I assume no two characters are equal in any way.
If we use numbers, R must be even cause at the end S + S = R ( Odd + Odd and Even + Even both equal Even, so R's got to be Even )
First conclusion: R is even ( Say, 2, 4, 6, 8, since solution set doesn't consist of any 0 ( i.e. 1-9 only, possibly ) )
Now CROSS
+ROADS
------
DANGER
As you'd notice, D is extra, like something carried over in addition. So D's got to be 1 since any sum of 1-9 will can can give a carry of 1 alone.
So, A will have a value of [2,3,4,5,6,7,8,9] cause D's already taken 1.
First value found, D = 1
C+R = Set of numbers ( 12,13,14,15,16,17,18 ( 19 not possible as 10 is not allowed ) )
Which equal only to sums of:
If A: C+R C+R
Case 2:
3+9 || 9+3
4+8 || 8+4
5+7 || 7+5
( Other combo's not possible cause C!=R )
Case 3:
4+9 || 9+4
5+8 || 8+5
6+7 || 7+6
Case 4:
5+9 || 9+5
6+8 || 8+6
Case 5:
6+9 || 9+6
7+8 || 8+7
Case 6:
7+9 || 9+7
Case 7:
8+9 || 9+8
Case 8:
Not Possible as ( 9,9 ) is the only case and C!=R ( Assumed at the beginning of this puzzle solving )
Thus from the above equation, as R is even, we have R = Possiblity( 4, 6, 8 ) and C = Possiblity( 4, 5, 6, 7, 8, 9 )
So far,
A = ( 2, 3, 4, 5, 7 ) [6's totally Odd for values of R]
D = 1
C = ( 4, 5, 6, 7, 8, 9 )
R = ( 4, 6, 8 )
Now since R is Even and definitely in [4, 6, 8]
From R + O = N, we get R = N - O
Thus,
If R: N-O
Case 4:
9-5
7-3
6-2
( Other combo's not possible as in our assumption, we use R = 4 )
Case 6:
9-3
8-2
Case 8:
Not possible as 1's already taken by D ( 9-1 being the only possible way )
Thus from the above,
N = ( 6, 7, 8, 9 )
O = ( 2, 3, 5 )
A = ( 2, 3, 4, 5 )
D = ( 1 )
C = ( 4, 5, 6, 7, 8, 9 )
R = ( 4, 6 )
Also, A doesn't have 7 since C + R = A but R has only ( 4, 6 ) while C is ( 4, 5, 6, 7, 8, 9 ) ( Adding C+R will never yeild 17 )
Now similarly, G = O + A
Thus, O = G - A
If O: G-A
Case 2:
7-5
6-4
5-3
Case 3:
8-5
7-4
5-2
( 6-3 not Possible as O!=A, thus I G can't be 6 and A can't be 3 )
Case 5:
9-4
8-3
7-2
This gives us two conclusions,
First, A is now ( 2, 4, 5 )
Second, G is ( 5, 7, 8, 9 )
Phew .. I need a break.
Continuing,
We now have:
A = ( 2, 4, 5 )
D = ( 1 )
C = ( 4, 5, 6, 7, 8, 9 )
R = ( 4, 6 )
O = ( 2, 3, 5 )
N = ( 6, 7, 8, 9 )
G = ( 5, 7, 8, 9 )
Good so far? I hope so, else my entire evening is lifeless.
Now since R = S + S and R has ( 4, 6 )
If R: S+S
Case 4:
2+2
Case 6:
3+3
Therefore, S = ( 2,3 )
Applying for E = S + D, we have:
E = 2+1 = 3 or
E = 3+1 = 4
Thus, E = ( 3,4 )
So this far,
D = ( 1 )
A = ( 2, 4, 5 )
C = ( 4, 5, 6, 7, 8, 9 )
R = ( 4, 6 )
N = ( 6, 7, 8, 9 )
O = ( 2, 3, 5 )
G = ( 5, 7, 8, 9 )
S = ( 2, 3 )
E = ( 3, 4 )
Not being able to proceed more as I did so long, I try to apply the smallest value sets to the solution of bigger sets.
If S is 2, R will be equal to 4 ( S + S ) and E will equal 3 ( S + D )
Thus, if R = 4 and E = 3 and S = 2 the only possible values of A will be ( 5 ) and O will be ( 5 ) too which makes this impossible since A cant equal O.
Thus S will NOT be 2 and is surely 3 instead.
Now if S is 3, we have:
E = S + D = 4
R = S + S = 6
Thus, we now have possible values as:
D = ( 1 )
A = ( 2, 5 )
C = ( 5, 7, 8, 9 )
R = ( 6 )
N = ( 6, 7, 8, 9 )
O = ( 2, 3, 5 )
G = ( 5, 7, 8, 9 )
S = ( 3 )
E = ( 4 )
Now A = C + R and has got to be either 12 or 15 since it gives D = 1 as carry.
Thus only 9 + 6 satisfies it for A = 5 and 12 isn't possible with the remaining values.
Hence, A = ( 5 ) and C = ( 9 )
[Proof: C = A - R = 15 - 6 = 9 not considering the Carry character. This proof is only additive to the above logical result]
So again, so far, removing duplicates as we've been doing,
D = ( 1 )
A = ( 5 )
C = ( 9 )
R = ( 6 )
S = ( 3 )
E = ( 4 )
N = ( 7, 8 )
O = ( 2 )
G = ( 7, 8 )
( As numbers 3, 5 and 9 are already used up. )
Now N = R + O = 6 + 2 = 8 and hence G = 7 ( Odd one out )
[Proof for G: G = O + A = 2 + 5 = 7, again an additive proof.]
Thus finally the solution is,
D = 1
A = 5
N = 8
G = 7
E = 4
R = 6
C = 9
O = 2
S = 3
Or in numeric order:
D = 1
O = 2
S = 3
E = 4
A = 5
R = 6
G = 7
N = 8
C = 9
Phew, took an hour. Hope am not wrong in my approach itself! If I am, I desperately need a life.
P.s. Am attaching the TXT format in case the formatting I typed this in isn't showing well for reading here.
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