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

SEND+MORE=MONEY

1) SEND + MORE = MONEY

    5   4    3    2    1
         S   E   N   D
+      M  O   R   E
        c3  c2  c1
----------------------
  M  O   N   E   Y

1. From Column 5, M=1, since it is only carry-over possible from sum of 2

    single digit number in column 4.

2. To produce a carry from column 4 to column 5 'S + M' is atleast 9 so

    'S=8or9' so 'S+M=9or10' & so 'O = 0 or 1'. But 'M=1', so 'O = 0'.

3. If there is carry from Column 3 to 4 then 'E=9' & so 'N=0'. But

    'O = 0' so there is no carry & 'S=9' & 'c3=0'.

4. If there is no carry from column 2 to 3 then 'E=N' which is

    impossible, therefore there is carry & 'N=E+1' & 'c2=1'.

5. If there is carry from column 1 to 2 then 'N+R=E mod 10' & 'N=E+1'

    so 'E+1+R=E mod 10', so 'R=9' but 'S=9', so there must be carry

    from column 1 to 2. Therefore 'c1=1' & 'R=8'.

6. To produce carry 'c1=1' from column 1 to 2, we must have 'D+E=10+Y'

    as Y cannot be 0/1 so D+E is atleast 12. As D is atmost 7 & E is

    atleast 5 (D cannot be 8 or 9 as it is already assigned). N is atmost 7

   & 'N=E+1' so 'E=5or6'.

7. If E were 6 & D+E atleast 12 then D would be 7, but 'N=E+1' & N would

    also be 7 which is impossible. Therefore 'E=5' & 'N=6'.

8. D+E is atleast 12 for that we get 'D=7' & 'Y=2'.


SOLUTION:

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

VALUES:
S=9
E=5
N=6
D=7
M=1
O=0
R=8
Y=2