Maths question

is it just the first round of the tournament, or does it go into quarter-finals, semis, etc?

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?

i havent used any stats to do q

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.

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.

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.

The factor theorem is a special case of the remainder theorem, they really should teach the remainder theorem first.

isnt it obvious anyway?

isnt it obvious anyway?

Only when you've actually learned the remainder theorem.

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.

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.

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