(I)     ABC  *    DE  =  CFGH
         +        *        -
(II)   JDHJ  +   DGC  =  JGKK
       ----------------------
(III)  JEDK  +  EBAH  =  FAGH
       (IV)      (V)     (VI)


Solution:
From (III.1) or (VI.1) follows K=0.
Now collect the other equations of the last digits
and other digits at the positions, where the state of the carry is known
(II.1)   J + C == 0
(II.2)   H + G + 1 == 0   with carry
(II.3)  2D + 1 == G       with carry
(III.2)  D + A == G
(IV.2)   B + H + 1 == D   with carry
(VI.3)   A + G == F
(IV.3)   A + D + 1 == E   with carry, as no carry in (III.2)
where a == b means a = b (modulo 10).
This system is underdetermined. Solve the last 6 equation depending on D.
  G ==  2D + 1   from (II.3)
  A ==   D + 1   with (III.2)
  H == -2D - 2   with (II.2)
  B ==  3D + 1   with (IV.2)
  F ==  3D + 2   with (VI.3)
  E ==  2D + 2   with (IV.3)
As E=B <=> D=1 we know that D >= 2.
Otherwise the first digit of (I) gives A*D = (D+1)*D <= C < 10
so only D = 2 is possible. Therefore A=3, H=4, G=5, E=6, B=7, and F=8.
The digits left for J and C are 1 and 9.

       379  *    26  =  9854
        +        *        -
      1241  +   259  =  1500
      ----------------------
      1620  +  6734  =  8354


References:

Hoer Zu, No 38, 2001-09-14