In a $3 \times 3$ board, nine houses must be painted such that in each column, each row, and each of the two diagonals, there are no two houses of the same color. What is the minimum number of colors needed for this?
(a) 3
(b) 4
(c) 5
(d) 6
(e) 7
The correct option is (c).

To satisfy the conditions of the problem, the five houses on the diagonals, marked with *, must have different colors. Therefore, we will need at least five distinct colors. Let's denote these five distinct colors by 1, 2, 3, 4, and 5, and determine how we can choose the colors for the remaining four houses to satisfy the given conditions. One way is shown on the right, as follows.

![](https://cdn.mathpix.com/cropped/2024_05_01_666a1aec7e447b0a6cbeg-048.jpg?height=166&width=200&top_left_y=1142&top_left_x=608)

| 1 |  | 4 |
| :--- | :--- | :--- |
|  | 3 |  |
| 2 |  | 5 |$\rightarrow$| 1 | 2 | 4 |
| :--- | :--- | :--- |
| 4 | 3 | 1 |
| 2 | 4 | 5 |

Thus, it is possible to paint the remaining four houses without using more colors. Therefore, five colors are sufficient. Below, we show three other ways to color the houses.

| 2 |  | 3 |
| :--- | :--- | :--- |
|  | 1 |  |
| 5 |  | 4 |$\rightarrow$| 2 | 4 | 3 |
| :--- | :--- | :--- |
| 4 | 1 | 2 |
| 5 | 2 | 4 |$\quad$| 1 |  | 2 |
| :--- | :--- | :--- |
|  | 4 |  |
| 3 |  | 5 |$\rightarrow$| 1 | 3 | 2 |
| :--- | :--- | :--- |
| 2 | 4 | 1 |
| 3 | 2 | 5 |


| 1 |  | 5 |
| :--- | :--- | :--- |
|  | 2 |  |
| 4 |  | 3 |$\rightarrow$| 1 | 3 | 5 |
| :--- | :--- | :--- |
| 3 | 2 | 4 |
| 4 | 1 | 2 |