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    (Original post by elpaw)
    is it just the first round of the tournament, or does it go into quarter-finals, semis, etc?
    The number of games refers only to the number of games in the league stage of the tournament.
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    (Original post by bono)
    But you would need a sort of nth term formula to calculate the number of games played per group, depending on the number of players in that group.

    That would be the first step? :confused:
    Yeah, that's pure maths
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    (Original post by lgs98jonee)
    i havent used any stats to do q
    i havent dunnit i hasten to add
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    In the original grouping, say we had n groups of a people, and so in the second there were n+1 groups of b people.
    So we have total number in tournament =
    na = (n+1)b
    And total number of games
    n(aC2) for the first grouping, and
    (n+1)(bC2) for the second.
    We are given that

    n(aC2) = 55 + (n+1)(bC2)

    Now, writing out the triangle thingy, u can see that aC2 are triangular numbers. The nth triangular number is given by the sum of all numbers from 1 to n. And after some manipulation, you get

    n(a^2-a)/2 = 55 + (n+1)(b^2-b)/2
    n(a^2 - a - b^2 + b) = 110 + b^2 - b

    Use na = (n+1)b, then you get

    b(n^2(1+b) + n + b) = 110

    Now, 110 = 2x5x11, so b must be either 2, 5, 11, 10, 22, 55, 110

    Try each case.


    I expect ive gone wrong in the algebra somewhere tho.
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    (Original post by JamesF)
    In the original grouping, say we had n groups of a people, and so in the second there were n+1 groups of b people.
    So we have total number in tournament =
    na = (n+1)b
    And total number of games
    n(aC2) for the first grouping, and
    (n+1)(bC2) for the second.
    We are given that

    n(aC2) = 55 + (n+1)(bC2)

    Now, writing out the triangle thingy, u can see that aC2 are triangular numbers. The nth triangular number is given by the sum of all numbers from 1 to n. And after some manipulation, you get

    n(a^2-a)/2 = 55 + (n+1)(b^2-b)/2
    n(a^2 - a - b^2 + b) = 110 + b^2 - b

    Use na = (n+1)b, then you get

    b(n^2(1+b) + n + b) = 110

    Now, 110 = 2x5x11, so b must be either 2, 5, 11, 10, 22, 55, 110

    Try each case.


    I expect ive gone wrong in the algebra somewhere tho.
    i did something similar but couldnt get any of the values to work...surely as there are 55 fewer matches with seconding grouping, b should be 5 or 11 but i am tried so am going to bed :-(
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    (Original post by JamesF)
    In the original grouping, say we had n groups of a people, and so in the second there were n+1 groups of b people.
    So we have total number in tournament =
    na = (n+1)b
    And total number of games
    n(aC2) for the first grouping, and
    (n+1)(bC2) for the second.
    We are given that

    n(aC2) = 55 + (n+1)(bC2)

    Now, writing out the triangle thingy, u can see that aC2 are triangular numbers. The nth triangular number is given by the sum of all numbers from 1 to n. And after some manipulation, you get

    n(a^2-a)/2 = 55 + (n+1)(b^2-b)/2
    n(a^2 - a - b^2 + b) = 110 + b^2 - b

    Use na = (n+1)b, then you get

    b(n^2(1+b) + n + b) = 110

    Now, 110 = 2x5x11, so b must be either 2, 5, 11, 10, 22, 55, 110

    Try each case.


    I expect ive gone wrong in the algebra somewhere tho.
    isnt it MINUS 55 not +??
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    I dunno, it doesnt work anyway.
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    You're close, 110 is on the right lines :d
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    damn, ive got a polynomial with T and n. (T:=total n:=number-per-league)

    now ive just got to extract the "n and T are whole numbers" solution
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    (Original post by elpaw)
    damn, ive got a polynomial with T and n. (T:=total n:=number-per-league)

    now ive just got to extract the "n and T are whole numbers" solution
    I've done Polynomials and Factor Theorem etc. in P2.
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    (Original post by bono)
    I've done Polynomials and Factor Theorem etc. in P2.
    The factor theorem is a special case of the remainder theorem, they really should teach the remainder theorem first.
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    (Original post by ZJuwelH)
    The factor theorem is a special case of the remainder theorem, they really should teach the remainder theorem first.
    isnt it obvious anyway?
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    (Original post by elpaw)
    isnt it obvious anyway?
    We haven't done that, we did Polynomials and Factor Theorem in P2.

    What is it by the way?
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    (Original post by elpaw)
    isnt it obvious anyway?
    Only when you've actually learned the remainder theorem.
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    (Original post by ZJuwelH)
    Only when you've actually learned the remainder theorem.
    Which is?
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    (Original post by bono)
    Which is?
    "When a polynomial f(X) is divided by (X-A) the remainder is f(A)"

    And the factor theorem:

    "If f(X) is a polynomial and f(A)=0, then (X-A) is a factor of f(X)"

    That is, the factor theorem is the remainder theorem with no remainder.
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    (Original post by ZJuwelH)
    "When a polynomial f(X) is divided by (X-A) the remainder is f(A)"

    And the factor theorem:

    "If f(X) is a polynomial and f(A)=0, then (X-A) is a factor of f(X)"

    That is, the factor theorem is the remainder theorem with no remainder.
    Will I be doing the remainder theorem in my next pure maths module then???
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    (Original post by bono)
    Will I be doing the remainder theorem in my next pure maths module then???
    For me it's in P2. Maybe P3 for you. It may involve long division which is a pain. The theorems from what I quoted aren't too clear on their own I know, but a few examples later all is easy. Don't worry about it.
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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. The theorems from what I quoted aren't too clear on their own I know, but a few examples later all is easy. Don't worry about it.
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    (Original post by elpaw)
    damn, ive got a polynomial with T and n. (T:=total n:=number-per-league)

    now ive just got to extract the "n and T are whole numbers" solution
    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'??
 
 
 
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