Wednesday, September 5, 2018

EXAMPLE OF CRYPTARITHMETIC PROBLEM


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

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