(I) HJ + AD = DF
+ + +
(II) BG + EC = DF
------------------
(III) AC + GE = HEK
(IV) (V) (VI)
Solution:
From (III) or (VI) follows H=1.
The only variable which can be zero is K.
As there are ten variables one must be zero. So K = 0.
Looking at the last digits of (VI) we get F=5.
Now collect the other equations of the last digit
(I) J + D == 5
(II) G + C == 5
(III) C + E == 0
(IV) J + G == C
(V) D + C == E
where a == b means a = b (modulo 10).
This system is underdetermined as
(I) + (II) - (III) = (IV) + (V).
But (VI) gives a further equation
20 D + 10 = 100 + 10 E that is
(VII) 2 D + 1 == E
Now combine (VII) - 2(V) + 2(III)
3 E + 1 == 0
which gives E=3. Substitute in the congruence
(III) gives C=7, then (II) gives G=8,
then (V) gives D=6 and (I) gives J=9.
Note that (VII) is not usefull to determine D.
The last variables are easily found.
19 + 46 = 65
+ + +
28 + 37 = 65
------------------
47 + 83 = 130
References:
Hg. Richard Zehl;
Denken mit Spass, Die besten trend-Nüsse
Orac Pietsch, Wien 1981, ISBN 3-85368-893-4
- Addition von Gleichungen, p52