Big Rubiks Cube

An 8 × 8 × 8 8 \times 8 \times 8 cube is painted red on 3 faces and blue on 3 faces such that no corner is surrounded by three faces of the same color. The cube is then cut into 512 unit cubes. How many of these cubes contain both red and blue paint on at least one of their faces?


The answer is 56.

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4 solutions

Andrew Caldwell
Mar 23, 2015

The solution in my mind for this is visual; as I can't draw diagrams on my phone, I'm not sure if I can convey it successfully, but I'll try:

Due to the symmetric nature of a cube, there's only one way to paint it that meets the terms of the question. A side, the opposite side and any one of the other sides. Of the twelve edges, two are where red sides meet, two are where blue sides meet and eight are where blue meets red.

Each edge has six edge cubes (plus two shared corner cubes), so there are 6x8=48 edge cubes with both colors. Add on the eight corner cubes for a solution of 56.

Count the dark colored cubes. Cube Pic Cube Pic

Atul Vaibhav - 6 years, 2 months ago

Can u explain me where we have to paint the cube so that it meets the terms of the question

Abhinav Choudhary - 6 years, 2 months ago

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Three faces must be colinear for this to work.

Joeie Christian Santana - 6 years, 2 months ago
Rezwan Arefin
Nov 26, 2015

Problem No.8

http://internetolympiad.org/archive/OMOWinter12/OMOWinter12Solns.pdf

Aditya Chauhan
Apr 17, 2015

I visualized so it is difficult to express it.

Then you shouldn't write this is as a solution but rather as a reply to somebody else's solution.

Shubham Bhargava - 5 years, 7 months ago
Towkir Hossain
Mar 26, 2015

The cube has 8 corners contain both red and blue paint on at least one of their faces. Moreover, It has 4 sides out of 12, do not support the condition and have same color on both faces. So, there remains only 8 sides support the condition . 8 sides have 6 sub cube if we exclude the corners. So, counting their availability, the total number of at least one distinct colored sub cube is (8*6) + 8 = 56

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