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Makematics

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I think I got question 1, 2, 5, 6 right. Hopefully anyway lol. And hoping my partial solutions to questions 3 and 4 gets me a couple marks.
yeah i did Q1 and Q2 properly, got somewhere with 3 and 5, and nowhere with 4 and 6. Q2 was also in the junior paper... Care to explain the solutions to 5 and 6? and 3 and 4 if you got them now :p sorry, i'm just really curious to know...
 
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RealiseNothing

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yeah i did Q1 and Q2 properly, got somewhere with 3 and 5, and nowhere with 4 and 6. Q2 was also in the junior paper... Care to explain the solutions to 4 and 6?
This is what I had for question 6, I hope it's actually right lol.

Basically we want any 4 to be able to unlock it, but any 3 can not. So if we give out 4 keys for each lock, then we will have 3 people without a key for each lock. So now we work out how many combinations this can be done in, which is just:



If we have all possible combinations of groups of 4 from these 7 people, and give all groups of 4 a key to a certain lock, then any group of 4 people will be able to unlock all locks and thus the vault as at least 1 person from the group of 4 will have key for any one lock. Also since all groups of 3 are thus accounted for, for any group of 3 we choose, there will be one lock which none of them can open.
 

sbhs2013

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This is what I had for question 6, I hope it's actually right lol.

Basically we want any 4 to be able to unlock it, but any 3 can not. So if we give out 4 keys for each lock, then we will have 3 people without a key for each lock. So now we work out how many combinations this can be done in, which is just:



If we have all possible combinations of groups of 4 from these 7 people, and give all groups of 4 a key to a certain lock, then any group of 4 people will be able to unlock all locks and thus the vault as at least 1 person from the group of 4 will have key for any one lock. Also since all groups of 3 are thus accounted for, for any group of 3 we choose, there will be one lock which none of them can open.
Yeah I used 7C4, but getting 35 ways of choosing four people who can open the lock at any time didn't make much sense to me at the time. Well, considering I had 5 minutes to do this last question haha

Are you saying you got 35 locks as your answer?
 

RealiseNothing

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Yeah I used 7C4, but getting 35 ways of choosing four people who can open the lock at any time didn't make much sense to me at the time. Well, considering I had 5 minutes to do this last question haha

Are you saying you got 35 locks as your answer?
Yep.
 

RealiseNothing

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Dammit. I was going to write that, but it seemed too absurd. Did you give an example of 35 locks and the distribution of keys?

Also did you get the other questions x0x0
I just said "all possible combinations of 4 people from 7, ie ABCD, ABCE, ABCF, ... , DEFG"

I'm pretty sure I got question 1, 2, 5, and 6 right, with partial solutions to 3 and 4 (came so close to getting 3, just couldn't get it out in the time limit :( )
 

sbhs2013

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I just said "all possible combinations of 4 people from 7, ie ABCD, ABCE, ABCF, ... , DEFG"

I'm pretty sure I got question 1, 2, 5, and 6 right, with partial solutions to 3 and 4 (came so close to getting 3, just couldn't get it out in the time limit :( )
My friend got k = n for Q3, and I got a modified version of k=n, or n+1
What were you going to get, if you continued?


LOL how many people in your school did this comp?
 

RealiseNothing

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sick one! i got this result too, although i did it by 'inspection', rather than solid mathematical reasoning, if you know what i mean...
Same here, I tried to prove it with my 4 page induction :p

Came so, so close as well, was disheartening having to give up and move on as time was running out.
 

Makematics

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Lawl what school?

I was going to just write k=n+1 but when n=2, k=2 :( And I was confused and lost and I saw the clock and cry, exams tomorrow, why'm I even on this website
i got that k=3 when n=2
n=2, , which is divisible by
 
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