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Part (b):

To determine how many different cubes of various colors could be visible from a single point in space on a colored cube made up of 343 small 1x1x1 cubes, let's first analyze the structure of the overall cube.

The cube consists of \( 343 = 7 \times 7 \times 7 \) small cubes. The entire cube itself is 7 cubes along each edge.

When viewed from a vantage point, we can typically see three faces of the cube (the top face and two adjacent side faces). We need to ensure that no two adjacent small cubes on these visible faces have the same color.

Given that there are 4 different colors available, we can assign colors to the small cubes while adhering to the restriction of not having adjacent cubes of the same color.

Let's set up a coloring scheme based on the structure:

1. Each face of the cube is made up of \( 7 \times 7 = 49 \) small cubes.

2. Only the outer layer of cubes is visible on these three faces. The three visible faces share some edges and corners at their intersections.

### Counting visible cubes

Since we are looking at three faces:

- The top face has 49 cubes.

- The two side faces also have 49 cubes each, but they share 7 cubes along the edge with the top face and share 7 cubes on the vertical edge between them.

From the above:

- Total cubes from three faces without accounting for overlaps: \( 49 + 49 + 49 = 147 \)

- Overlap from the shared edge between the top and each side face (7 cubes each): \( -7 - 7 \)

- Overlap from the corner cube (counted in both side faces and on top): \( +1 \)

So, the total unique visible small cubes:

\[
147 - 14 + 1 = 134 \text{ cubes}
\]

### Conclusion on the maximum number of visible colors

To maximize the number of visible colors while respecting the adjacent color restriction, we can use a checkerboard-like pattern on the faces.

Given that there are 4 colors and color assignments can be chosen judiciously, you can easily keep up with the requirement of different colors on adjacent cubes.

Based on the checkerboard coloring on such patterns, essentially you can effectively show all 4 colors visible from a single viewing point while adhering to the adjacent color restriction, especially since you're viewing three faces.

### Answer

Thus, at most **4 cubes of different colors** can be visible from a single point in space while satisfying the color adjacency condition.

Analyzing the Cube

Understanding the Constraints:

The cube is composed of 343 small cubes (7x7x7).

Each small cube has one of four colors.

No two consecutive cubes on any side can have the same color.

Visualizing the Cube: Imagine the cube as a stack of 7 layers, each layer being a 7x7 square.

Determining the Maximum Number of Cubes per Color

Strategy: We'll start by considering one face of the cube and then extend the logic to the entire cube.

Considering One Face:

On a single face (a 7x7 square), the maximum number of cubes of a single color would be achieved by alternating colors in a checkerboard pattern.

In this pattern, there would be 25 cubes of one color and 24 cubes of another.

Extending to the Entire Cube:

To ensure no consecutive cubes on any side have the same color, we can alternate colors between layers.

This means that half the layers would have one color pattern, and the other half would have the opposite color pattern.

Calculation:

Total cubes = 343

Half of the cubes = 343/2 = 171.5

Since we can't have half a cube, we'll round down to 171.

Therefore, the maximum number of smaller cubes of each color possible is 171.

The answer is -4n + 8.

The integers that work are 1, 5, 10, 17, 22, 25, 38, 40, 47, 55, 82, 91, 97, 105, 122, 145, 156, and 160.

The answer is 5/42.

To determine the value of \( k \) for which the system of equations has no solutions, we can represent the system in matrix form as follows:

\[
\begin{bmatrix}
1 & 1 & 3 \\
-4 & 2 & 5 \\
k & 0 & 1
\end{bmatrix}

\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}

=

\begin{bmatrix}
10 \\ 7 \\ 3
\end{bmatrix}

\]

To analyze the conditions under which the system has no solution, we can use the concept of the rank of the coefficient matrix compared to the augmented matrix. Specifically, a system will have no solutions if the rank of the coefficient matrix is less than the rank of the augmented matrix.

First, we set up the augmented matrix:

\[
\left[\begin{array}{ccc|c}
1 & 1 & 3 & 10 \\
-4 & 2 & 5 & 7 \\
k & 0 & 1 & 3
\end{array}\right]
\]

Next, we will perform row operations to bring this matrix to row echelon form.

1. **Row 1** stays the same.

2. **Row 2** can be modified by adding 4 times Row 1 to it:

\[
R_2 = R_2 + 4R_1 \\
\Rightarrow -4 + 4(1) = 0, \\
2 + 4(1) = 6, \\
5 + 4(3) = 17, \\
7 + 4(10) = 47 \\
\Rightarrow R_2 = [0, 6, 17 | 47]
\]

3. Now, we will modify **Row 3**. To eliminate the first term in Row 3, we can do the following. Let's say we perform:

\[
R_3 = R_3 - kR_1
\]

This gives:

\[
R_3 = [k - k, 0 - k, 1 - 3k | 3 - 10k] \\
\Rightarrow R_3 = [0, -k, 1 - 3k | 3 - 10k]
\]

The new augmented matrix looks like this:

\[
\left[\begin{array}{ccc|c}
1 & 1 & 3 & 10 \\
0 & 6 & 17 & 47 \\
0 & -k & 1 - 3k & 3 - 10k
\end{array}\right]
\]

Next, we need to see when the rank of the coefficient matrix differs from the rank of the augmented matrix. For that, the third row has to be a nontrivial linear combination of the first two rows such that it produces a contradiction.

From the second row of the system, we can conclude that, as long as \( k \neq 0 \), Row 3 provides a new pivot. Thus, we must examine what happens when \( k = -3 \):

If \( k = -3 \)

\[
R_3 = [0, 3, 10 | 27] \quad \text{(after substituting \( k = -3 \))}
\]

Now, Row 3 should be checked against the other equations. If we solve with the last row resulting in a contradiction, we find inconsistent results leading to no solutions.

Thus, the value of \( k \) which results in this system having no solutions is:

\[
\boxed{-3}
\]