Help with math puzzle!!!?

Question

Help please. I haven't had a math class in 5 years and now I am being required to take K-8 mathematical concepts. Does anyone know how to solve this cryptarithm? I'm about ready to pull my hair out!!!

ABCD * 4 =DCBA

Answer 1

2178 * 4 = 8712.  This is in fact the only solution, but that takes a little work to prove.

Answer 2

I've seen a similar puzzle.

I'll try to show you the steps into solving it.

We start at A*4. Notice that the answer is still 4 digits. so A*4 must be either 4 or 8. That means A = 1 or 2.

Now look at D*4=A, How can we get 1 as the last digit in 4x table? We can't!. So A must be = 2! and since A*4=D, D=8

So far we have 2BC8*4 = 8CB2

Now B*4 cannot have any carryover. So the only choice is B=1.

Next, we know D*4=8*4=32 we have to carry over the 3.

What multiple of 4 in the 4xtable that let you add 3 to make an number that ends with a 1? 4*7=28 so C=7.

A=2, B=1, C=7, D=8

Answer 3

2178 * 4 = 8712

if

ABCDE X 4= EDCBA

ABCDE=?

Answer:  21978

Answer 4

A has to be <3 because 3*4 = 12 so RHS shall be 5 digit number.

A cannor be 1 as from RHS A has to be even

so A has to be 2

now B can be either 1 or 3 or 5 or 7 but B < 5 because 4*25 = 100 that is 5 digit.

So AB = 21 or 23

if AB = 23 DC >= 92

do D = 9

which is not possible as 4*8 is 2 ending but 4*9 is not

AB = 21

So D = 8

so the number = 4*(2108+10C) = 8032+100C

or 8432+40C = 8012+ 100C

60C = 420 C =7

so number = 2178*4 = 8712