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9/14/2017 - An Abstract Way To Equalize

  • #21

    it maybe a draw !!!

  • #22

    exsiting

  • #23
    aHorseWithNoName wrote:

    Come to my intelligent forum

    2+2=5 can it be true?

    Let a + b = c
    Note that 5a - 4a = a
    Note that 5a - (2 + 2)a = a

    Then 5a - (2 + 2)a + 5b - (2 + 2)b = 5c - (2 + 2)c
    Rearranging,

    5a + 5b - 5c = (2 + 2)a + (2 + 2)b - (2 + 2)c

    Factorising,

    5(a + b - c) = (2 + 2)(a + b - c)
    Cancelling the common factor (a + b - c), and removing the brackets as they are now redundant,

    5 = 2 + 2

    Hope this helps.

  • #24

    happy.png

  • #25

    impressively at real

  • #26

    Wonderful concept.

    John 8:21

  • #27
    fatalphenom wrote:

    As a newer player, I find no matter what this ending in a stalemate.

    I am certainly no expert myself, but a stalemate? I was thinking more in the lines of a draw? But I might be mistaken happy.png 

     

  • #28

    I don't get the point of that result

    Clearly White could Checkmate Black.

  • #29

    Cool!

  • #30

    Good one

  • #31

    The first move has to be a queen check because all other moves lead to instant mate.  There are only four queen checks.  Consider each one.  Narrowing it down like this helps a lot.

  • #32

    Draw

  • #33

    That white King looks very vulnerable and my Queen can only give check but do I wish to loose the Queen?.   It might be a light bulb for me. but not just yet.   King to g5 comes to mind as a breather but alas "incorrect"

  • #34

    Why doesn't white start off with Qe3?

  • #35

    NİCE ONE

  • #36
    Gil-Gandel wrote:
    aHorseWithNoName wrote:

    Come to my intelligent forum

    2+2=5 can it be true?

    Let a + b = c
    Note that 5a - 4a = a
    Note that 5a - (2 + 2)a = a

    Then 5a - (2 + 2)a + 5b - (2 + 2)b = 5c - (2 + 2)c
    Rearranging,

    5a + 5b - 5c = (2 + 2)a + (2 + 2)b - (2 + 2)c

    Factorising,

    5(a + b - c) = (2 + 2)(a + b - c)
    Cancelling the common factor (a + b - c), and removing the brackets as they are now redundant,

    5 = 2 + 2

    Hope this helps.

    @ Gil-Gandel:

    The fallacy is that if a+b=c, then a+b-c=0, so when you write

    "Factorising,

    5(a + b - c) = (2 + 2)(a + b - c)"

    what you are really saying is that 5 x 0 = (2+2) x 0, which is true, but you can't cancel the 0s to leave 5 = 2+2 because 5x0/0 does not equal 5 and (2+2)x0/0 does not equal (2+2).

    This took about the same amount of time for me to solve as the chess puzzle.

  • #37

    e3 is covered by the Rook on b3.   @marco_lapegna_29

     

  • #38
    Grey_Goose wrote:

    That white Queen looks very vulnerable and my Queen can only give check but do I wish to loose the Queen?.   It might be a light bulb for me. but not just yet.   King to g5 comes to mind as a breather but alas "incorrect"

    Kg5 Qxg4#.  Don't forget which way his pawns are going happy.png.

  • #39

    happy.pngA very good puzzles for starters but not for me it only made me do some ramdon moves to do it

  • #40
    Gil-Gandel wrote:
    aHorseWithNoName wrote:

    Come to my intelligent forum

    2+2=5 can it be true?

    Let a + b = c
    Note that 5a - 4a = a
    Note that 5a - (2 + 2)a = a

    Then 5a - (2 + 2)a + 5b - (2 + 2)b = 5c - (2 + 2)c
    Rearranging,

    5a + 5b - 5c = (2 + 2)a + (2 + 2)b - (2 + 2)c

    Factorising,

    5(a + b - c) = (2 + 2)(a + b - c)
    Cancelling the common factor (a + b - c), and removing the brackets as they are now redundant,

    5 = 2 + 2

    Hope this helps.

    gil, that's why you need to come to my forum. to learn!

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