(I) AB * BC = DAD
* : *
(II) EC : F = F
-----------------
(III) BCD * G = BBEA
(IV) (V) (VI)
Solution:
(II) F >= 4
(II.1) F != 5 => F != 0 (mod 5)
(II.1) F * F == C (mod 5)
(V.1) F * G == C (mod 5)
Therefore F*F == F*G (mod 5). As F != 0 (mod 5) we can cancel F
so we get F == G (mod 5).
(IV) A * E <= B => E < B
So the product (II) EC is larger than the product (V) BC.
So G = F + 5. As G <= 9 we get F <= 4.
Therefore F = 4 and G = 9.
(II) C = 6 and E = 1.
(V) B = 3
(I.1) D = 8
(III.1) A = 2
23 * 36 = 828
* : *
16 : 4 = 4
-----------------
368 * 9 = 3312
References:
Jürgen Köller;
Cryptogramme,
www.mathematische-basteleien.de