(I) AB * CDC = BABE
* + -
(II) AFD + GHJE = GGDC
----------------------
(III) HFKH + GJEF = KDAH
(IV) (V) (VI)
Solution:
(III.1) F = 0
(III.3) J + 1 = D there must be a carry
(III.2) K + E = A + 10
(III.4) H + G = K
(V.1) C + E = 10
(V.2) D + J + 1 == E
(II.1) D + E = C + 10
(II.3) A + H = G
(VI.1) C + H == E
D
J == D - 1
E == 2D
C == -2D
H == 4D
Now A = G = K (modulo 2). As C, E, F, H are even
A, G, K must be odd otherwise we had too much even ones.
Now D and J are an even-odd pair so the last number left B
is odd.
As A*A < 9 the only odd number left is A = 1.
Then H = 2 as H < 2*2 and must be even.
Now G = A+H = 3 and K = H+G = 5.
Then (IV) looks like 1B * 10D = 2052. The odd digits
left for B are 7 and 9. Only B = 9 is possible.
As we get D = 8 we can compute the missing digits.
19 * 484 = 9196
* + -
108 + 3276 = 3384
----------------------
2052 + 3760 = 5812
References:
Hoer Zu, No 10, 1999-03