(I) ISE + EID = ALK
- + -
(II) RAS - KL = RUM
----------------
(III) ERR + MDR = LLM
(IV) (V) (VI)
Solution:
(VI.2) shows U = 0 or U = 9. Now no other letter can be zero.
As there are no leading digits I, E, A, R, K, M, L cannot be zero.
S is not zero from (IV.1). D is not zero from (I.1).
Therefore U = 0.
No other letter can be recogniced directly.
Assume R is known then
(III.1) M == 2R
(VI.1) K == 2M == 4R
(V.3) E == M - 1 == 2R - 1 (with carry)
(I.1) D == K - E == 2R + 1
(IV.1) S == E - R == R - 1
(II.1) L == S - M == -R - 1
(VI.3) A == L + R == -1 (no carry)
Therefore A = 9.
(II.2) K = A - 1 = 8.
As K == 4R (mod 10) we can conclude 3 = 4R (mod 5) that is 3 + R = 0 (mod 5).
Therefore R must be 2 or 7. As L > E from (III.3) only R = 2 is possible.
Than M = 4, E = 3, D = 5, S = 1, L = 7.
The last letter I = 6.
613 + 365 = 978
- + -
291 - 87 = 204
----------------
322 + 452 = 774
References:
H. Müller-Merbach;
The Role of Puzzles in Teaching Combinatorial Programming,
In: B. Roy (ed)., Combinatorial Programming: Methods and
Applications, 379-386, D. Reidel Publ. 1975
Zeit Magazin 19/1972