(I) ABC * DE = CFGH
+ * -
(II) JDHJ + DGC = JGKK
----------------------
(III) JEDK + EBAH = FAGH
(IV) (V) (VI)
Solution:
From (III.1) or (VI.1) follows K=0.
Now collect the other equations of the last digits
and other digits at the positions, where the state of the carry is known
(II.1) J + C == 0
(II.2) H + G + 1 == 0 with carry
(II.3) 2D + 1 == G with carry
(III.2) D + A == G
(IV.2) B + H + 1 == D with carry
(VI.3) A + G == F
(IV.3) A + D + 1 == E with carry, as no carry in (III.2)
where a == b means a = b (modulo 10).
This system is underdetermined. Solve the last 6 equation depending on D.
G == 2D + 1 from (II.3)
A == D + 1 with (III.2)
H == -2D - 2 with (II.2)
B == 3D + 1 with (IV.2)
F == 3D + 2 with (VI.3)
E == 2D + 2 with (IV.3)
As E=B <=> D=1 we know that D >= 2.
Otherwise the first digit of (I) gives A*D = (D+1)*D <= C < 10
so only D = 2 is possible. Therefore A=3, H=4, G=5, E=6, B=7, and F=8.
The digits left for J and C are 1 and 9.
379 * 26 = 9854
+ * -
1241 + 259 = 1500
----------------------
1620 + 6734 = 8354
References:
Hoer Zu, No 38, 2001-09-14