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FairyJoe
Post Posted: Tue Sep 04, 2012 11:31 pm   Post subject:
FairyJoe
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Post Posted: Tue Sep 04, 2012 11:31 pm   Post subject:
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I think you got it right, but you made a typo in your answer at a critical point, so I'm not sure. 9 * 10^17? What about the second part?
And I'm going to add another part, how does the solution generalise to a number system base n?
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charlie
Post Posted: Wed Sep 05, 2012 7:14 am   Post subject:
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Post Posted: Wed Sep 05, 2012 7:14 am   Post subject:
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edited that silly typo (was posted, half asleep on my phone at 3am...)

I think the smallest diff between them is 1.01x10^176

general rule eh?

hmm. maybe where L is number of digits, and n is base then

number of palindromes = (n-1)*n^(floor(L/2))
smallest difference between palindromes = 1.01*n^(ceiling(L/2))

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KRiSPY
Post Posted: Wed Sep 05, 2012 11:40 am   Post subject:
KRiSPY
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Post Posted: Wed Sep 05, 2012 11:40 am   Post subject:
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charlie wrote:
edited that silly typo (was posted, half asleep on my phone at 3am...)

I think the smallest diff between them is 1.01x10^176

general rule eh?

hmm. maybe where L is number of digits, and n is base then

number of palindromes = (n-1)*n^(floor(L/2))
smallest difference between palindromes = 1.01*n^(ceiling(L/2))


I got the same answer for the number of palindromes: 9*10^175
But I believe the smallest difference is: 11

I don't have a proof for it, but I believe the closest pair of digits will be the largest value starting with n and the lowest value starting with n+1

ie.
2000...0002 - 1999...9991 = 11
3000...0003 - 2999...9992 = 11
4000...0004 - 3999...9993 = 11
...
9000...0009 - 8999...8993 = 11

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FairyJoe
Post Posted: Wed Sep 05, 2012 12:06 pm   Post subject:
FairyJoe
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Post Posted: Wed Sep 05, 2012 12:06 pm   Post subject:
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Oh man, this is awesome. Great work.
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Tanryn
Post Posted: Thu Sep 06, 2012 10:27 am   Post subject:
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Post Posted: Thu Sep 06, 2012 10:27 am   Post subject:
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KRiSPY is a lot more elegant than my answer would have been as I was basing it on assumption.

Basically I assumed the number of numbers in the middle didn't matter so I did 202-191 = 11 and 2002-1991 = 11 as a confirmation and assumed everything else would all be the same.

for the other bit (351-1)/2 = 175 (As everything's mirrored around the centre number so you take away that one and half the rest. Then 9 (as the first 0 would be ignored) x10^175.


I know other people have answered this already, but this is the first time I've got back to my computer, stupid cleaning up lab that has been broken by students.
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FairyJoe
Post Posted: Thu Sep 13, 2012 7:58 am   Post subject:
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Post Posted: Thu Sep 13, 2012 7:58 am   Post subject:
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Solution to last week's problem has been posted, and it's pretty much what we got here, including Krispy's solution to the minimum difference.


New puzzle is up. Check it out.
http://www.youtube.com/watch?v=2mWsNGawjeU&feature=player_embedded

For those who can't be bothered watching the video, here's the problem:
Quote:
Imagine a corridor 1m wide by 2.5m high. It goes along for a bit then turns 90deg into another corridor 1m wide by 2.5m. What's the longest curtain rod you can get down this corridor in one piece?


My quick working out gives an answer of a 3.77m long rod. That's assuming the width of the rod is zero, in reality it will be a little shorter than that. But I'm not 100% confident in my answer, do you get the same?
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Coach
Post Posted: Thu Sep 13, 2012 8:30 am   Post subject:
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Post Posted: Thu Sep 13, 2012 8:30 am   Post subject:
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Curtain rods can have a bit of give, or are we to assume it's completely rigid?
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Tanryn
Post Posted: Thu Sep 13, 2012 11:26 am   Post subject:
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Post Posted: Thu Sep 13, 2012 11:26 am   Post subject:
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I'm not confident with my answer either and I get 2.87m (or a touch shorter due to the width of the, assumed rigid) pole. I've probably missed something obvious though.
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Kootwam
Post Posted: Thu Sep 13, 2012 12:02 pm   Post subject:
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Post Posted: Thu Sep 13, 2012 12:02 pm   Post subject:
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It's a trick question! Electric rods don't make smoke!!!

No really, I get the same as you Joe, though I feel that it may be longer as (assuming we calculated the same way) we are only looking at 2 planes? Not sure hmmmmm :)
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SuperRatBum
Post Posted: Thu Sep 13, 2012 1:34 pm   Post subject:
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Post Posted: Thu Sep 13, 2012 1:34 pm   Post subject:
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Kootwam wrote:
It's a trick question! Electric rods don't make smoke!!!

No really, I get the same as you Joe, though I feel that it may be longer as (assuming we calculated the same way) we are only looking at 2 planes? Not sure hmmmmm :)


I also get 3.77m when it goes from the bottom corner to the top so I assume it would be about 5 to 10 cm shorter than the length you suggest Joe to account for width. I double checked my answer with google sketch-up and it checks out there also.
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charlie
Post Posted: Thu Sep 13, 2012 1:48 pm   Post subject:
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Post Posted: Thu Sep 13, 2012 1:48 pm   Post subject:
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I get the same answer as Joe, and I even did some analysis to make sure that the angle I used produced a minimum! (thanks wolframalpha.com, my graphics calc battery is dead/redundant)
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Tanryn
Post Posted: Fri Sep 14, 2012 8:56 am   Post subject:
Tanryn




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Post Posted: Fri Sep 14, 2012 8:56 am   Post subject:
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While 3.77m is the longest length a straight line can be I'm convinced you'd get wedged with one end on the skirting board, one on the cornices and your pole stuck on the pivot point, infact, I'm not sure you would even get into that position.

If you shave it back a few cm you might be able to do it, assuming you don't mind scraping the paint on the pivot point of the corner of the wall, but having moved house a few times and knowing that standard window frames for a wide window is about 2.7m I'm going with my first answer of 2.87m as being more applicable to the situation.

(Okay, I'll allow you might want fancy curtains and you want them hung well back from the window to give more room for the sunlight in which case you could have up to 3.3m, but if you get larger than that then the curtains are going to be way too expensive. Also, you won't want to get them made at Spotlight as they'll take forever and then end up running out of material and you'll have to change your mind as to what curtains you want halfway through and it won't turn out as you originally envisaged it and then they'll never give you the $100 store credit that they promised you)
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SuperRatBum
Post Posted: Sat Sep 15, 2012 11:37 am   Post subject:
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Post Posted: Sat Sep 15, 2012 11:37 am   Post subject:
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I had a look at 10cm increments from the 3.77m as you go through the corner. The shortest area is exactly on that figure and for each 10cm you move round the corner you gain a 1 cm gap so I think it would work. But as you say the question is a bit limited as they needed to give you more specifications about the curtain rail.
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FairyJoe
Post Posted: Sat Sep 15, 2012 6:38 pm   Post subject:
FairyJoe
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Post Posted: Sat Sep 15, 2012 6:38 pm   Post subject:
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SuperRatBum wrote:
they needed to give you more specifications about the curtain rail.

Given that it's from a pure-mathematics website, and the nature of the previous problem, I think it's safe to assume the problem is obviously simplified. The question is getting at what's the smallest theoretical length of a 0width, rigid rod in a frictionless environment. I believe the appropriate physics analogy is "Imagine all cows are spherical..."
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Tanryn
Post Posted: Mon Sep 17, 2012 8:29 am   Post subject:
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Post Posted: Mon Sep 17, 2012 8:29 am   Post subject:
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FairyJoe wrote:
I believe the appropriate physics analogy is "Imagine all cows are spherical..."


They would be an absolute bugger to milk...
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