(I)     AB  *   B  = CDD
         :      *      :
(II)    CB  :   B  =   E
        ----------------
(III)    E  *   D  =  BD
       (IV)    (V)   (VI)

Solution:
(II)     E != 1
(V)      0 < B <= 3
(II.1)   B == B*E (mod 5)  =>  1 == E (mod 5)
Therefore E=6. Then
(II)     B == 6B (mod 10)  =>  0 == 5 B (mod 10)
That is B is even and we get B=2.
(V)      D=4
(II)     C=1
(IV)     A=7


  72  *   2  = 144
   :      *      :
  12  :   2  =   6
  ----------------
   6  *   4  =  24



References:

Walter Lietzmann;
  Lustiges und Merkw\"urdiges von Zahlen und Formen,
  Verlag Vandenhoeck & Ruprecht,
  G\"ottingen 1955, 8. Auflage, ISBN 3-525-39112-9
  Chap II.5: Vergilbte Manuskripte (p128-131, 21 problems)
  Chap II.13: Anagramme, Kryptogramme, Geheimschriften und dergleichen
  (p191-192, 10 problems)