DONALD+GERALD=ROBERT

2) DONALD + GERALD = ROBERT
Given 'D=5' (If not given assume D=5 at initial stage)


   6    5    4    3   2    1
   D   O   N   A   L   D
+ G   E   R   A   L   D
   c1  c2  c3  c4  c5
-----------------------------
   R   O   B   E   R   T

1. 'D=5' is assumed, so 'D+D=T' therefore 'T=0' & 'c5=1'.

2. In column 5 'O+E=O' as 'T=0' so E cannot be 0, therefore 'E=9'.

    'O+9=O' is possible if 'c2=1'. Therefore 'c2=1'.

3. In column 3 'A+A=9', but addition of any 2 same number is

    always even, given that addition is 9 which is possible when

    there is carry. Therefore 'c4=1', so 'A=4'.

4. Remaining numbers to be assigned are {1,2,3,6,7,8} to {O,N,R,B,L,G}.

5. We have 'E=9' & 'c2=1' so from column 5 we get 'c1=1'. Also from

    column2 we have 'L+L+c5=R' where 'c5=1' therefore R is odd so R can
    be 1or3or7. As 'D+G' does not generate carry shown in column 6 so R
    cannot be 1or3. Therefore 'R=7' & 'G=1'.

6. We have 'R=7' so from column 2 we have 'L+L+1=17', therefore 'L=8'.

7. From column 3 we get that 'A+A+c4=E' and so there is no carry,

    therefore 'c3=0'.

8. From column 4 we get 'N+R+c3=B' we have R=7 & 'c3=0',
    so 'N+7=10+B', therefore 'N=B+3'. {2,3,6} are remaining to be
    assigned so to satisfy the constraint 'N=B+3' we get 'B=3' & 'N=6'.

9. And remaining 'O=2'.



SOLUTION:

   5   2   6   4   8   5
+ 1   9   7   4   8   5
------------------------
   7   2   3   9   7   0

VALUES:
D=5
O=2
N=6
A=4
L=8
G=1
E=9
R=7
B=3
T=0