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The Soc for People of 'GRDCT2008' Mk VI Watch

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    Why do we only have 3 active ''leaders''? :beard: :hmmmm:
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    (Original post by natty_d)
    Why do we only have 3 active ''leaders''? :beard: :hmmmm:
    Your musical compatibility with natty_d is VERY HIGH :party:
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    (Original post by natty_d)
    Why do we only have 3 active ''leaders''? :beard: :hmmmm:
    I suppose RoadWarrior and led aren't particularly active :p:

    Then again, we've never really had a situation where there hasn't been a leader around when one is "needed". :erm:

    To be fair, the leaders are only needed to add or remove members from the Soc (and I suppose, make more leaders), and these aren't exactly pressing matters. :erm:
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    if i dont get into Birmingham ill, ill .......... :cry:



    :sad:
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    PS writing :sigh: :stupido: :sigh:
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    (Original post by GHOSH-5)
    PS writing :sigh: :stupido: :sigh:
    :console:
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    I'm a lazy mare. I'm a lazy mare. :hoppy:
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    (Original post by Kuriso)
    :ditto:

    I hate posts made of ditto or this :shifty:


    You only just got round to doing this :wtf:

    :five:

    and yes as having to mod in there would have been ever harder :moon:
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    :five:

    at least the thread's in chat now :rip:


    internet shopping :sigh:
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    (Original post by clad in armour)
    if i dont get into Birmingham ill, ill .......... :cry:



    :sad:
    Are you applying for medicine?
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    (Original post by bob9001)
    Are you applying for medicine?
    :yep: :yep: :yep: :yep: :yep: :yep: :yep: :yep: :yep:
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    Stuck on maths, but I think it might be out of my league, but I thought the idea itself was quite interesting, so I'm sharing it with you. I was thinking back years ago, when we were quite young, and learning multiplication (times) tables. My seemingly simple question is: given a standard n by n multiplication table, how many different numbers are in it? I've made no progress on the sequence itself (starting 1, 3, 6, 9, 14, 18).

    I considered the following question: Let us define the set  A_n = \{ n, 2n, 3n, \cdots n^2 \} . Can we find a general expression for the cardinality of the set B, where

     B = A_n \cap \left( \displaystyle \bigcup_{i=1}^{n-1} A_i \right)

    but there doesn't seem an easy way to find that, and I can't see a nice number theory route coming without it becoming too general for me to cope with it. :confused:

    Meh.
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    (Original post by clad in armour)
    :yep: :yep: :yep: :yep: :yep: :yep: :yep: :yep: :yep:

    :woo:
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    (Original post by clad in armour)
    :yep: :yep: :yep: :yep: :yep: :yep: :yep: :yep: :yep:
    ahh same
    good luck

    what are your other choices other than Birmingham?
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    (Original post by GHOSH-5)
    Stuck on maths, but I think it might be out of my league, but I thought the idea itself was quite interesting, so I'm sharing it with you. I was thinking back years ago, when we were quite young, and learning multiplication (times) tables. My seemingly simple question is: given a standard n by n multiplication table, how many different numbers are in it? I've made no progress on the sequence itself (starting 1, 3, 6, 9, 14, 18).

    I considered the following question: Let us define the set  A_n = \{ n, 2n, 3n, \cdots n^2 \} . Can we find a general expression for the cardinality of the set B, where

     B = A_n \cap \left( \displaystyle \bigcup_{i=1}^{n-1} A_i \right)

    but there doesn't seem an easy way to find that, and I can't see a nice number theory route coming without it becoming too general for me to cope with it. :confused:

    Meh.

    Take it to the (other) pros:top:


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    :p: I'm not looking for help at all :p: But I thought it was an interesting question of a very simple idea.
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    You could get an answer though [unless you already know:confused: ]

    the mind baffles so I'll just shush now
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    (Original post by GHOSH-5)
    Stuck on maths, but I think it might be out of my league, but I thought the idea itself was quite interesting, so I'm sharing it with you. I was thinking back years ago, when we were quite young, and learning multiplication (times) tables. My seemingly simple question is: given a standard n by n multiplication table, how many different numbers are in it? I've made no progress on the sequence itself (starting 1, 3, 6, 9, 14, 18).

    I considered the following question: Let us define the set  A_n = \{ n, 2n, 3n, \cdots n^2 \} . Can we find a general expression for the cardinality of the set B, where

     B = A_n \cap \left( \displaystyle \bigcup_{i=1}^{n-1} A_i \right)

    but there doesn't seem an easy way to find that, and I can't see a nice number theory route coming without it becoming too general for me to cope with it. :confused:

    Meh.
    :moon:

    On another note, guess who wasted today?

    And also Ghosh your posts on their thread are :toofunny:
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    (Original post by Malsy)
    You could get an answer though [unless you already know:confused: ]

    the mind baffles so I'll just shush now
    From reading around on the subject a little, I doubt there are more than three or four people that could comment on the answer. I'm not even sure if anyone has found a definite answer (but I do know that it is bounded, and the bound itself isn't anything neat).
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    NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO OOOOOOOOOOOOOOOOOOOOOOO http://www.thestudentroom.co.uk/show....php?t=1014964

    :cry:
 
 
 
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