(I) ABBC : DE = FGD
- + *
(II) FBHI - FGKD = FD
----------------------
(III) EHGI - FBCG = CCEK
(IV) (V) (VI)
Solution:
(IV.3) H = 0 or H = 9
(II.3) B = G + 1
(II.2) 10 + H = K + F + c with c in {0,1}.
Now H cannot be 9 otherwise K=F=9.
So H=0 from (IV.3).
(V.2) C < D
(VI) F*F <= C
(I) F*D <= A
So F*(F*F+1) <= F*D <= A <= 9.
Therefore only F=1 is possible.
(VI) C < (F+1)^2 = 4
(IV.1) C == 2*I (modulo 10)
As C is even we get C = 2.
(III.4) E = F + C + 1 = 4
(IV.4) A = E + F = 5
(II.1) I == 2*D (modulo 10)
As I is even we have I=6 from (IV.1).
Further D=3 as D