Mike Lindell Media Corp (MLMC)

[fibinacci]

I suppose the most obvious answer would be:

A=0
B=0
C=0
D=0

0000 x 4 = 0000

Somehow I doubt this is the answer you're looking for. However, assuming A, B, C, and D are positive integers 0 thru 9, then I see no other solution.

A must be 0, 1, or 2. Otherwise there will be a carry.
This means 4xD = 0, 1, or 2 plus an optional carry.
Since multiples of 4 are always even, you are left with D = 0 or 2 plus an optional carry. Since 4 is already greater than 0 or 2, you must have a carry. This leaves D = 0, 3, 5, or 8.
Chosing 3 or 8 leads to a carry of 1 or 3 respectively. This means 4C + carry = B. Since A is 2 in this case, 4xB can not have a carry otherwise 4 x A /= D. Therefore, B must be 0, 1, or 2. If B is 0, then C = carry/4 = 1/4 or 3/4. If B is 1, then C = 0 or C = -1/2. If B is 2, then C = 1/4 or -1/4. Since all of these are fractions, the problem falls apart. Going back to D... we can't select 5 because the largest carry provided by 4xB is 3 (4x9 = 6 carry 3). With A being 0, 4xA + carry /= D. So what does selecting D = 0 do? This leads us to ABCD = 0BC0. 4xB can not have a carry, so B is 0, 1, or 2. This means C = 0, 3, 5, or 8. 4x8 = 2 carry 3, so B = 2. But 4x2 + 3 leads to 1 carry 1, so that doesn't work. 4x5 = 0 carry 2. But 4x0 + 2 /= 5. 4x3 = 2 carry 1, but 4x2 + 1 /= 3. So C must be 0 and B must be 0. So the only plausible answer is 0000.