Face Colouring Problem

Consider a cube. It has 6 faces. Each face is adjacent to 4 other faces and opposite to 1 face. We want to find the minimum number of colors required to paint all the faces such that no two adjacent faces have the same color. Let's try to paint the cube with the minimum number of colors: (A) Paint one face with Color (A) (B) The four faces adjacent to the first face must have a different color from Color (A) Let's paint them with Color (B) (C) Now consider the face opposite to the first face (painted with Color 1). This face is adjacent to the four faces painted with Color (B) Therefore, it cannot be painted with Color (B) However, it is not adjacent to the first face (painted with Color 1), so it can be painted with Color (A) So far, we have used 2 colors. Let's see if this works for all adjacent faces. - The first face (Color 1) is adjacent to four faces (Color 2). The condition is satisfied. - Each of the four faces (Color 2) is adjacent to the first face (Color 1) and two other faces (also Color 2). This violates the condition that no two adjacent faces have the same color. Therefore, 2 colors are not sufficient. Let's try with 3 colors: Color A, Color B, and Color C. (A) Paint the top face with Color A. (B) Paint the bottom face with Color B. (The top and bottom faces are not adjacent). (C) The four side faces are adjacent to both the top and the bottom faces. They must be painted with a color different from Color A and Color B. So, paint all four side faces with Color C. Now let's check the adjacent faces: - The top face (Color A) is adjacent to four side faces (Color C). The condition is satisfied. - The bottom face (Color B) is adjacent to four side faces (Color C). The condition is satisfied. - Any two adjacent side faces (both Color C) violate the condition that no two adjacent faces have the same color. This arrangement does not work. Let's try another arrangement with 3 colors. (A) Paint one face with Color (A) (B) Paint one of the adjacent faces with Color (B) (C) Paint the face opposite to the first face with Color (B) (It is not adjacent to the first face or the second face). (D) The remaining four faces are adjacent to both Color 1 and Color (B) We can paint two opposite of these four faces with Color 3, and the remaining two opposite faces with Color (A) Let's visualize this: - Top: Color 1 - Bottom: Color 2 - Front: Color 2 - Back: Color 1 - Left: Color 3 - Right: Color 3 Checking adjacent faces: - Top (A) is adjacent to Front (2), Back (A), Left (3), Right (3). Condition satisfied. - Bottom (2) is adjacent to Front (2), Back (A), Left (3), Right (3). Condition violated (Bottom and Front have the same color). Let's try another approach: Paint opposite faces with the same color. We have 3 pairs of opposite faces. We can paint each pair with a different color. - Top and Bottom: Color 1 - Front and Back: Color 2 - Left and Right: Color 3 Now check adjacent faces: - Top (A) is adjacent to Front (2), Back (2), Left (3), Right (3). Condition satisfied. - Bottom (A) is adjacent to Front (2), Back (2), Left (3), Right (3). Condition satisfied. - Front (2) is adjacent to Top (A), Bottom (A), Left (3), Right (3). Condition satisfied. - Back (2) is adjacent to Top (A), Bottom (A), Left (3), Right (3). Condition satisfied. - Left (3) is adjacent to Top (A), Bottom (A), Front (2), Back (2). Condition satisfied. - Right (3) is adjacent to Top (A), Bottom (A), Front (2), Back (2). Condition satisfied. This arrangement uses 3 colors and satisfies the condition. Therefore, the minimum number of colors required is (C)