Problem 1:
(I) A + B = CD
+ + +
(II) CE + F = CF
---------------
(III) CA + CC = DE
(IV) (V) (VI)
Problem 2:
(I) A + B = CD
+ + +
(II) CE + F = CF
---------------
(III) CG + CC = HE
(IV) (V) (VI)
Solution 1:
(V) C = 1
(II.1) E = 0
(III.1) A = 9
(III) D = 3
(I.1) B = 4
(VI.1) F = 7
9 + 4 = 13
+ + +
10 + 7 = 17
---------------
19 + 11 = 30
Solution 2:
(V) C = 1
(II.1) E = 0
(IV.1) A = G
(III.1) A = 9
(III) H = 3
(I.1) B = D + 1
(VI.1) D + F = 10
So assuming that different letters may be the same digit
we get solutions for D in {1,2,3,4,5,6,7,8}.
References:
J. Petigk, Mathematik in der Freizeit, 1998, p15