(I)     AB  *   A  = CCD
         -      *      +
(II)    AE  :   F  =   G
        ----------------
(III)    B  *  CF  = CHE
       (IV)    (V)   (VI)


Solution:
(IV.1)  E = 0
(II.1)  5 is in {F, G}
(III.1) 5 is in {F, B}
Therefore F = 5.
(V)     A is odd, but not 1.
(II)    G is even and A = G/2 <= 8/2 = 4.
Therefore A = 3. Then G = 6.
(VI.1)  D + G = 10  => D = 4
(V)     C = 1
(VI.2)  H = C + 1   => H = 2
(III)   B = 8

        38  *   3  = 114
         -      *      +
        30  :   5  =   6
       ------------------
         8  *  15  = 120 



References:

- Johannes Lehmann;
  Mathe mit Pfiff,
  Manz Verlag 1977, ISBN 3-7863-0383-5
  Chap: Kryptarithmetik p71-72
  Problem 25

- Johannes Lehmann;
  Mathe mit Herz,
  Urania Verlag 1991, ISBN 3-332-00479-4, new edition of "Mathe mit Pfiff"
  Chap: Kryptarithmetik p81-83
  Problem 25