$$
\begin{array}{r}
\text { FORTY } \\
\text { TEN } \\
+\quad \text { TEN } \\
\hline \text { SIXTY }
\end{array}
$$

In the above problem, each letter represents a digit from $0,1,2,3,4,5,6,7,8,9$, and different letters represent different digits.
Solution (The column numbers are counted from right to left)
From the first column, we see $\mathbf{Y}+\mathrm{N}+\mathbf{N}=\mathrm{Y}$, so $\mathrm{N}=0$ or $N=5$. But $N \neq 5$, otherwise, from the second column, we would have $T+2 E+1$ $=T$, which is impossible. Similarly, $E=0$, or $E=5$, but since $N=0$, we have $E=5$.

From the fourth column, a number should be carried over to the fifth column. This number can only be 1, because the sum of the digits in the fourth column (the digit represented by the letter 0 plus the number carried over from the third column) is no more than 11. So $I=0$ or $I=1$, but since $N=0$, we have $I=1$, and the digit $0=9$.

The sum of the digits in the third column is no less than 21, because the number represented by $\mathbf{X}$ must be greater than 0 and 1, and the number carried over from the second column to the third column is no more than 1, so $\mathrm{T}=7$ or $\mathrm{T}=8$ or $\mathrm{T}=9$. But since the letter $\mathrm{O}=9$, we have $\mathrm{T}=7$ or $\mathrm{T}=8$.

If $T=7$, then $R=8, X=3$. But the digits represented by $F$ and $S$ are different from 1, and these adjacent digits can only be 2 and 3 or 3 and 4. Thus, $X \neq 3$, so $T=8$, and thus $R=7$, $\mathrm{X}=4$.

Now, the only remaining pair of adjacent digits is 2 and 3, so $F=2, S=3$.
Finally, $\mathrm{Y}=6$.
In summary, the only solution to this problem is:
$$
\begin{array}{r}
29786 \\
+\quad 850 \\
\hline 31486
\end{array}
$$