(I)    ABC :  CD  =   EF 
        -     *       +
(II)   FDF +  EG  =  AHD
       -----------------
(III)   BB + FGA  =  AIJ
       (IV)   (V)    (VI)


Solution:
(IV.3)   A = F + 1   (with carry)
(IV.2)  10 = D + 1   (with carry)
Therefore we get D = 9.
(VI.1)   J = F - 1   as F != 0
(III.1)  B = J - A + 10 = 8
(I.1)    C == 9 F (mod 10)  =>  C + F == 10 F == 0 (mod 10) 
That is C = 10 - F
(IV.1)   8 + F = C + 10  as C != 9 = D.
Substitute C and we get F = 6 and back substitution gives
C = 4, J = 5, A = 7.
(II.1)   G = 9 - F = 3
(III)    I = 2
(I)      E = 1
(II)     H = 0


       784 :  49  =   16 
        -     *       +
       696 +  13  =  709
       -----------------
        88 + 637  =  725


References:

Bernhard Berchtold:
   Puzzle 9: Kryptogramm
   Januar 1999
   http://www.mathematik.ch/puzzle/             Puzzle collection
   http://www.mathematik.ch/puzzle/puzzle9.php  Puzzle
   http://www.mathematik.ch/puzzle/solut9.php   Solution