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# Maths question watch

1. (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.
2. (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?
Yeah, that's pure maths
3. (Original post by lgs98jonee)
i havent used any stats to do q
i havent dunnit i hasten to add
4. 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.
5. (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 :-(
6. (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 +??
7. I dunno, it doesnt work anyway.
8. You're close, 110 is on the right lines :d
9. 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
10. (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.
11. (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.
12. (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?
13. (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?
14. (Original post by elpaw)
isnt it obvious anyway?
Only when you've actually learned the remainder theorem.
15. (Original post by ZJuwelH)
Only when you've actually learned the remainder theorem.
Which is?
16. (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.
17. (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???
18. (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.
19. (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.
20. (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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