# Solution: A Very Tricky Octahedron Counting Problem

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**Published on Aug 2, 2018**- Let's color the vertices of a regular octahedron.

Congratulations to Hizami Anuar, staffehn, Laura Kuttnig, Minh Cong Nguyen, FaTalCaT FL, Prof Bits, and attyfarbuckle for successfully solving this math challenge question! Hizami Anuar was the first person to solve the question.

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Thanikachalam Mannagatti2 months agoWhy cannot I have coloring 2 red, 2 yellow and 2 blue?

Akhilesh Kumar10 months agoSir ! I am from india and i am quirious to know about ......what is the probability that a randomly chosen real number from real number line is irrational?if it is not possible to give exact value can we have a feel that which one are contributing more to the real number line i.e. rational/irrational? I think an eligent video from your side will help me.

Zachary Hunter9 months agomost of the number line is irrational, and most irrational numbers are transcendental

Atharva #breakthrough10 months agothat question cant be answered satisfactorily. See classes and types of infinities.

João Matos10 months ago^{+1}The Octahedron of TRANSCENDENCE!

Adlet Irlanuly10 months ago^{+1}LetsSolveMathProblems,

Could you help me with this identity

(2x-1)^20-(ax+b)^20=(x^2 + px + q)^10? . find all real numbers a , b ,p ,q so that the identity true for all x values

Problematic(puzzle channel)10 months ago^{+5}Dang this is an awesome math channel. Instantly subscribed!!

Problematic(puzzle channel)10 months agoYa Exactly!! They deserve more subscribers than they have right now!

Atharva #breakthrough10 months ago^{+1}i know right. this channel is so underrated. this dude is smart. it is hard to find good channels on youtube that do actual hard and challenging problems.

LetsSolveMathProblems10 months ago^{+2}I appreciate your heartwarming compliment. Welcome to the channel! =)

Jonathan Liu10 months ago^{+4}Incredible USclip channel!

dvir yacovi10 months agoI think you have missed 1 option in the 3 colors case. If we call the 6 vertexes as: up, down, and the 4 vertexes of the middle squere as north, south, east and west, so if we color east and down as red, north, south and west as green, abd up and north as blue, it's still a solution, because we actaully have 3 edges that thier's vertexes are at the same color, and in each face there is 1 of these edges.

LetsSolveMathProblems10 months ago^{+1}Consider the triangle formed by south (green), east (red), and up (blue)--that does not fit our requirement. I also note that we have proven that 3-color cases we want must have a monochromatic square, but your example does not.

Jay Chen10 months agoPólyas enumeration theorem should get the job done