(I) ABCD + EFGG = HEBG
: - +
(II) CI + EHKI = EFCH
----------------------
(III) CKB * GH = IKFD
(IV) (V) (VI)
Solution:
(I.1) D = 0
(V.2) K = 9
(III.1) B * H == 0 (mod 10) => 5 is in {B, H}
(IV.1) B * I == 0 (mod 10) => 5 is in {B, I}
Therefore B = 5 and H and I are even.
(II.1) H == 2*I (mod 10)
(V.1) G == H + I == 3*I (mod 10)
(III) C * G <= I => G < I
(VI.4) H + E <= I => H < I
Now try the possible even values for I.
I H==2I G==3I
2 4 6
4 8 2
6 2 8
8 6 4
Only for I=8 we have H