(I) ATU + IAS = IITE - - : (II) NEG : IOG = E ------------------ (III) PAU - NS = PPA (IV) (V) (VI) Solution: (IV.1) G = 0 (I) I = 1 (I.3) A + I + c = 11*I with c<=1 => A + c = 10 => A=9, c=1 (VI.1) E == 9 E (mod 10) => 2E == 0 (mod 5) => E == 0 (mod 5) But E=0 in (VI) is impossible. Therefore E=5. Now divide in (VI) to get the first digit of the quotient. (VI) 5 P <= 11 < 5(P+1) => P = 2 Then only T is left undetermined in (VI). (VI) T = 4 (IV.3) P + N + 1 = A (there must be a carry) This determines N = 6. (V.2) N + O = A => O = 3 Now all but S and U are found. (I) U + S = 15 (III) U - S = -1 This linear system has the solution U=7, S=8. Note we did not used the assumption that each digit is encoded by a single letter. 947 + 198 = 1145 - - : 650 : 130 = 5 ---------------- 297 - 68 = 229 Ordering the letters by their encoded digit we get 0123456789 GIPOTENUSA The mathematical term in question is hypotenuse. References: Boris A. Kordemsky; Köpfchen muss man haben, Aulis Verlag Deubner Köln, 1975, ISBN 3-7614-0315-1 Problem 241.1 with explained solution W. N. Bolchowitinow, B. I. Koltowoi, I. K. Lagowski; Spass für freie Stunden - Rätsel Spiele Denkaufgaben, Verlag für die Frau, Leipzig, 1980, 2. Auflage Problem 195