

Because anything times 4 is even, you have that A must be 2.

So now your problem is:

2BCD*4=DCB2

I'll add more help in a second.

EDIT: Someone beat me to it.

If A, B, C, and D are all variables then the only possible solution is that they multiply to give you 0.