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• nutankhaire01

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Puzzle: DONALD + GERALD = ROBERT Each letter represents a distinct digit.

What is the mapping

Solution (not necessarily unique): c1 c2 c3 c4 c5 D O N A L D G E R A L D ---------------R O B E R T

carries

Steps: 1. T is even ( D + D) 2. D >= 5 c5 == 1 D = 5 c4 == 1 L = 5 c3 == 1 A 9

8. c1 + D + G D + G L >= 5 13. By (9): c4 == 1, E == 9 ==> A = 4 (A cannot be 9 because E is already 9) ==> c3 = 0

14. Three instance of R and of D. Make assumption. Aside: simplified version of this puzzle gives clue "R is odd". 14a. R is odd ==> c5 == 1, D >= 5 Choices sorted for this additional information. c5 == 1 : T == 0 D == 5 T == 2 D == 6 T == 4 D == 7 : unavailable: (13): D == 4 T == 6 D == 8 T == 8 D == 9 : unavailable: (9): E == 9 c5 == 0 (not possibilities with the additional clue): T == 2 D == 1 T == 4 D == 2 : unavailable: (13): D == 4 T == 6 D == 3 T == 8 D == 4 : unavailable: (13): D == 4 Iterate through the possbilities: ==> D=5, T=0, c5=1 15. By (14): D == 5 and (10): c1 == 1 ==> R == c1 + D + G == 1 + 5 + G ==> R == G + 6 16. By (15), R == R == R == R ==

6 7 8 9



G G G G

== == == ==

0 1 2 3

17. Unavailable values: By (14): 0 By (9): 9 R == 7 G == 1 R == 8 G == 2 is odd")

(invalidated by additional clue that "R

18. c3 + N + R == B with carry c2 By (9): c2 == 1 By (13): c3 == 0 mod10(N + R) == 1 rem10(N + R) == B 19. Assume two possible values for R from (17). First, assume R == 7, and hence G == 1. By (18), N >= 3. N == 3 B == 0 : unavailable value (14) N == 4 unavailable value (13) N == 5 unavailable value (14) N == 6 B == 3 N == 7 unavailable value (19) (value assumed here for R) N == 8 B == 5 : unavailable value (14) N == 9 unavailable value (9) ==> N = 6, B = 3 20. By (12): L >= 5, Unavailable values: 5, 6, 7, 9

==> L = 8 21. By unavailable values: ==> O = 2 Done. Two assumptions (14) and (19) that led to a consistent solution. Alternate branches not explored. ====================================================== Adding restrictions (sorted by variable): A: 13). 4 B: 19). 3 D: 14). 5 E:

7). 0 or 9 9). 9

G: 19). 1 L: 12). >= 5 20). 8 N: 19). 6 O: 21). 2 R: 5). not 0 19). 7 T: 14). 0 c1: 10). 1 c2: 9). 1 c3: 13). 0 c4: 9). 1 c5:

14). 1 ====================================================== Final assignment: 0 1 2 3 4 5 6 7 8 9 T G O B A D N R L E A 4

B 3

D 5

E 9

G 1

L 8

N 6

O 2

R 7

T 0