Puzzle: DONALD + GERALD = ROBERT
Each letter represents a distinct digit. What is the mapping
Solution (not necessarily unique):
c1 c2 c3 c4 c5 carries
D O N A L D
G E R A L D
----------------
R O B E R T
Steps:
1. T is even ( D + D)
2. D >= 5 <=> c5 == 1
D <= 4 <=> c5 == 0
3. L >= 5 <=> c4 == 1
L <= 4 <=> c4 == 0
4. A >= 5 <=> c3 == 1
A <= 4 <=> c3 == 0
5. c5 == 1 <=> R is odd
c5 == 0 <=> R is even ( L + L)
R not 0 (because R and L cannot both be 0)
6. c4 == 1 <=> E is odd
c4 == 0 <=> E is even ( A + A)
E not 0 (because E and A cannot both be 0)
7. c2 == 0 <=> E == 0
c2 == 1 <=> E == 9 and c3 + N + R > 9
8. c1 + D + G <= 9 (no carry)
R <= D
R <= G
9. By (6) and (7), E must odd and hence 9.
By (7) it is either 0 or 9, but if it is 0, then it is even
and by (6) it cannot be even and 0.
==> c4 = 1, E = 9, c2 = 1
O is the last variable to be solved for - it can be anything
10. By (9): E == 9, c2 == 1
==> c1 = 1
11. By (9) and (8),
==> D + G <= 8
12. By (9): c4 == 1
==> 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 == 6 <=> G == 0
R == 7 <=> G == 1
R == 8 <=> G == 2
R == 9 <=> G == 3
17. Unavailable values:
By (14): 0
By (9): 9
R == 7 <=> G == 1
R == 8 <=> G == 2 (invalidated by additional clue that "R is odd")
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
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Final assignment:
0 1 2 3 4 5 6 7 8 9
T G O B A D N R L E
A B D E G L N O R T
4 3 5 9 1 8 6 2 7 0