Forty, Ten, Ten

Worked solution

Work column by column from the right, writing $c_1, c_2, \dots$ for the carries. *Units:* $Y + N + N = Y + 10c_1$, so $2N = 10c_1$ and $N$ is 0 or 5. *Tens:* $T + 2E + c_1 = T + 10c_2$, so $2E + c_1 = 10c_2$. If $N = 5$ then $c_1 = 1$ and $2E + 1$ would be an odd multiple of 10, impossible. So $N = 0$, $c_1 = 0$, and $2E$ is a multiple of 10 with $E \ne 0$: $E = 5$, $c_2 = 1$. *Ten-thousands:* $F + c_4 = S$ with $F \ne S$, so $c_4 = 1$. *Thousands:* $O + c_3 = I + 10$, where $c_3 \le 2$. This needs $O = 9$ and then $I = c_3 - 1$; $I = 0$ is taken by $N$, so $c_3 = 2$ and $I = 1$. *Hundreds:* $R + 2T + 1 = X + 20$, i.e. $R + 2T = X + 19$. The unused digits are now 2, 3, 4, 6, 7, 8 for $F, S, R, T, X, Y$, with $S = F + 1$. Since $X \ge 2$ and $R \le 8$, $2T \ge 13$, so $T$ is 7 or 8. If $T = 7$, then $R = X + 5$ forces $X = 3$, $R = 8$, leaving 2, 4, 6 for $F, S, Y$, with no two consecutive for $S = F + 1$. If $T = 8$, then $R = X + 3$ gives $(X, R) = (3, 6)$, leaving 2, 4, 7 (again no consecutive pair), or $(X, R) = (4, 7)$, leaving 2, 3, 6: $F = 2$, $S = 3$, $Y = 6$. So the unique solution is $29786 + 850 + 850 = 31486$, and SIXTY is **31486**.