Mike Lindell Media Corp (MLMC)

[prof]

fibinacci
You were so close, and then you stopped.
You were correct that A must be 0,1, or 2 - otherwise the answer would have 5 digits ( EDCBA), and not 4 (DCBA).
0 does not work - one sees that quickly. If A is 1, then the answer must end in 1 - and we can find no multiple of 4 which ends in 1, so 1 cannot work. So A must be 2, making D 8.
On to B abd C.
B must also be a small number - 2 is already taken. Let's try 1. The B in the answer also has to be 1. Remember! We are carrying 3 from the 4x8=32. So, 4xB + 3 = a number ending in 1. so 4x2 or 4x7 would work. We already have used 2, so let's try 7. It works very nicely, combining 1 with 7.

4 x ABCD = DCBA

4 x 2178 = 8712

Best to you,
prof