coca
cola
====
oasis
When two rows are being added, the carry-over can only be equal to zero or one.
The o from oasis represents the carry-over from column 4. The leading digit is assumed not to be zero. Therefore o = 1.
Let co = carry-over.
c1ca
c1la
====
1asis
addition of column 4
2c + co = 1a >= 10 ---- (1)
but 2a, from column 1, = s, or s + 10, (Since 2a = s, or s + 10, then s is even ---- (2))
and if a = 0, then s would be zero. i.e. a = s! (but all characters should be different digits)
Therefore a not equal to zero.
Therefore, 2c + co > 10
or, 2c > 9 (co could be 1)
Therefore c > 5 (c must be an integer)
i.e. c = 6, 7, 8, 9
Therefore 2c = 12, 14, 16, 18
But, from (1) 2c + co = 1a
1a - co = 2c = 12, 14, 16, 18
a - co = 2, 4, 6, 8
But co = 0, since addition in column 3 cannot give a sum greater than 9.
Therefore, a = 2, 4, 6, 8
and so 2a = 4, 8, 12, 16
but 2a = s, or 1s, from column 1.
Therefore s = 4, 8, 2, 6
Now from column 3, 2*1 + co = s
and s is even, therefore co = 0, and s = 2
but 2a = s, or 1s, from column 1. i.e. 2a = 2 or 12
Therefore a = 1 or 6
c1ca
c1la
====
1a2i2
But a is even from the addition in column 4. Therefore a = 6
c1c6
c1l6
====
162i2
The addition in column 4 now gives c = 8
8186
81l6
====
162i2
The addition in column 1, 8 + 8 = 16, gives co = 1.
The addition in column 2 is 8 + l + co = i
8 + l + 1 = i
9 + l = i
We showed, when getting s = 2, that the co from this column was zero.
I.e. 9 + l < 10
Therefore l = 0, and i = 9.
Finally,
8186
8106
====
16292