Silliness Quiz - Page 7

SolRaven · 12-12-2010, 11:10 PM

Quote:

Originally Posted by SneakySnake

Would it be at midday? It seems to take all the daylight hours to complete the journey. We could imagine two suns rising, travelling, and setting. At what point would they pass?

Methinks the time has come for TGS to post a full and rational explanation.

please!

I agree. How can you prove it even if it's possible? And how to prove a location? Midday would prove a time but not a location.

GraceKrispy · 12-13-2010, 01:21 AM

Ok, here's my scientific answer: midday, slightly below midway, and I know because the monk calls me on his droid x and says "Hey, Grace, so I just passed the point I was at exactly at this time a few days ago! cool, eh? I know because I left my Bible here because it was getting too heavy to carry up and I have the passages memorized anyway. Don't forget to bring my other brown cloak and a bell when you pick me up tonight!"

And if I don't answer, he'll swype it to me, and probably get a few words wrong, so I may bring Nell instead, and perhaps some brown coal instead.

SneakySnake · 12-13-2010, 01:44 AM

Gracie, I think that you are the winner. Hopefully TGS will get out of bed and confirm it.

Cheers

TGS · 12-13-2010, 04:27 AM

Ah, the question was, can you prove that there is a place where the monk is at the same time of day on the way up as on the way down - not to say what that place is. Now, there are mathematical solutions to the problem involving Brouwer's Fixed Point Theorem, which are virtually useless for actually understanding the solution. Much better is to use your intuition - imagine that a monk starts from the bottom of the mountain at sunrise, and another monk starts from the top of the mountain at sunrise. Now you can simply "see" that these two monks will pass each other and, hence, be at the same place on the path at the same time - though you don't know where or when, (but that's not what you are asked). From this you can also simply "see" that if the journeys take place on different days this makes no difference to whether the monks pass the same place at the same time of day.

Hope that helps. Apparently Arthur Koestler is responsible for this puzzle.

Javed · 12-13-2010, 05:01 AM

My answer is no. I can not prove it, he could have dematerialized at some stage.

SneakySnake · 12-13-2010, 05:31 PM

I can only prove it with my imagination. Or a mathematical equation. Or a diagram. All require spatial concepts that a mere snake cannot understand.

poohbear_nc · 12-13-2010, 06:27 PM

Quote:

Originally Posted by TGS

Ah, the question was, can you prove that there is a place where the monk is at the same time of day on the way up as on the way down - not to say what that place is. Now, there are mathematical solutions to the problem involving Brouwer's Fixed Point Theorem, which are virtually useless for actually understanding the solution. Much better is to use your intuition - imagine that a monk starts from the bottom of the mountain at sunrise, and another monk starts from the top of the mountain at sunrise. Now you can simply "see" that these two monks will pass each other and, hence, be at the same place on the path at the same time - though you don't know where or when, (but that's not what you are asked). From this you can also simply "see" that if the journeys take place on different days this makes no difference to whether the monks pass the same place at the same time of day.

Hope that helps. Apparently Arthur Koestler is responsible for this puzzle.

I knew it involved math!

TGS · 12-14-2010, 05:13 AM

Quote:

Originally Posted by poohbear_nc

I knew it involved math!

Well, only in the same way that the question, which of the two letter "A"s bellow is the larger, involves maths:

A A

SneakySnake · 12-14-2010, 05:31 PM

neither! because you have used capitals in your answer choices and a lowercase a in your question! Hee hee

My turn!

Which of these numbers does not belong in the following list? :...

1, 4, 7, 9, 11, 14.

nikkie · 12-15-2010, 01:43 AM

14 does not belong. All the others have only have only one unique divisor (not including the identity).

1x1=1
2x2=4
7*1=7
3*3=9
11*1=1
7*2=14 X

SneakySnake · 12-15-2010, 04:05 AM

Quote:

Originally Posted by nikkie

14 does not belong. All the others have only have only one unique divisor (not including the identity).

1x1=1
2x2=4
7*1=7
3*3=9
11*1=1
7*2=14 X

That's a good answer, but not the one I was looking for.
it is a lot simpler!

SneakySnake · 12-15-2010, 04:06 AM

To fix you up. How about I ask it again.
only the list =
1, 4, 7, 9, 11, 17.

TGS · 12-15-2010, 04:07 AM

9? (is too short a post apparently)

SneakySnake · 12-15-2010, 04:08 AM

Quote:

Originally Posted by TGS

9? (is too short a post apparently)

Yes 9 is the answer. Do you know why?

TGS · 12-15-2010, 04:10 AM

Intuitively, with no workings out, it might have something to do with the addition of 3?

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12-13-2010, 01:21 AM	#92
GraceKrispy It's Dr. Penguin now! Posts: 3,909 Karma: 4705733 Join Date: Jan 2009 Location: (USA) Device: iPad mini, Samsung Note 3, Sony PRS-650 (rarely used now)	Ok, here's my scientific answer: midday, slightly below midway, and I know because the monk calls me on his droid x and says "Hey, Grace, so I just passed the point I was at exactly at this time a few days ago! cool, eh? I know because I left my Bible here because it was getting too heavy to carry up and I have the passages memorized anyway. Don't forget to bring my other brown cloak and a bell when you pick me up tonight!" And if I don't answer, he'll swype it to me, and probably get a few words wrong, so I may bring Nell instead, and perhaps some brown coal instead.

12-13-2010, 01:44 AM	#93
SneakySnake I'm watching you! Posts: 6,113 Karma: 22344652 Join Date: Mar 2010 Location: Sunny Coast Qld, OZ Device: Sony PRS-900(unused lately) iPadAir2, want me Kindle Oasis	Gracie, I think that you are the winner. Hopefully TGS will get out of bed and confirm it. Cheers

12-13-2010, 04:27 AM	#94
TGS Country Member Posts: 9,058 Karma: 7676767 Join Date: Feb 2010 Location: Denmark Device: Liseuse: Irex DR800. PRS 505 in the house, and the missus has an iPad.	Ah, the question was, can you prove that there is *a place* where the monk is at the same time of day on the way up as on the way down - not to say what that place is. Now, there are mathematical solutions to the problem involving Brouwer's Fixed Point Theorem, which are virtually useless for actually understanding the solution. Much better is to use your intuition - imagine that a monk starts from the bottom of the mountain at sunrise, and another monk starts from the top of the mountain at sunrise. Now you can simply "see" that these two monks will pass each other and, hence, be at the same place on the path at the same time - though you don't know where or when, (but that's not what you are asked). From this you can also simply "see" that if the journeys take place on different days this makes no difference to whether the monks pass the same place at the same time of day. Hope that helps. Apparently Arthur Koestler is responsible for this puzzle.

12-13-2010, 05:01 AM	#95
Javed Grand Sorcerer Posts: 10,270 Karma: 1126878541 Join Date: Oct 2009 Device: Astak Pocket PRO, iPod Touch, PRS-650	My answer is no. I can not prove it, he could have dematerialized at some stage.

12-13-2010, 05:31 PM	#96
SneakySnake I'm watching you! Posts: 6,113 Karma: 22344652 Join Date: Mar 2010 Location: Sunny Coast Qld, OZ Device: Sony PRS-900(unused lately) iPadAir2, want me Kindle Oasis	I can only prove it with my imagination. Or a mathematical equation. Or a diagram. All require spatial concepts that a mere snake cannot understand.

12-14-2010, 05:31 PM	#99
SneakySnake I'm watching you! Posts: 6,113 Karma: 22344652 Join Date: Mar 2010 Location: Sunny Coast Qld, OZ Device: Sony PRS-900(unused lately) iPadAir2, want me Kindle Oasis	neither! because you have used capitals in your answer choices and a lowercase a in your question! Hee hee My turn! Which of these numbers does not belong in the following list? :... 1, 4, 7, 9, 11, 14.

12-15-2010, 01:43 AM	#100
nikkie Guru Posts: 614 Karma: 73700 Join Date: Oct 2009 Location: WA, USA Device: Android, Kindle Paperwhite, lots of ancient readers	14 does not belong. All the others have only have only one unique divisor (not including the identity). 1x1=1 2x2=4 71=7 33=9 111=1 72=14 X

12-15-2010, 04:06 AM	#102
SneakySnake I'm watching you! Posts: 6,113 Karma: 22344652 Join Date: Mar 2010 Location: Sunny Coast Qld, OZ Device: Sony PRS-900(unused lately) iPadAir2, want me Kindle Oasis	To fix you up. How about I ask it again. only the list = 1, 4, 7, 9, 11, 17.

12-15-2010, 04:07 AM	#103
TGS Country Member Posts: 9,058 Karma: 7676767 Join Date: Feb 2010 Location: Denmark Device: Liseuse: Irex DR800. PRS 505 in the house, and the missus has an iPad.	9? (is too short a post apparently)

12-15-2010, 04:10 AM	#105
TGS Country Member Posts: 9,058 Karma: 7676767 Join Date: Feb 2010 Location: Denmark Device: Liseuse: Irex DR800. PRS 505 in the house, and the missus has an iPad.	Intuitively, with no workings out, it might have something to do with the addition of 3?

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