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    (Original post by lgs98jonee)
    that is wot i got too :-(

    i got n(g^2+g-2gn)=110

    where g-no of groups
    n-total no of teams

    is the above wrong 'theone'??
    I'm not going to say, you're both on the right lines, you and james, you just need to finish it off I think, unless you've made a mistake in your algebra somewhere . I'm not certain, if by teams you mean no. of competitors, then I think you've got your algebra wrong.
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    (Original post by ZJuwelH)
    For me it's in P2. Maybe P3 for you. It may involve long division which is a pain..
    I prefer 'algebraic juggling', its a less messy method.
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    (Original post by elpaw)
    I prefer 'algebraic juggling', its a less messy method.
    How's that work?
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    (Original post by theone)
    I'm not going to say, you're both on the right lines, you and james, you just need to finish it off I think, unless you've made a mistake in your algebra somewhere . I'm not certain, if by teams you mean no. of competitors, then I think you've got your algebra wrong.
    Yea, think im almost there, but i know ive gone wrong somewhere because i know the answer, and the equation i got doesnt work for it.
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    i keep getting answers but am not sure that they are right? could u pm ur answers to me james?
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    i have a formula and answers which were easily deduced from equation...i didnt use nCr or anything like that though...i can not see where any of ur stuff came in...if my equation and deuctions are correct (have sent them to two people), then i can see how it should be done (not my way obviously :-()
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    In each league the players are paired together in every possible combination. So you use nCr to find the number of games, ie. the total number of possible pairings.
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    (Original post by JamesF)
    In each league the players are paired together in every possible combination. So you use nCr to find the number of games, ie. the total number of possible pairings.
    you dont have to use nCr, you just have to use the sigma r formula, i.e. 1/2 n(n+1) (but you dedeuced that anyway, when you said they were triangular numbers)
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    (Original post by elpaw)
    you dont have to use nCr, you just have to use the sigma r formula, i.e. 1/2 n(n+1) (but you dedeuced that anyway, when you said they were triangular numbers)
    yeah except u use 1/2n(n-1) instead
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    (Original post by lgs98jonee)
    yeah except u use 1/2n(n-1) instead
    can i post my answer?
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    (Original post by lgs98jonee)
    can i post my answer?
    so x= no of leagues
    a= no of people in total

    no games before = a/x(a/x-1)/2
    no of games after = a/(x+1)(a/(x+1)-1)/2

    let x+1 = b
    x=m

    => a/m*(a/m-1)/2=a/b*(a/b-1)/2 +55
    =>a/m*(a-m)/m=a/b*(a-b)/b+110
    =>ab*(a-m)=am*(a-b)+110bm
    =>a^2b-abm=a^m-amb+110bm
    =>a^2b=a^2m+110bm
    =>a^2b-a^2m=+110bm
    =>a^2(b-m)=+110bm
    =>a^2(x+1-x)=+110(x(x+1))
    =>a^2=110(x(x+1)

    anyway x and a are both natural numbers...so therefore 110(x(x+1)) must be a perfect square..i.e 110=x(x+1) or 110x=(x+1) but after trying different combinations, it is clear that the only one that gives natural number solutions is 110=x(x+1) so x=10 and thus a=110...no doubt 'theone' or sum1 else will clarify why my answer is the right one as i have probably jumped the gun here :-)

    clearly 110x=x+1 can not be right as that gives x=1/109 and 55x=2x+2 gives x=2/53 so clearly there must be no x with the 110 or any coefficient larger than 1 else u get a fraction as answer? maybe this clarifies it?
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    (Original post by lgs98jonee)
    yeah except u use 1/2n(n-1) instead
    you dont have to. it's just the way you define n.
 
 
 
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