Answer:
Pulling apart the structure above,
?A? x {2-9} = ??AA
?A? x {2-9} = ??A
?A? x A = ?A??
?A? x {2-9} = ????
Where {2-9} is a single digit (can't be 0 or 1).
A is also not 1 or 0 or whatever number is used for {2-9}.
The first term is the most interesting, if we just look at the
tens and ones terms we end up with: A? x {2-9} = AA
I found four cases where this is true:
37 x 9 = 333 (A = 3 and, ?A? = ?37, etc)
43 x 8 = 344 (A = 4 and, ?A? = ?43, etc)
48 x 3 = 144 (A = 4 and, ?A? = ?48, etc)
84 x 7 = 588 (A = 8 and, ?A? = ?84, etc)
So A is 3, 4, or 8.
There must be a number {2-9} such that ?AY = ??A (Y known),
and ALSO multiplying ?A? x A results in a four digit number
so that number {2-9} must be less than A.
?X? can not be ?37 (9 x 7 is 63 but 9 is not less than 3).
?X? can not be ?43 (either 2 nor 3 times ?43 can be ??4).
?X? can be ?48 (3 times ?48 can be ??4).
?X? can be ?84 (2 and 7 times ?84 can be ??8).
So A is 4, or 8.
In order for ?48 to work, there needs to be a number N such that
(N48 x 4 = ?4??) and N48 x 3 = ??4 ... and there is no such number.
So A is 8.
N84 x 8 = ?8??
N=4 are N=9 are the only numbers which make this work.
N84 x (2 or 7) = ??8 (2 or 7 because of the ending 8)
In order to have 3 digits N must be 4.
?A? x A = ?A?? is actually: 484 x 8 = 3872
?A? x ? = ??A is actually: 484 x 2 = 968
?A? x ? = ???? is actually: 484 x (3,4,5,6,9) (because it can't have any 8s).
So,
?A? x ??A? = ????A?? (the entire problem, is...)
484 x 728N = ????8?? (with N = 3,4,5,6, or 9)
N = 9 because that's the only number with an 8 there.
So the entire problem is: 484 x 7289 = 3527876
