(I) AB * C = CD - + : (II) E : B = F ---------------- (III) F + G = H (IV) (V) (VI) Solution: The letters can be reconstructed in a simple way. (IV) A = 1 (I) B * C = D (II) B * F = E (VI) C < F Now B,C,F are not zero. The numbers {2,3,4} must be {B,C,F}. Otherwise we had a product which is at least 10. Only the assignment B=2, C=3, F=4 is valid. Thus the problem is solved. 12 * 3 = 36 - + : 8 : 2 = 4 ---------------- 4 + 5 = 9 References: Walter Lietzmann; Lustiges und Merkw\"urdiges von Zahlen und Formen, Verlag Vandenhoeck & Ruprecht, G\"ottingen 1955, 8. Auflage, ISBN 3-525-39112-9 Chap II.5: Vergilbte Manuskripte (p128-131, 21 problems) Chap II.13: Anagramme, Kryptogramme, Geheimschriften und dergleichen (p191-192, 10 problems)