(I) ABC * DC = EAEF
* + -
(II) BF + AEGG = AEBF
------------------
(III) FFEE + AEDC = DBHG
(IV) (V) (VI)
Solution:
(II.1) G = 0
Now the equations (II) and (V) are useless.
The most influential variable is C.
(I.1) F == C * C
(IV.1) E == C * F == C*C*C
(III.1) C + E == 0 (*)
(III) F + A <= D
(VI) D + A <= E
(III)+(VI) F + 2A <= E
Therefore F < E. (**)
Now try all possible C and check these equations
C F E C+E (*) (**)
0 0 0 0 ok fail
1 1 1 2 fail fail
2 4 8 0 ok ok
3 9 7 0 ok fail
4 6 4 8 fail fail
5 5 5 0 ok fail
6 6 6 2 fail fail
7 9 3 0 ok fail
8 4 2 0 ok fail
9 1 9 2 fail ok
So only C=2 gives no contradition with (*) or (**).
(IV) A * B <= F = 4
Therefore we have {A,B} = {1,3} as 2 and 4 are used.
Another way to determine B is to use the last two digits of (IV).
(IV.2) E == (C+F)*B (modulo 10) => 3 == B (modulo 5).
Now (IV) is encoded as 132 * 34 = 4488.
Furthermore we get (I) as 132 * 62 = 8184.
132 * 62 = 8184
* + -
34 + 1800 = 1834
------------------
4488 + 1862 = 6350
References:
- Torsten Sillke;
2003-11