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This is an unusual year

  • 06-02-2011 9:29pm
    #1
    barney4001
    Registered Users, Registered Users 2 Posts: 1,447 ✭✭✭


    This year we will experience 4 unusual dates...1/1/11, 1/11/11, 11/1/11 and 11/11/11......AND go figure this out...

    Take the last two digits of the year you were born plus the age you will be this year, it will equal 111......


Comments

  • #2
    CabanasBoy
    Closed Accounts Posts: 186 ✭✭CabanasBoy


    barney4001 wrote: »
    This year we will experience 4 unusual dates...1/1/11, 1/11/11, 11/1/11 and 11/11/11......AND go figure this out...

    Take the last two digits of the year you were born plus the age you will be this year, it will equal 111......

    Well, if you add your age to the last two digits of the year you were born it'll always give you the present year:D It just happens to be 2011 this year. What happens if you were born in 1899 or 2001 etc?


  • #3
    Thomas_S_Hunterson
    Closed Accounts Posts: 6,151 ✭✭✭Thomas_S_Hunterson


    During every day in 2011, the time will, at some stage, be 11:11. Twice.


  • #4
    Yakuza
    Registered Users, Registered Users 2 Posts: 5,141 ✭✭✭Yakuza


    CabanasBoy wrote: »
    Well, if you add your age to the last two digits of the year you were born it'll always give you the present year:D It just happens to be 2011 this year. What happens if you were born in 1899 or 2001 etc?

    That only works for folks born in 20th century (or to be pedantic, for folks born between 1900-1999 (let's leave the whole "2000 is not in the 21st century" debate for another thread)). For someone born in 1899, the answer will be (99+112 = 211). For someone born in 2001, the answer is (1 + 10 = 11). I thought the Y2K problem had been sorted years ago :)

    A truer statement would be:
    (Birth year (mod 100) + age attained in current year (mod 100)) (mod 100) = current year (mod 100) (Only to be used as your specialist subject in Mastermind; the bleedin' obvious :))


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