Posers and Puzzles

Posers and Puzzles

  1. my head
    Joined
    03 Oct '03
    Moves
    671
    19 Jan '06 23:49
    John the Typical Art Student is in a color theory class. his his teacher tells him to make an 8 by 8 grid, and color each cell a singel unique color, using only three colored pencils: red, yellow and blue. she advises that he play about with them a little before hand to get a feel for how they work on the paper. to his desmay he finds that he can only get two consistand shades, light and medium, out of each pencil, which really bums him out because he thought he was being smart when he bought the cheap pencils. he is pretty sure he can not do the assignment with these pencils, untill he finds that when he layers the pencils, each order of application produces a different color. can John compleat his grid?
  2. Standard memberPBE6
    Bananarama
    False berry
    Joined
    14 Feb '04
    Moves
    28719
    20 Jan '06 00:04
    Originally posted by fearlessleader
    John the Typical Art Student is in a color theory class. his his teacher tells him to make an 8 by 8 grid, and color each cell a singel unique color, using only three colored pencils: red, yellow and blue. she advises that he play about with them a little before hand to get a feel for how they work on the paper. to his desmay he finds that he can only ...[text shortened]... s the pencils, each order of application produces a different color. can John compleat his grid?
    I'm not totally clear on the rules you've stated, but I think John can complete the grid.

    An 8x8 grid requires 64 unique colours. John has 3 coloured pencils, and 2 shades for each colour. Also, the order of application is important. So:

    1. Using one pencil at a time, John can create (2 shades) * (3 pencils, choose 1) = 6 colours

    2. Using two pencils at a time, John can create (2 shades) * (3 pencils, choose 1) * (2 shades) * (2 pencils, choose 1) = 24 colours

    3. Using three pencils at a time, John can create (2 shades) * (3 pencils, choose 1) * (2 shades) * (2 pencils, choose 1) * (2 shades) * (1 pencil, choose 1) = 48 colours

    And 6 + 24 + 48 = 78 colours, so John can complete the grid.
  3. Standard memberTheMaster37
    Kupikupopo!
    Out of my mind
    Joined
    25 Oct '02
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    20443
    24 Jan '06 09:47
    3 pencils, 2 shades, is equal to 6 colors.

    He can use a single pencil on a cell: 6 possibilities

    He can use two pencils on a cell: 30 possibilities (including different orders)

    He can use three pencils on a cell: 120

    Four pencils: 360

    Five: 720

    Six: 720

    He can leave the cell blank: 1

    1+720+720+360+120+30+6 = 1957 possible colorisations on a cell.

    64 cells, so yeah, he can do it.
  4. Standard memberXanthosNZ
    Cancerous Bus Crash
    p^2.sin(phi)
    Joined
    06 Sep '04
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    25076
    24 Jan '06 10:32
    Originally posted by TheMaster37
    3 pencils, 2 shades, is equal to 6 colors.

    He can use a single pencil on a cell: 6 possibilities

    He can use two pencils on a cell: 30 possibilities (including different orders)

    He can use three pencils on a cell: 120

    Four pencils: 360

    Five: 720

    Six: 720

    He can leave the cell blank: 1

    1+720+720+360+120+30+6 = 1957 possible colorisations on a cell.

    64 cells, so yeah, he can do it.
    I don't think you can call three pencils with 2 thicknesses each equal to 6 different colours.

    R1 + R2 wouldn't be a new shade but it would be R2 wouldn't it? Otherwise he wouldn't have been limited to two shades per pencil in the first place.
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