Cryptarithm

N: We see that $N+N$ needs to end in $0$ hence it can either be $0$ or $5$ . If $N=5$ , then the $1$ that would be carried over would be added to the $T+E+E$ column.Then $E+E+1$ would have to end in $0$ but we know that that can't happen, hence $N=0$ .

E: Since $N = 0$ , thus $E=5$ as $E+E$ also needs to end in $0$ but we know that every digit is distinct.Hence we have found two of the digits.

O:Now,when we add $5+5+T$ ,there would be a carry over of $1$ .Now,we see that $O\neq I$ ,hence there has to be some carry over when we calculate $1+R+T+T$ ,it can either be $2X$ OR $1X$ ,assuming it to be $1X$ ,then $O+1=\overline{1I}$ as $F\neq S$ hence it can only be $F+1=S$ as $2$ can't be carried over.But the max. value of $O+1=10$ hence $I=0$ but that is not possible as the numbers are distinct.So, $1+R+T+T=\overline{2X}$ hence $2$ would be carried to $O$ hence $O+2=\overline{1I}$ ,as $O\neq 8$ , $O=9$ hence, $I=1(9+2=1\boxed{1})$ .Now,the sum becomes, $\ \ F\ 9\ R\ T\ Y\\ +\ T\ 5\ \ 0\\ +\ T\ 5\ \ 0\\ =\ S\ 1\ X\ T\ Y$

T:Now,we had that, $1+R+T+T=2X$ ,now we observe that their max. value can be $8+8+7+1=24$ as $T\neq 8$ ,and their least possible value is $21$ as $X\neq 0$ hence $1+R+T+T$ can be $21,22,23,24$ ,we have to exclude $21$ as $1$ has already been used.So we are left with $22,23,24$ ,now if $T<7$ ,then the max value of our sum would be $21$ hence we have that $T=8$ . $$ Rest of the digits:Hence we have used up the digits, $0,1,5,8,9$ and $F+1=S$ now, $S \neq 0,5,9,1,8,2,6$ hence we have that $S={3,4,7}$ .Now,digits that are left should for $F$ and $S$ ,need to have two consecutive digits,this only takes place when $R=7$ and the two consecutive digits left would be $2,3$ and hence $Y=6$ .Hence we have found all the digits.And done!

Summarizing the above, the only solution set is $F = 2, O = 9, R = 7, T = 8, Y = 6 , E = 5, N = 0, S = 3, I = 1, X = 4$ .

This solutions deduces letters in the order N, E, O, I, T, R, X, F, S and Y

Please scroll for the final solution

FINDING N AND E

The two Ns in the units column must add to 0 so that the final result is equal to Y

This means that N can be either 5 or 0

By the same logic, E+E=0 (mod 10)

If N were 5, a 1 would be carried over, making it impossible for the two Es to add up to 0 (mod 10)

Therefore, N=0 must be true

And thus E=5

FINDING O AND I

O and I cannot be distinct without having numbers carried over from the previous summation

Likewise F and S

The maximum that can be carried over from the hundreds to thousands column is a 2

If a 1 was carried over, O=9 to carry over a unit to the 10000s - however, this renders I as 0

But N=0

Therefore a 2 must be carried over

For a 1 to be carried over to the final column O>8

Again, if O=8 then I=0 which cannot be true

Thus O=9

And therefore I=1

FINDING T

We have already shown that a 2 must be carried over from R+T+T

We can also see that a 1 must be carried over from 5+5+T (no numbers are carried from the previous 0+0+Y)

Therefore R+T+T+1>20

The maximum value of R+T+T+1 is 24 (T=8, R=7)

and the minimum value is 22 (we have to exclude 21 and 20 as I=1 and N=0)

and R+T+T+1 = 22 or 23 or 24

If T=6 the maximum value of R+T+T+1=21, therefore T=7 or T=8

However, also note that F+1=S, meaning we need two consecutive numbers for F and S

If T=7 then R=8 and X=3

The numbers remaining consist of 2, 4 and 6, none of which are consecutive - thus, T cannot equal 7

Leaving only T=8

FINDING R and X

Since T=8 and R+T+T+1 = 22 or 23 or 24

Leaving R = 5 or 6 or 7

Since E=5, R = 6 or 7

As F and S are consecutive numbers, from the remaining numbers F=2 and S=3 OR F=3 and S=4

Thus, no other letter can represent 3, otherwise F and S cannot be consecutive

We can see that if R=6 then R+T+T+1=23, thus X=3 which is not possible

By elimination, R=7

Now R+T+T+1=24 therefore X=4

FINDING F AND S AND Y

The remaining digits are 2, 3 and 6

Since F+1=S, F=2 and S=3

Finally, Y=6

FINAL ANSWER

F=2, O=9, R=7, T=8, Y=6, E=5, N=0, S=3, I=1, X=4

FORTY = 29876 TEN = 850 SIXTY = 31486

FORTY + TEN + SIXTY = 29876 + 850 + 31486 = 62212

This one was tough, but I got there by brute force and ignorance!

Because Y + 2N = Y plus a carryover of 0 or 1, N must be 0 or 5. However, if N = 5, then T + 2E +1 cannot equal T plus a carryover. Therefore N = 0, and by the same logic, E = 5.

Now consider the first three columns. We know that the second column must generate a carryover in order for F < S, and similarly for O < I - so S = F +1.

Because N = 0, there is only one case that would provide a carryover for the first column, and that is O = 9 and I = 1. That means that R + 2T + 1 (the carryover from column 4) > 21 to provide a carryover of two.

There are three cases that meet this criterion - R = 6, T = 8; R = 7, T = 8; and R = 8, T = 7.

However, because F or S must = 3, the only valid case is R = 7, T = 8, and thus X = 4.

By elimination, F = 2, S = 3 and Y = 6.

The final sum is therefore 29786 + 850 + 31486 = 62122

Step 1. E + E (+carry) and N + N must be 0 mod 10 to allow the T and Y to come down to the sum.

         With matching digits there are only two possibilities, 0 and 5.

         The carry from N + N is at most 1.  
         For any choice of E, E + E + 1 is odd and not 0 mod 10.
         Therefore there can be no carry from N + N thus N = 0 and E = 5 is forced.

Step 2. At the other end, F and O must have a carry coming in to make distinct digits in the sum.

         The only choice is O = 9 and I = 1 with a carry of 2. 
         ( 8 with carry 2, or 9 with carry 1 would have I=0 duplicating N.)

Step 3. The carry to F is limited to 1. Thus S = F+1, that is, F and S are consecutive digits.

looking at the digits used so far only 2,3,4, and 6,7,8 remain available to provide consecutive digits.

Step 4. The sum R + T + T must produce a carry of 2 for O and I to work.

   That means that both values must come from the 6-8 range
   and the consecutive F,S must come from the 2-4 range.

Step 5. Evaluate X for each choice of R and T.

  R = 7 and T = 8 work, with a carry of 1 add up to 24, X= 4.

 (the others either duplicate a digit in X, or take out 3 leaving no consecutive digits for F and S.)

  Thus X = 4, F = 2 and S=3.

Step 6. This puzzle has ten digits, all distinct, nine have been solved leaving only 6 for Y.

Matching the ten characters FORTY, EN, SIX into {0, 1, 2,....9}

we know S = F+1 and remaining (2,3,4) and (6,7,8)

S,F should be in (2,3,4) as we need two elements in (6,7,8) for (R+T+T) >= 20

3 must be used by (S,F), hence R+T+T+1 = 22 or 24 ; X=2 or 4

I cheated and wrote a quick Python program to just check all possible combinations until one with each digit being an unique non-negative integer yielded a solution. Took about 0.3 sec machine time with very sloppy code. Turing and Bletchley Park live on!