Solution for Alphabet Math

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      ABJ
    -----
ECA|FDBHJ
    CGG
    ---
    AGAH
    AAEA
    ----
     KDDJ
     KDBH
     ----
       AJ

1. B > A
   Because ECA x A has 3 digits and ECA x B has 4 digits.

2. J > A
   Because ECA x A has 3 digits and ECA x J has 4 digits.

3. H = 0
   Because J - H = J.

4. J = 2, 4, 6, or 8 and A = 5
   or
   A = 2, 4, 6, or 8 and J = 5
   Because A x J has a one's digit of H, and H = 0.

5. J = 5
   Because: A x A has a one's digit of G. If A was 5, A x A would have a one's digit of A. This plus #4 shows J = 5.

6. A = 2 or 4
   From #2, #4, and #5.

7. A = 2 and B = 6
   For now, this is an educated guess.
   Because: It fits with #1. Plus we know A x B has a one's digit of A and we know that ECA x A is a 3 digit number. So I'll guess the smaller option for A (from #6, 2 rather than 4). This gives A = 2 and B = 6. 2 x 6 = 12, which fits with A x B having a one's digit of A.

8. E = 3
   If #7 is correct we have: ECA x B = AAEA or EC2 x 6 = 22E2.
   The division of 22E2 by 6 has a first digit of 3 (22/6) therefore E = 3.

9. C = 7
   If #7 and #8 are correct: ECA x B = AAEA or 3C2 x 6 = 2232. Therefore C is equal to 7. With these values the assumptions made in #7 and #8 are validated.

10. G = 4
    Because ECA x A = CGG or 372 x 2 = 744.

11. K = 1 and D = 8
    Because ECA x J = KDBH or 372 x 5 = 1860.

12. F = 9
    Because CGG + AGA = FDB or 744 + 242 = 986

which results in:

      265
    -----
372|98605
    744
    ---
    2420
    2232
    ----
     1885
     1860
     ----
       25

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