(I) ABC : CD = EF
- * +
(II) FDF + EG = AHD
-----------------
(III) BB + FGA = AIJ
(IV) (V) (VI)
Solution:
(IV.3) A = F + 1 (with carry)
(IV.2) 10 = D + 1 (with carry)
Therefore we get D = 9.
(VI.1) J = F - 1 as F != 0
(III.1) B = J - A + 10 = 8
(I.1) C == 9 F (mod 10) => C + F == 10 F == 0 (mod 10)
That is C = 10 - F
(IV.1) 8 + F = C + 10 as C != 9 = D.
Substitute C and we get F = 6 and back substitution gives
C = 4, J = 5, A = 7.
(II.1) G = 9 - F = 3
(III) I = 2
(I) E = 1
(II) H = 0
784 : 49 = 16
- * +
696 + 13 = 709
-----------------
88 + 637 = 725
References:
Bernhard Berchtold:
Puzzle 9: Kryptogramm
Januar 1999
http://www.mathematik.ch/puzzle/ Puzzle collection
http://www.mathematik.ch/puzzle/puzzle9.php Puzzle
http://www.mathematik.ch/puzzle/solut9.php Solution