The year 2011

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    Mar 25, 2011 7:12 AM GMT
    Take the year you were born (87 for me) and add that to the age you will become in 2011. EVERYONE born before 2000 will add up to 111...I can make no sense of it!!
  • TheAlchemixt

    Posts: 2294

    Mar 25, 2011 11:01 AM GMT
    I was born in 84 in 2011 I will be 27. 84 + 27= 111.

    So kanaka in 2011 you should be 24 because

    87 + 24 = 111. I hope that made sense. Your buddy - Ryan
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    Mar 25, 2011 11:13 AM GMT
    Lol Ryan. Your addition made perfect sense. What I don't understand is how everyone in the world, 12 and older, share 111.
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    Mar 25, 2011 11:28 AM GMT
    it's perfectly logical. the numbers added "fills" up last century (100) plus the eleven years you've lived in this century.
    next year the "mystery" number will be 112 for everyone who was born last century.
    edit: All it describes is that from 1900 - 2011 is a span of 111 years. any point (birth year) where you divide that timeline the number of years on each side added will be 111, easy.

    (.... whoever came up with this "weird fact" must have a lot of time on his hands or have strange interests LOL)
  • TheAlchemixt

    Posts: 2294

    Mar 25, 2011 11:37 AM GMT
    Oh okay Kanaka. I misunderstood you then lol but I think judo guy knows what he is talking about.vicon_biggrin.gif
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    Mar 25, 2011 11:43 AM GMT
    hahah perfect! I was studying all night and drifted a bit lol.
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    Mar 25, 2011 11:54 AM GMT
    Kanaka24 saidhahah perfect! I was studying all night and drifted a bit lol.


    Get some rest and then back to studying. We need someone to solve world peace, cure health problems, solve world finance issues, etc.. and have somebody hot like yourself do it.

    Do it! icon_biggrin.gif
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    Mar 25, 2011 11:57 AM GMT
    epic-fail-math-fail.jpg

    That was a bit of a math fail Kanaka icon_eek.gif
  • turtleneckjoc...

    Posts: 4685

    Mar 25, 2011 11:59 AM GMT
    Worked for me too.

    Born in '54 and I will be 57 this year. Comes to 111.
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    Mar 25, 2011 12:04 PM GMT
    a total fail! too many line integrals and diverging parametric equationsicon_lol.gif "skewing" my addition skills
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    Mar 25, 2011 12:27 PM GMT
    92+19=111
  • alphatop

    Posts: 1955

    Mar 25, 2011 12:30 PM GMT
    75+36=111
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    Mar 25, 2011 12:46 PM GMT
    You're welcome....
    1 110 111
    2 109 111
    3 108 111
    4 107 111
    5 106 111
    6 105 111
    7 104 111
    8 103 111
    9 102 111
    10 101 111
    11 100 111
    12 99 111
    13 98 111
    14 97 111
    15 96 111
    16 95 111
    17 94 111
    18 93 111
    19 92 111
    20 91 111
    21 90 111
    22 89 111
    23 88 111
    24 87 111
    25 86 111
    26 85 111
    27 84 111
    28 83 111
    29 82 111
    30 81 111
    31 80 111
    32 79 111
    33 78 111
    34 77 111
    35 76 111
    36 75 111
    37 74 111
    38 73 111
    39 72 111
    40 71 111
    41 70 111
    42 69 111
    43 68 111
    44 67 111
    45 66 111
    46 65 111
    47 64 111
    48 63 111
    49 62 111
    50 61 111
    51 60 111
    52 59 111
    53 58 111
    54 57 111
    55 56 111
    56 55 111
    57 54 111
    58 53 111
    59 52 111
    60 51 111
    61 50 111
    62 49 111
    63 48 111
    64 47 111
    65 46 111
    66 45 111
    67 44 111
    68 43 111
    69 42 111
    70 41 111
    71 40 111
    72 39 111
    73 38 111
    74 37 111
    75 36 111
    76 35 111
    77 34 111
    78 33 111
    79 32 111
    80 31 111
    81 30 111
    82 29 111
    83 28 111
    84 27 111
    85 26 111
    86 25 111
    87 24 111
    88 23 111
    89 22 111
    90 21 111
    91 20 111
    92 19 111
    93 18 111
    94 17 111
    95 16 111
    96 15 111
    97 14 111
    98 13 111
    99 12 111
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    Mar 25, 2011 2:04 PM GMT
    Kanaka24 saidTake the year you were born (87 for me) and add that to the age you will become in 2011. EVERYONE born before 2000 will add up to 111...I can make no sense of it!!



    Let 1900 + x = the year you were born, where 0 < or = x < or = 99.

    Then your age in 2011 is 2011 - (1900 + x) = 111 - x.

    Therefore, the year you were born (in the 20th century) + your age
    = x + (111 - x)
    = 111


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    Mar 25, 2011 2:10 PM GMT
    To add onto Judo guys observation... it has to do with our numeral system, which is based on the number 10.. that is... after 9 you begin with the first numeral (1) and add a 0..... after 99 you add two zeros to the 1 etc.

    Its the same reason why any multiple of 9 (18, 27, 36, 108 etc) always adds up to 9... its the highest number in our system

    If you were to use another numeral system, say the Mayan one, with cycles of 20 being the base... you would get different patterns.. any multiple of 19 would add up to 19 there

    .....


    Interestingly enough, the old german peoples used a system that went up to twelve.... which is why german languages have different names for "eleven" and "twelve" than the following "ten" or "teen"-based numbers... (thirteen = three-ten... fourteen = four-ten etc.)
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    Mar 25, 2011 2:19 PM GMT
    Pato_Rico saidTo add onto Judo guys observation... it has to do with our numeral system, which is based on the number 10.. that is... after 9 you begin with the first numeral (1) and add a 0..... after 99 you add two zeros to the 1 etc.

    Its the same reason why any multiple of 9 (18, 27, 36, 108 etc) always adds up to 9... its the highest number in our system

    If you were to use another numeral system, say the Mayan one, with cycles of 20 being the base... you would get different patterns.. any multiple of 19 would add up to 19 there

    .....


    Interestingly enough, the old german peoples used a system that went up to twelve.... which is why german languages have different names for "eleven" and "twelve" than the following "ten" or "teen"-based numbers... (thirteen = three-ten... fourteen = four-ten etc.)


    Actually this doesn't have anything to do with our numeral system. It just has everything to do with the fact that if you have two numbers and you add 1 to one of the numbers and subtract 1 from the other that the result of adding the two numbers will be the same as it was before

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    Mar 25, 2011 2:39 PM GMT
    Pato_Rico saidIts the same reason why any multiple of 9 (18, 27, 36, 108 etc) always adds up to 9... its the highest number in our system


    Take any 4-digit number ABCD.

    ABCD
    = 1000A + 100B + 10C + D
    = 999A + 99B + 9C + (A + B + C + D)
    = 9 (111A + 11B + C) + (A + B + C + D)

    Therefore ABCD is divisibe by 9 if and only if A + B + C + D is divisible by 9.


    The same argument can easily be extended to a number consisting of any number of digits. Just use 10^3 instead of 1000, etc.


    Pato_Rico saidIf you were to use another numeral system, say the Mayan one, with cycles of 20 being the base... you would get different patterns.. any multiple of 19 would add up to 19 there


    Take any 4-digit number ABCD in the 20-numeral system.

    ABCD
    = 20^3*A + 20^2*B + 20C + D
    = (20^3 - 1)A + (20^2 -1)B + 19C + (A + B + C + D)
    = 19 (421A + 21B + C) + (A + B + C + D)

    Therefore ABCD is divisibe by 19 if and only if A + B + C + D is divisible by 19.


    Note that 20^n - 1 = 19(20^(n-1) + 20^(n-2) + ... + 20 + 1) is divisible by 19 for all positive integers n.


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    Mar 25, 2011 7:34 PM GMT
    Ummmm....k
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    Mar 25, 2011 8:04 PM GMT
    Try this one:

    REAL
    + JOCK
    GAGA

    Where: L=C+K, O=E+R, C=7, G=6, and R=2. There is a unique solution.
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    Mar 25, 2011 9:50 PM GMT
    *sigh* I'm so old- thanks for reminding me.
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    Mar 25, 2011 10:04 PM GMT
    okaaayyyy....last night i was waaayyy spaced out off your addy, scotty :-( crazy shit would not leave my mind!!!
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    Mar 25, 2011 10:26 PM GMT
    chocmandms saidTry this one:

    REAL
    + JOCK
    GAGA

    Where: L=C+K, O=E+R, C=7, G=6, and R=2. There is a unique solution.


    I randomly guessed A=9 and worked from there.
    2398
    +4571
    6969
    69.jpg

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    Mar 25, 2011 10:56 PM GMT
    math.... eh.... i can add by 45 lbs really easily... but other than that.... meh icon_confused.gif
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    Mar 25, 2011 11:57 PM GMT
    While we're on the subject of 2011, what is the sum of all integers from 1 to 2011? A fun thing to do Saturday night.
    1 + 2 + 3 + ..... + 2011

    Well the good news is you don't have to spend your whole evening and then some figuring it out. First write it forwards then backwards (system not letting me line up the columns - imagine 1 in the first row is directly above 2011 in the second row, 2 directly above 2010, etc):
    1 + 2 + 3 + ..... + 2011
    2011 + 2010 + 2009 + ..... + 1

    Now if we add up each column we get:
    2012 + 2012 + 2012 + ..... + 2012

    Because we added both rows, we have double our answer.
    There are 2011 columns, each with a sum of 2012. So our answer is
    2012 x 2011 / 2 = 2,023,066.
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    Mar 26, 2011 5:02 AM GMT
    monochrome said
    Pato_Rico saidIts the same reason why any multiple of 9 (18, 27, 36, 108 etc) always adds up to 9... its the highest number in our system


    Take any 4-digit number ABCD.

    ABCD
    = 1000A + 100B + 10C + D
    = 999A + 99B + 9C + (A + B + C + D)
    = 9 (111A + 11B + C) + (A + B + C + D)

    Therefore ABCD is divisibe by 9 if and only if A + B + C + D is divisible by 9.


    The same argument can easily be extended to a number consisting of any number of digits. Just use 10^3 instead of 1000, etc.


    Pato_Rico saidIf you were to use another numeral system, say the Mayan one, with cycles of 20 being the base... you would get different patterns.. any multiple of 19 would add up to 19 there


    Take any 4-digit number ABCD in the 20-numeral system.

    ABCD
    = 20^3*A + 20^2*B + 20C + D
    = (20^3 - 1)A + (20^2 -1)B + 19C + (A + B + C + D)
    = 19 (421A + 21B + C) + (A + B + C + D)

    Therefore ABCD is divisibe by 19 if and only if A + B + C + D is divisible by 19.


    Note that 20^n - 1 = 19(20^(n-1) + 20^(n-2) + ... + 20 + 1) is divisible by 19 for all positive integers n.





    Note also that the more generalized result is that any number written in the k-numeral system is divisible by k-1 if and only if the sum of its digits is divisible by k-1.