Solution: A Very Tricky Octahedron Counting Problem

  • 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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Comments • 14

  • Thanikachalam Mannagatti

    Why cannot I have coloring 2 red, 2 yellow and 2 blue?

  • Akhilesh Kumar
    Akhilesh Kumar Year ago

    Sir ! 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 Hunter
      Zachary Hunter Year ago

      most of the number line is irrational, and most irrational numbers are transcendental

    • Atharva #breakthrough
      Atharva #breakthrough Year ago

      that question cant be answered satisfactorily. See classes and types of infinities.

  • João Matos
    João Matos Year ago +1

    The Octahedron of TRANSCENDENCE!

  • Adlet Irlanuly
    Adlet Irlanuly Year ago +1

    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)

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    • Problematic(puzzle channel)
      Problematic(puzzle channel) Year ago

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    • Atharva #breakthrough
      Atharva #breakthrough Year 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.

    • LetsSolveMathProblems
      LetsSolveMathProblems  Year ago +2

      I appreciate your heartwarming compliment. Welcome to the channel! =)

  • Jonathan Liu
    Jonathan Liu Year ago +4

    Incredible USclip channel!

  • dvir yacovi
    dvir yacovi Year ago

    I 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.

    • LetsSolveMathProblems
      LetsSolveMathProblems  Year 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 Chen
    Jay Chen Year ago

    Pólyas enumeration theorem should get the job done