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Posers and Puzzles

Posers and Puzzles

  1. 31 Jul '09 02:34
    Said a chess master playing for free, "My age I've divided by three. The result is a square. Yet my age, I declare, is six times a cube. How can that be?
  2. Standard member wolfgang59
    Infidel
    31 Jul '09 07:45
    Originally posted by petrovitch
    Said a chess master playing for free, "My age I've divided by three. The result is a square. Yet my age, I declare, is six times a cube. How can that be?
    We are looking for integer solutions for 3.x^2 = 6.y^3

    ie x^2 = 2.y^3

    A set of solutions is

    x = 2^(3n-1) and y=2^(2n-1)

    The only solution giving a reasonabe age for our master is n=1

    x=4
    His age is 3 * 4^2 = 48 = 6 * 2^3
  3. 10 Sep '09 03:09
    48 is the only reasonable age. 2x2x2x6, which fits the criteria for 1/3 being a square (4x4=16).
    The next number is 3x3x3x6=168. 1/3rd of that is 56, which is not a square.
    I doubt there are any chess masters much over 80, let alone older than 168. If there were, would they play for free?
  4. Standard member wolfgang59
    Infidel
    10 Sep '09 06:44
    we are assuming the age is measured in years.

    What answers are there if the age is measured in months?
  5. Standard member TheMaster37
    Kupikupopo!
    11 Sep '09 10:07
    Originally posted by wolfgang59
    we are assuming the age is measured in years.

    What answers are there if the age is measured in months?
    X^2 = 2 * Y^3

    Should Y have any factor F other than 2, then F^6 divides Y^3. For the powers of 2; Y^3 contains an odd number of factors 2 (3, 9, 15, ...).

    Y^3 = 2^(6n+3) * F^6

    So Y = 2^(2n+1) * F^2
    And X = 2^(3n+2) * F^3 with n >= 0 and F = 1 or F > 2.

    For F = 1 and n = 0 we get X=4 and Y=2 for an age of 48, already given by previous posters.