Maths question

Year 12s: what will be the first way you research your future options? >>

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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 :biggrin:

. I'm not certain, if by teams you mean no. of competitors, then I think you've got your algebra wrong.

Reply 61

elpaw

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.

Reply 62

makesomenoise

elpaw

I prefer 'algebraic juggling', its a less messy method.

How's that work?

Reply 63

JamesF

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 :biggrin:

. 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.

Reply 64

lgs98jonee

i keep getting answers but am not sure that they are right? could u pm ur answers to me james?

Reply 65

lgs98jonee

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 :-()

Reply 66

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.

Reply 67

elpaw

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)

Reply 68

lgs98jonee

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

Reply 69

lgs98jonee

yeah except u use 1/2n(n-1) instead

can i post my answer?

Reply 70

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?