Question

Consider the cube shown below with its 8 corners labelled a, b, c, d, e, f, g, and h. The figure is representative.

All corners are to be colored such that any two corners that are connected by an edge must be of different colors. The minimum number of colors required to achieve this is ________

• A. 8
• B. 4
• C. 3
• D. 2

Hint:

Label each corner by how many of its 3 coordinates are 1, then color by whether that count is even or odd, like a 3D checkerboard.
Check whether any single edge of the cube ever joins two corners that share the same parity.

Answer 1

The Correct Option is D

Step 1: Try the smallest possible answer first.
Instead of building a general rule, test whether just 2 colors, say red and blue, can color the cube legally, since checking a small case first is often faster than a general argument for a concrete shape like a cube.

Step 2: Color one face and see what the opposite face must do.
Take the top face with corners $a, b, d, c$ going around it. Color $a$ red, then $b$, joined to $a$, must be blue, then $d$, joined to $b$, must be red again, and $c$, joined to both $a$ and $d$, must be blue. So the top face alternates red, blue, red, blue as you go around it, which works because a square has an even number of corners.

Step 3: Extend this to the bottom face.
Each bottom corner is joined to exactly one top corner directly above it. So each bottom corner simply takes the opposite color to the top corner it sits under: the corner below $a$ becomes blue, the corner below $b$ becomes red, the corner below $d$ becomes blue, and the corner below $c$ becomes red.

Step 4: Check the bottom face's own edges.
Going around the bottom face in the matching order gives blue, red, blue, red, which alternates correctly with no two adjacent bottom corners sharing a color. Every vertical edge also joins a red top corner to a blue bottom corner or a blue top corner to a red bottom corner. So all 12 edges of the cube are checked, and every one joins two different colors.

Step 5: Confirm 2 is the least possible.
A single color would force both ends of every edge to match, which breaks the rule the moment the cube has even one edge. Since 2 colors already color the whole cube with no violation, 2 is both achievable and the smallest number that can work.

Final Answer:
Coloring the cube like a 3D checkerboard, alternating colors across every edge, uses only 2 colors and cannot be done with fewer, matching option (D). \[ \boxed{2} \]