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Posers and Puzzles

Posers and Puzzles

  1. Subscriber FreakyKBH
    Acquired Taste...
    30 Jan '11 01:25
    I do not possess the keenest of mathematical minds, so I turn to those of you who can turn this into English for me. For the life of me, I cannot figure out the formula--- even though I know it's going to turn out to be the simplest of statements.

    When the two-year birth year of the parent is reached by the age of the child, the parent's age has arrived at the two-digit birth year of the adult.


    I noticed it when my dad (born in '42) turned 64, and I (born in '64) turned 42. Turns out, the same result follows any scenario I can devise.

    I can't figure it out. Any help?
  2. 30 Jan '11 02:57
    Originally posted by FreakyKBH
    I do not possess the keenest of mathematical minds, so I turn to those of you who can turn this into English for me. For the life of me, I cannot figure out the formula--- even though I know it's going to turn out to be the simplest of statements.

    [quote]When the two-year birth year of the parent is reached by the age of the child, the parent's age has ...[text shortened]... ut, the same result follows any scenario I can devise.

    I can't figure it out. Any help?
    Parent is born in XY

    Parent has child when B years old, this is Year XY + B

    XY years later, the child is XY old, and the parent is XY+B years old which is also the year when the child was born.

    i.e. adding the Year the parent was born to how old the parent was when they had the child gives the year when the child was born, pretty self evident really

    It doesn't give a very good equation because its essentially:
    XY + B = XY + B

    which contains no information about the values of XY and B and so simplifies to the rather meaningless
    0 = 0

    And it obviously works for any numbers, with the constaint that, as it uses 2 digit years to equal age, the parents age cannot be more than 99 on the final date. (XY + B < 99)

    Eg it works here:
    parent Born in 1880
    has child when 19, this is year 1899
    When child is 80 parent will be 19+80 = 99, so that's ok

    but it doesn't work here:
    parent born in 1880
    has child when 20, this is year 1900
    when child is 80, parent will be 100, the parent's age is now not two digits so it can't equal a 2-digit year.
  3. 30 Jan '11 14:45
    it shouldnt always work...
  4. 30 Jan '11 14:46
    what about: when the child has reached the age that your mother was when you were born she will be exactly twice your age.
  5. 31 Jan '11 09:40
    Originally posted by Banana King
    what about: when the child has reached the age that your mother was when you were born she will be exactly twice your age.
    Ok
    Mother was M at the time of your birth, and you were Zero, the Child is C

    When the child is M, M-C years have passed, so you are M-C years old

    so M = 2(M - C)

    M = 2C

    There are a lot of answers, the mother's age has to be an even number.

    So, for instance Your mother was 20, and the child was 10 at the time of your birth

    When the child is 20 you will be 10 which satisfies the problem.

    Or were your conditions in addition to the first conditions?
  6. 31 Jan '11 12:43
    im sorry i didnt mean to add an unnecessary variable.
    i meant...when YOU are the age of your mother when she was when you were born she will be twice your age.
  7. Standard member Agerg
    The 'edit'or
    04 Feb '11 01:05 / 8 edits
    non-maths
    people with 6 digit phone numbers, for example, (neglecting area code) are often amazed by this one:

    1. Consider the first three digits of your home phone number (ignore area code)
    2 Multiply by 8
    3. Add 1
    4. Multiply by 250
    5. Add twice the last 3 digits of your phone number
    6. Subtract 250
    7. Divide number by 2
    8. What have you got?


    changing steps 4,5,6 slightly if more than 6 digits