cinelli wrote:FORTY
TEN
TEN
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SIXTY
Spoiler...
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Adding twice "EN" leaves the bottom two digits unchanged, so twice EN must be a multiple of 100. Since E and N must be distinct digits, the only possibility is EN = 50.
Adding the carry out from RTY + TEN + TEN to FO produces SI. That carry out is at most 2, and it:
* cannot be 0 because that would leave FO unchanged;
* cannot be 1 because in order for adding it to change both digits of FO, O would have to be 9, which implies that I must be 0, duplicating N.
So the carry out from RTY + TEN + TEN must be 2. For adding it to change both digits of FO, O must be 8 or 9, and 8 is not possible because it again implies that I must be 0. So O is 9 and I is 1. This means we've now got:
F9RTY
T50
T50
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S1XTY
The carry into the middle column is 1 and we've just established that the carry out from it is 2, so R+T+T+1 = 20+X, or R+T+T = 19+X. R+T+T is at most 7+8+8 = 23, since the digit 9 has already been used, and 19+X is at least 19+2 = 21, since the digits 0 and 1 have already been used. So R+T+T is 21, 22 or 23. T cannot be <= 6, since that would imply R >= 9, and so T can only be 7 or 8. That allows us to enumerate the six possibilities for R+T+T and T, from which R and X can be deduced:
R+T+T T R X OK?
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21 7 7 2 No - duplicates 7
21 8 5 2 No - duplicates 5
22 7 8 3 Yes - remaining digits are 2,4,6
22 8 6 3 Yes - remaining digits are 2,4,7
23 7 9 4 No - duplicates 9
23 8 7 4 Yes - remaining digits are 2,3,6
Finally, from the leftmost column note that S = F+1, so two of the remaining digits must be adjacent. That only happens for the last possibility, so it must be that one, with S=3, F=2, and by elimination Y=6:
29786
850
850
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31486
Gengulphus