(I)     ABCDE  -  FGEE  = AHBDE
           :        -        -
(II)       DI  *   HDB  =  BKDB
        -----------------------
(III)     CDE  +  FKCD  =  KDGD
         (IV)      (V)     (VI)


Solution:
(I.1)   E = 0
(VI)    A = 1
(VI.2)  G = 9  (as there is a carry)
(IV')   DI * CD = ABCD
(IV'.1) I * D = D (mod 5)
(II.1)  I * B = B (mod 5)
As not both B and D can be 5 we can divide by B or D
and get I = 1 (mod 5). As A=1 is used already we get
 I=6
Furthermore B and D are even.
(II)    D * H <= B
The smallest two not used numbers are 2 and 3.
Therefore 2*3 <= B but 6 is used. The only even number left is
 B=8.
(VI.1)  D = 2
(III.2) C = 7
(VI.3)  K = 5
(V.3)   H = 3
(III.4) F = 4


        18720  -  4900  = 13820
           :        -        -
           26  *   328  =  8528
        -----------------------
          720  +  4572  =  5292



References:

- Johannes Lehmann;
  Kurzweil durch Mathe,
  Urania Verlag 1988, ISBN 3-332-00239-2
  Mathematische Schülerbücherei MSB Nr. 77
  Chap: Heiterer Stundenplan, p87, 181
  Problem 10