1. Joined
    08 May '07
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    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 memberwolfgang59
    howling mad
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    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. Joined
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    28635
    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 memberwolfgang59
    howling mad
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    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 memberTheMaster37
    Kupikupopo!
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    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.
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