(I)      ABC *   DC = EAEF
          *      +      -
(II)      BF + AEGG = AEBF
        ------------------
(III)   FFEE + AEDC = DBHG
        (IV)    (V)   (VI)


Solution:
(II.1)  G = 0
Now the equations (II) and (V) are useless.
The most influential variable is C.
(I.1)   F == C * C
(IV.1)  E == C * F == C*C*C
(III.1) C + E == 0               (*)
(III)   F + A <= D
(VI)    D + A <= E
(III)+(VI) F + 2A <= E
Therefore  F < E.                (**)
Now try all possible C and check these equations
  C   F   E  C+E   (*)   (**)
  0   0   0   0     ok   fail
  1   1   1   2    fail  fail
  2   4   8   0     ok    ok
  3   9   7   0     ok   fail
  4   6   4   8    fail  fail
  5   5   5   0     ok   fail
  6   6   6   2    fail  fail
  7   9   3   0     ok   fail
  8   4   2   0     ok   fail
  9   1   9   2    fail   ok
So only C=2 gives no contradition with (*) or (**).
(IV)    A * B <= F = 4
Therefore we have {A,B} = {1,3} as 2 and 4 are used.
Another way to determine B is to use the last two digits of (IV).
(IV.2)  E == (C+F)*B  (modulo 10)  =>   3 == B  (modulo 5).
Now (IV) is encoded as 132 * 34 = 4488.
Furthermore we get (I) as 132 * 62 = 8184.


    132 *   62 = 8184
     *      +      -
     34 + 1800 = 1834
   ------------------
   4488 + 1862 = 6350



References:

- Torsten Sillke;
  2003-11