40 + 10 + 10 = 60

\[\begin{array} { } & F & O & R & T & Y \\ & & & T & E & N \\ + & & & T & E & N \\ \hline \\ & S & I & X & T & Y \end{array}\]

Starting from the last column, column 0. It is clear that $N$ can either be $0$ or $5$ so that when $Y+N+N = Y$ or $1Y$ . But if $N=5$ , then $E$ in column 1 has to be $0$ and then $T+E+E+1 \ne T$ . Therefore, $N=0$ and $E=5$ .

Now column 4, we note that $S$ must be $=F+1$ and $O$ can only be $8$ or $9$ . But if $O$ is $8$ then $I$ must be $0$ which clashes with $N=0$ . Therefore $O=9$ and $I=1$ , and $R+T+T+1 \ge 20$ . With $O=9$ , $R$ and $T$ can only take on $6$ , $7$ or $8$ . The possible cases are:

\[\begin{array} {} R = 6 & T = 8 & X = 3 \\ R = 7 & T = 8 & X = 4 \\ R = 8 & T = 7 & X = 3 \end{array}\]

Now let us put all three options and the known codes in a table below.

Knowing that $S=F+1$ , we find that we can only fit $F$ and $S$ in when $X=4$ , $R=7$ and $T=8$ . Therefore, $F=2$ and $S=\boxed{3}$ .

\[\Rightarrow \begin{array} {} & 2 & 9 & 7 & 8 & 6 \\ & & & 8 & 5 & 0 \\ + & & & 8 & 5 & 0 \\ \hline \\ & 3 & 1 & 4 & 8 & 6 \end{array}\]

This is my first solution that I am writing, so, please feel free to suggest any modifications.

This is a bit too long solution:

$\text{F O R T Y}$ $\text{+ T E N}$ $\text{+ T E N}$ $\text{---------}$ $\text{S I X T Y}$

For units digit:

$Y + N + N = 10 + Y$ $\implies N = 5\tag{1}$

or

$Y + N + N = Y$ $\implies N = 0\tag{2}$

Y + N + N = 20 + Y is not possible

For tens digit:

$T + E + E = T$ $\implies E = 0$ $\text{(NOT POSSIBLE AS N AND E SHOULD BE DISTINCT)}$

or

$T + E + E + 1 = 10 + T$ $\implies 2E = 9$ $\text{(NO SOLUTION)}$

or

$T + E + E + 1 = 20 + T$ $\implies 2E + 1 = 20$ $\text{(NOT POSSIBLE AS max(E) = 9)}$

or

$T + E + E = 10 + T$ $\implies E = 5\tag{4}$

Thus: $\boxed{\text{E = 5 and N = 0}}$

Hundred digit:

$1 + R + T + T = X$ $\text{(NOT POSSIBLE AS THIS MEANS O = I)}$

SO,

$1 + R + 2T = 10 + X$ $\implies R + 2T = 9 + X \tag{5}$

or

$1+R+2T = 20 + X$ $\implies R + 2T = 19 + X\tag{6}$

Thousand digit:

$O + 1 = 10 + I$ $\implies O = 9 + I$ $\implies I = 0,O = 9$ $\text{Not possible as digits should be unique}$ or

$O + 2 = 10 + I$ $\implies O = 8 + I$ $\implies I = 0,O = 8$ $\text{Not possible}$ $I = 1,O = 9 \tag{7}$

Thus: $\boxed{\text{I = 1, O = 9, 1+R+2T = 20 + X, E = 5, N = 0}}$

Thus, possible (T,R,X) = (7,8,3), (8,6,3), (8,7,4)

Ten Thousand digit:

$F + 1 = S$

Thus:

$F = 2$ $\implies S = 3$ $\implies (T,R,X) = (8,7,4)\tag{8}$

$F = 3$ $\implies S = 4$ $\text{No value of (T,R,X) suits these set of values}$

$F = 6$ $\implies S = 7$ $\text{No value of (T,R,X) suits these set of values}$

$F = 7$ $\implies S = 8$ $\text{No value of (T,R,X) suits these set of values}$

Thus, answer:

$\boxed{\text{(F,S,O,I,T,R,X,E,N,Y) = (2,3,9,1,8,7,4,5,0,6)}}$

$:)$

FORTY + TEN + TEN =SIXTY

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