wrong's and right's

WRONG
+WRONG
-----------------
R I G H T
Lets start with what we know:

....................WR0NG
..................+WR0NG
---------------------------------
...................R I G H T

1--The most we can have as a carryover, "co", is 1 since the highest sum is 9+9 = 18
2--G must be 1 as 0 + 0 = G can only result from the 1 carryover from N + N
3--T must be 2 as 1 + 1 = 2
4--N can only be 7, 8, or 9 as 3 or 4 won't leave the required carryover to make G = 1, 5 would make H = 0 which can't be, and 6 would make H = 2 which can't be.
5--W can only be 3 or 4 with a carryover from the previous R + R
6--R cannot be 6 or I = 2 which cannot be so it must be 7, 8, or 9 to produce the required carryover.

So where are we?
....................W..R..0..N..1
..................+W..R..0..N..1
---------------------------------------
....................R...I...1..H..2

7--It would appear that W can be 3 or 4 and R 7, 8, or 9
8--What combinations of R and W will make W + W + 1 = R
9--R = 7 and W = 3 works while W = 4 does not work. R = 8 and W = 3 or 4 does not work. R = 9 and W = 4 works. (Another way of looking at this is to say that R + R = 10 + I and W + W + 1 = R from which you get W = (10 + I - 2)/4. The only values of I, i.e., 4, 6, or 8, that yield viable results are 4 and 8 from R = 7 or 9)

So where are we now?
....................4...9..0..N..1 Case 1
..................+4...9..0..N..1
---------------------------------------
....................9...8..1..H..2
or
....................3...7..0..N..1 Case 2
..................+3...7..0..N..1
---------------------------------------
....................7..4...1..H..2

10--In Case 1, N cannot be 7, 8, or 9 so is eliminated.
11--In Case 2, N can be 8 or 9 so we have two possible answers.
 
Two wrongs don't make a right, but two Wrights make an airplane. :wink:

Eliz.
 
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