Solution

We begin by noting the range of values that can be taken by some of the alphabets.

Clearly, F is not equal to 1, and is at least 2. Moreover, the product is a five-digit number, i.e. smaller than 100000. So A must be smaller than 5.

Now we can try A = 1, 2, 3, 4. We find that some lead to contradictions, leaving only one set of solutions.

(1)       A = 1

Now the unit digit of E  F is A, i.e. 1. The only possibilities are F = 3, E = 7 or F = 7, E = 3. That is, 1???7  3 = 7????1 or 1???3  7 = 3???1, both of which are not possible.

(2)       A = 2

Then F is either 3 or 4.

Suppose F = 3. Since the unit digit of E  F is A, E must be 4. Then 2???4  3 = 4???2, which is not possible.

Thus F = 4. Then E must be 8. Now ABCDE < 90000  4 = 22500, so B is either 0 or 1.But EDCBA is divisible by 4. Now ???02 is not divisible by 4. So B is 1. The carry from the unit digits to the tens is 3. Thus the unit digit of D  F is 8. It follows that D = 2 or 7. But A = 2, so D = 7. Finally, it can be deduced that C = 9. This completes the solution.

(3)    A = 3

Then F must be 2 or 3. But A = 3, so F must be 2. The unit digit of E F is A. But then E F = E 2 is even, A is odd, a contradiction.

(4)    A = 4

Then F must be 2. Since the unit digit of E  F is A, i.e. 4, E must be 2 or 7. But F = 2, so E = 7. Then 4???7  2 = 7???4, which is not possible.

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