Solve the following cryptarithm and find all solutions. Replace the same letters with the same digits, different letters with different digits.

$$
\begin{array}{r}
O S E L \\
S E L \\
E L \\
L \\
\hline 10034
\end{array}
$$

(M. Volfová)
We will start the discussion from the units: The sum of four $L$ ends with the digit 4, which is only possible for $L=1 (4 \cdot 1=4)$ or for $L=6 (4 \cdot 6=24)$. If $L=1$, then the sum of the tens, i.e., the sum of three $E$, would end with the digit 3. This would only be possible for $E=1$. However, this is not allowed because then two different letters $L$ and $E$ would be replaced by the same digit 1. Therefore, $L$ must be 6.

For $L=6$, we have $4 \cdot L=4 \cdot 6=24$; the two (tens) are added to the three $E$ and we need to get a number that ends with the digit 3. This means that $3 \cdot E$ ends with the digit 1, which is only possible for $E=7$.

For $E=7$, we have $2+3 \cdot E=2+3 \cdot 7=23$; the two is added to the higher order, i.e., to the two $S$, and we need to get a number that ends with the digit 0. This means that $2 \cdot S$ ends with the digit 8, which is only possible if $S=4$ or $S=9$.

a) If $S=4$, then $2+2 \cdot S=2+2 \cdot 4=10$; the one is added to $O$ and we need to get 10. Therefore, $O$ must be 9 and the solution is

![](https://cdn.mathpix.com/cropped/2024_04_17_df9261f4862336f3ce8cg-4.jpg?height=262&width=163&top_left_y=1431&top_left_x=975)

b) If $S=9$, then $2+2 \cdot S=2+2 \cdot 9=20$; the two is added to $O$ and we need to get 10. Therefore, $O$ must be 8 and the solution is

$$
\begin{array}{r}
8976 \\
976 \\
76 \\
6 \\
\hline 10034
\end{array}
$$

Note. If the work contains only one solution, evaluate it as 2 - good.