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Maths question help - oxford exam tomorrow

POSTED BY: James... 3 weeks, 2 days ago

Got my oxford university entrance test tomorrow and hate this question on an old paper. Getting really stressed now haha and would reallly appreciate an answer. Got to be without calculators too.

Question......

There are 1000 doors in a line. All are initially closed.

There are also 1000 students.

The first changes the state of every door, ie, opens them all.

The second person changes the state of every second door, ie, will close all the even numbers.

The third person changes the state of every third door, ie, if the third is open, they will close it, if the 6th is closed they will open it.

So on and every student changes the state of every door their number is a factor of.

You have to find how many doors are open after the 1000th person has been.

Does anyone know a way to do this without writing everything out incase something similar comes up tomorrow?

I think the second part said what if there were 2000 people and 2000 doors or something so there must be a formula of some sort for it.

Cheers
James

I'm back off to revise :'(

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#2

03-11-2009, 18:12
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03-11-2009, 18:17
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54 million?
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#4

03-11-2009, 18:17
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That cant be right as the first door will be opened by the first person and not touched by anyone else?
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#5

03-11-2009, 18:20
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1x1, 2x2, 3x3...... Google is your pal
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#6

03-11-2009, 18:21
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great question....would probably be able to work it out if i had the whole exam time for that question alone!
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#7

03-11-2009, 18:27
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If you can't work it out, break it down to a simple problem and then extend the rule and check. 1000 is just to make life difficult for the initial thoughts. I assume it's a sum to infinity type question as the 1000 person only has the ability to change the status of one door, whilst person 2 changes 500.

Unfortunately being able to answer that question tonight will not help your chances in the entrance exam tomorrow. I wouldn't get wound up about it - you should look at trying some more pattern based maths.
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#8

03-11-2009, 18:28
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work it out for the first few cases. then work out the nth rule case. then add in 1000 for n

= easy

but do it yourself.
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03-11-2009, 19:06
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Am I right in thinking it would be 31 as it looks like it changes at every square number?

As in, 1 door is open for 1,2 and 3.

when you get to 4-8 doors/people 2 are left open,

3 left open for 9 and 10 which I assume would go to 15 and change to 4 from 16-24, 5 from 25-35 and so on.

And because 1000 lies between the square numbers 961 (31^2) and 1024 (32^2) there will be 31 doors left open for 961-1023 number of doors/people?
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03-11-2009, 19:20
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1/4 past 2
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#11

03-11-2009, 19:26
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James...:
Am I right in thinking it would be 31 as it looks like it changes at every square number?

As in, 1 door is open for 1,2 and 3.

when you get to 4-8 doors/people 2 are left open,

3 left open for 9 and 10 which I assume would go to 15 and change to 4 from 16-24, 5 from 25-35 and so on.

And because 1000 lies between the square numbers 961 (31^2) and 1024 (32^2) there will be 31 doors left open for 961-1023 number of doors/people?
Sounds correct to me. Floor root n.
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#12

03-11-2009, 19:27
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Cheers, its made my night haha! when you say floor root n, does it mean root n rounded down to the nearest interger?
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03-11-2009, 19:28
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James...:
Cheers, its made my night haha! when you say floor root n, does it mean root n rounded down to the nearest interger?
Thats what I meant, its not the correct maths terminology though, just programmer speak! What are you hoping to study?

NB Not positive thats right but it certainly seems to make sense! [and Shengis/Google agrees! ]
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#14

03-11-2009, 19:30
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#15

03-11-2009, 19:54
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Thanks alot! I've applied for the 3 year BSci Mathematics course at Oxford, Bath, Bristol, Sheffield and Liverpool. Love the subject and cant wait to get to Uni now!!!
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