(I) ABC : C = DE
- * +
(II) AE + AF = DD
----------------------
(III) BGA - FHC = IJ
(IV) (V) (VI)
Solution:
First of all I show that there is no standard solution.
That means no leading zero and different letters represent
different digits. Assume there were a solution.
Then F>=1 from (III) and further B>=2 from (III).
Now A>=3 from (IV). With (II) we see D>=2A>=6.
But now (VI) shows I >= 2D >= 12 which is too big for a
digit in a decimal number system.
So we are looking for a non standard solution.
When I say A==B, that means A=B (mod 10).
(IV.1) E + A == C
(III.1) J + C == A
(IV.1) + (III.1) E + J == 0
(VI.1) E + D == J
(IV.1) + (III.1) + (VI.1) 2E + D == 0 => D is even
(II.1) E + F == D
(IV.1) + (III.1) + (VI.1) + (II.1) 3E + F == 0
(III) B >= F
(IV) A >= B
(II) D >= A
D >= A >= B >= F
(II.2) 2A = D no carry as D>=F
(VI) D <= 4 and therefore A <= 2.
There are three cases for A.
Case A=0, D=0: Then E=F=0 from (II) and I=J=0 from (VI).
Then all others letters get zero from (III).
Case A=1, D=2: Then E is 4 or 9 from 2E + D == 0.
But this contradicts (II).
Case A=2, D=4: Then E is 3 or 8 from 2E + D == 0.
But this E=8 contradicts (II).
The case E=3 gives the only solution.
215 : 5 = 43
- * +
23 + 21 = 44
----------------------
192 - 105 = 87
References:
J. Petigk,
Mathematik in der Freizeit,
Aulis Verlag Deubner & Co, Köln, 1998,
p151