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Sixty

cinelli
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Sixty

#53006

Postby cinelli » May 12th, 2017, 1:46 pm

FORTY
TEN
TEN
-----
SIXTY

Cinelli

Gengulphus
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Re: Sixty

#53114

Postby Gengulphus » May 12th, 2017, 7:42 pm

cinelli wrote:
FORTY
TEN
TEN
-----
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
-----
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?
----------------------------------------------------
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
-----
31486

Gengulphus

cinelli
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Re: Sixty

#53208

Postby cinelli » May 13th, 2017, 12:45 pm

Quite right, Gengulphus. The solution is unique which is always the aim with this sort of puzzle.

Cinelli


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