(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