Since $Y+N+N$ ends in $Y$, $N$ must be 0 or 5. But if $N=5$ then $T+E+E+1$ ends in T, which is impossible, so $N=0$ and $E=5$. Since $F \neq S$ we must have $O=9, R+T+T+1>10$, and $S=F+1$. Now $I \neq 0$, so it must be that $I=1$ and $R+T+T+1>20$. Thus $R$ and $T$ are 6 and 7, 6 and 8, or 7 and 8 in some order. But $X$ can't be 0 or 1 since those are taken, and $X$ cannot be 3 since $F$ and $S$ have to be consecutive, so it must be that $R+T+T+1$ is 21 or 23. This is satisfied only for $R=7, T=8$, so $F=2, S=3$, and $Y=6$. This $S I X T Y=\mathbf{31486}$.