E: 01/03 £200 firebox.com gift voucher (help)

bella165

brains needed

i have no idea so dont go asking me

Mystic_Charm

Wish I could help, but it's baffled me as well. Lisa x

Munchkim

The first answer is 6 years.

To help with the tie-breaker this is how I got 6 years:

As you multiply the diameter of the orbit, the circumference increases by the same multiple so Planet 1 takes a year to come back, Planet 2 takes a year and a half to come back and Planet 3 takes 2 years to come back so the first time they are all back aligned will be in 6 years.

clairesims1978

I think its 4 years but am not 100% and I think the tiebreaker is 1+N

clairesims1978

ahhh poo I read it that 1 was half the diameter of 2 not 3... scrap my contribution.. munchkim is right !

m0jo

the answer is 4

(reference to clock times, hope it makes explaining easier)

year 1
planet 1 12.00
planet 2 09.00
planet 3 06.00

year 2
planet 1 12.00
planet 2 06.00
planet 3 12.00

year 3
planet 1 12.00
planet 2 03.00
planet 3 06.00


year 4
planet 1 12.00
planet 2 12.00
planet 3 12.00

Easy, good luck all

fraz_babe

http://www.ecmselection.co.uk/high_iq_enter_and_win/entry_form.html for the entery form. Thank you for the answer x

dawnylou

I'm really sorry to bump - but I don't understand. Can anyone explain this to me as they would a child? lol I have spent ages looking at it now, but just can't it round my head ad I really want to understand!! lol

whitty999

me also sorry for bumping but really dont understand the tiebreaker bit!
sorry for being dumb blame it on my medication!!

1984ReturnsForReal_2

The answer is 4 years

Planet 1 goes around once per year.
Planet 2 goes 3/4 per year.
Planet 3 goes around once every two years.

bigbloke45

:rotfl: But, we still need a formula for the tie breaker!

Check out the previous comps om the left hand side and their answers! I could hardly follow the workings.

For example;

brainbuster no. 24 - Guided Bus

This competition closed on Sunday 25 November 2007.

the brainbuster

"The Melchester Guided Bus system has become increasingly popular following the introduction of a controversial congestion charge scheme.

The local council has therefore decided to add further stations on the guided bus route.

Every station on the guided bus route sells tickets to every other station.

When more stations are added, 46 sets of additional tickets will be required.

How many stations will be added to the guided bus system - and how many stations are there at present?"

the solution

Here is the answer as provided by our winner:

There are currently 11 stations and 2 stations are to be added.

The working:

In a system as described containing N stations, the number of different single tickets is:

(Eqn 1): N(N-1) = N^2 - N

If M stations are added then the new number of tickets will be:

(Eqn 2): (N+M)(N+M-1) = N^2 + NM - N + MN + M^2 - M

The difference will therefore be:

(Eqn 2) - (Eqn 1) = M^2 + M(2N-1)

Which we are told is equal to 46 giving us

(Eqn 3): M^2 + (2N-1)M -46 = 0

Using the quadratic formula, to solve for M, the discriminant is (2N-1)^2 + 184.

We know from the problem that M and N must be positive integers, so the discriminant must be a perfect square.

From the fact that the discriminant must be positive, we can deduce that N must be greater than 6.

Substituting N=7 into (Eqn 3) and solving gives M = 2.89.

This is an upper bound on M as can be seen by examination of (Eqn 3). This suggests two possible solutions for M: 1,2.

Substituting M = 2 into (Eqn 3) gives N = 11. Substituting M = 1 into (Eqn 3) gives N = 23.

The problem does state that "stations" (plural) are added, so the actual solution is that there are currently 11 stations and 2 stations are to be added.


I guess they won't be getting my CV!