Math Puzzles for 1st Grade - Free Printable

This problem involves coloring a grid of circles using three colors (red, yellow, and blue) such that no two adjacent circles share the same color. This is a classic graph coloring problem where each circle represents a vertex in a graph, and edges connect adjacent circles.

Step-by-Step Solution:

#### Understanding the Problem:
1. We have a 3x3 grid of circles.
2. Each circle must be colored either red, yellow, or blue.
3. No two adjacent circles (connected by a line) can have the same color.
4. We need to find 4 different valid colorings.

#### Approach:
To solve this, we can use a systematic approach:
1. Start by assigning colors to the top-left circle.
2. Proceed row by row or column by column, ensuring that adjacent circles do not have the same color.
3. Use the three available colors (red, yellow, blue) to fill in the grid while adhering to the rules.

#### Solution for One Valid Coloring:
Let's start with one possible valid coloring:

```
Row 1: Red, Yellow, Blue
Row 2: Blue, Red, Yellow
Row 3: Yellow, Blue, Red
```

Visual representation:
```
R Y B
B R Y
Y B R
```

#### Explanation:
- In the first row, we choose Red, Yellow, and Blue.
- In the second row, we alternate colors to ensure no two adjacent circles are the same. Starting with Blue ensures it is different from the Red above it.
- In the third row, we continue the pattern, starting with Yellow to ensure it is different from the Blue above it.

#### Finding Additional Valid Colorings:
We can generate more valid colorings by permuting the colors or shifting the pattern. Here are three more valid colorings:

1. Coloring 2:
```
Y B R
R Y B
B R Y
```

2. Coloring 3:
```
B R Y
Y B R
R Y B
```

3. Coloring 4:
```
R B Y
Y R B
B Y R
```

#### Verification:
For each coloring, check that:
- No two adjacent circles have the same color.
- All circles are colored using only red, yellow, and blue.

#### Final Answer:
Here are the four different valid colorings:

1.
```
R Y B
B R Y
Y B R
```

2.
```
Y B R
R Y B
B R Y
```

3.
```
B R Y
Y B R
R Y B
```

4.
```
R B Y
Y R B
B Y R
```

Thus, the final answer is:

\boxed{
\begin{array}{ccc}
R & Y & B \\
B & R & Y \\
Y & B & R \\
\end{array}
,
\begin{array}{ccc}
Y & B & R \\
R & Y & B \\
B & R & Y \\
\end{array}
,
\begin{array}{ccc}
B & R & Y \\
Y & B & R \\
R & Y & B \\
\end{array}
,
\begin{array}{ccc}
R & B & Y \\
Y & R & B \\
B & Y & R \\
\end{array}
}