A five-digit and a four-digit number add up to 33,190. If, however, we add the numbers formed by writing the digits in reverse order, we get 48,400. What are these numbers?
Let the digits of the two original numbers be denoted by $A, B, C, D, E$ and $F, G, H, K$, respectively. Then the addition problems are:

(1) $\begin{array}{lllll}A & B & C & D & E\end{array}$

|  | $F$ | $G$ | $H$ | $K$ |
| :---: | :---: | :---: | :---: | :---: |
| 3 | 3 | 1 | 9 | 0 |$(=x)$

(2) $\begin{array}{lllll} & E & D & C & B & A\end{array}$

|  | $K$ | $H$ | $G$ | $F$ |
| :---: | :---: | :---: | :---: | :---: |
| 4 | 8 | 4 | 0 | 0 |

Since $x < 10^4$, the digit $A$ in the ten-thousands place of (1) must be 2 or 3. From the units place of (2), we have $A + F = 10$, so $F$ must be 8 or 7. Therefore, there must be a carry of 1 from the thousands place to the ten-thousands place in (1), since the number $BCDE$ is less than $10^4$. Thus, $A + 1 = 3$, so $A = 2$ and $F = 8$.

Using the same reasoning starting from (2), the digit $E$ must be 3 or 4, and $K$ must be 7 or 6. There is no carry from the thousands place in (2), so $E = 4$ and $K = 6$. Substituting these values, the problem simplifies to:

| $\left(1^{\prime}\right)$ | $B$ | $C$ | $D$ |
| :---: | :---: | :---: | :---: |
|  |  | $G$ | $H$ |
|  | 5 |  | 8 |

![](https://cdn.mathpix.com/cropped/2024_05_02_cf7f4d96c2dfd567b3aag-1.jpg?height=146&width=426&top_left_y=694&top_left_x=833)

From this, $B$ can only be 4 or 5, so $G = 9 - B$ is also 4 or 5. Therefore, from ( $1^{\prime}$ ), $C + G \geq 10$, and thus $B = 4$ and $G = 5$. From ( $2^{\prime}$ ), $D = 1$, and from ( $1^{\prime}$ ), $H = 7$. Then, from ( $1^{\prime}$ ), $C$ can only be 6, which also satisfies ( $2^{\prime}$ ). Therefore, the two original numbers are 24614 and 8576.

Note. More precisely, we should have said: $A$ can only be 2, $F$ can only be 8, and so on, if the problem has a solution at all. The solution hinges on whether ( $2^{\prime}$ ) remains valid when $C = 6$ as derived from ( $1^{\prime}$ ). ( $C$ was determined in two different ways.)