All Torque
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Ian
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1 - (this one was on telly, read this - http://liv.facebook.com/note.php?note_id=7159716842 )
5 + 3 x 0 = 0
If you don't employ BODMAS and to the 5 + 3 first (wrong), then multiply that result by zero, you get zero. Done properly the answer is 5.
10 - 3 x 2 = 4
Done without BODMAS you'll get 14.
No idea about the others as I'm not familiar with the notations.
6
13 portions - 4000 / 13 = 307, making accomodation 1846, business 1230, retail 615 and park land 307. I think. All approx.
7
2:5 out of 7 is 6.86:17.14 out of 24.
So 6.86 hours of night
17.14 hours of day.
10.2 x 6.86 = 69.97 planes at night.
26.4 x 17.14 = 452.49 planes in the day.
69.97 + 452.49 = 522.46 planes in 24 hours.
Hourly average = 522.46 / 24 = 21.7
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All Torque
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Ian, what was that 'locker maths question' thingy you told us at Blackpool?
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Ian
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10
Volume of a cylinder is
Area of the base x height
pi r^2 h
3.14 x (9.8 / 2)^2 x h = 450
3.14 x 4.9^2 x h = 450
3.14 x 24.01 x h = 450
75.39 h = 450
h = 450 / 75.39
= 5.97 metres.
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Ian
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quote:
Originally posted by All Torque
Ian, what was that 'locker maths question' thingy you told us at Blackpool?
100 lockers in a corridor, all shut. When you pass a locker you must toggle its state - if it is closed, open it. If it is open, close it.
Walk past all them. Toggle (open) them all. Return to the start.
Close the first. Skip the next. Close the third. Skip the next. Close the fifth. And so on. Return to the start.
Open the first. Skip the next two. Close the fourth. Skip the next two. And so on. Return to the start.
Increase the number that you skip by one every time you return.
When you have toggled the first and skipped all but the last, you are done.
Is the last locker open or closed?
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All Torque
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My GF told me the answer to this but I can't remember it or figure out the formula to the question 
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Ian
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I could give you clues, its actually not difficult when you crack the method. Might give you a while longer on it.
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All Torque
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Its divisions of 100, or something like that. If the amount of times you can open doors is even the last door is open, and vice versa
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Ian
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Yes but you need to know the rule. The problem is based upon the maths, which if you know you can see the relationships within the numbers.
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All Torque
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Can you please explain the answer and how you'd find it? 
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Ian
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When I first heard this question, I wondered how I was going to remember the state of every locker in the corridor.
This is the wrong approach. The question asks for the state of the last one.
Factoring out the other lockers means we're left with a simple problem - the answer to which simply depends on when, and indeed how many times, the last locker is toggled.
So consider this more simple scenario.
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All Torque
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Thats what I meant about numbers in 100, how many numbers don't go into 100... like 17.
Maybe I'm making it sound too tricky
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All Torque
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Open on 1
Closed on 2
Left closed on 3
Open on 4
etc...
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Ian
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To find the rule, don't worry about the state. Just find when it happens.
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Haimsey
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I so could have aced A-level maths, i just didnt want to bother making a mockery of the exam so kept hush and walked off with my B at GCSE
Marcy Marc 
White Sport Progress Thread
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All Torque
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Is it a fraction? Like... it doesnt have to be 100... maybe just 10. In which case its 1/10 and...
*feels like an idiot*
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Ian
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Not a fraction no. There are only whole numbers involved.
Make a list.
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JonnyJ
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Can i post the answer, i think i know it 
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All Torque
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Now I'm using google...
"Which numbers have an odd number of divisors? That's the answer to
this problem. Just to help you along, here are the locker numbers up
to 100 that are left open:
1,4,9,16,25,36,49,64,81,100.
See if you can describe these numbers in a different way from "having
an odd number of divisors." Think about multiplying numbers together.
When you understand how to describe them, you will see that 31 of the
100 lockers are still open (without having to work it all out!)."
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JonnyJ
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Well that ruined my big moment 
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All Torque
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:Boggle:
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All Torque
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Jonny, give this a shot:
Imagine a light bulb connected to n switches in such a way that it lights only when all the switches are closed. A push button opens and closes each switch, but you have no way of knowing which push opens and which closes. What is the smallest number of pushes required to be certain that you will turn on the light regardless of how many switches are set at the outset?
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Eck
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Ian the last one is closed.
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All Torque
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*Bins laptop*
Thanks Eck Now care to explain why?
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Ian
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Too much.
The list is
1
2
4
5
10
20
25
50
100
These are the factors of 100, ie. those which combined with another whole number can be multiplied to make 100. So you end up with:
1 x 100
2 x 50
4 x 25
5 x 20
10 x 10
20 x 5
25 x 4
50 x 2
100 x 1
Which is 9 factors, which is odd, which means the door will be in the opposite state to which you started. Perfect squares have an odd number of factors as those numbers have equal median factors, ie. a number in the middle of the list which multiplies by itself.
That list you posted 1,4,9.. is a list of perfect squares. These all have an odd number of factors. These doors are all open.
Other numbers have an even number of factors. These doors are all closed. Example, door 24:
1 x 24
2 x 12
3 x 8
4 x 6
6 x 4
8 x 3
12 x 2
24 x 1
Eight factors - closed door.
[Edited on 10-01-2008 by Ian]
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