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## Brain Teasers Q&A Game

 Quote by davee123 Should be right about 4:54:33. So... What happened to the octagon, hexagon, etc., question? I still would like to hear the answer to that one? DaveE
Post # 23.

 Quote by Gokul43201 Post # 23.
Hm. I assumed that since Mordin never confirmed it as a correct answer, but instead merely posted another question instead, that his initial question was never correctly guessed, and so he decided to make it easier. I guess I also didn't really like the arbitrarity of the question, since the secondary polygons were effectively random... Made me want to believe there was a better answer, I guess :(

DaveE
 It was Neutrino's question and he said it was correct.

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 Quote by davee123 Should be right about 4:54:33. DaveE
That's spot on, but could we have an explanation please....

P.S. Gokul, what happened to your post?
 Woohoo, that feels good. I solved it! Although DaveE found the answer before me(DaveE gets to ask the next question),I'd like to compare my technique to DaveE's. I treated the clock like a unit circle. For the hour hand the function is $$y=Sin(\frac{2\pi}{12}x)$$ For the minute hand the function is $$y=Sin(2\pi*x)$$ (Of course the hands spin the other way, but that doesn't matter, the angles between them are the same) If we add these two functions together, the zero's represent points at which the arms form some of the 90^ angles (which we don't want) and 180^ angles (which we do) So I add the two functions and get $$y=Sin(\frac{2\pi}{12}x)+Sin(2\pi*x)$$ which has a zero at 4.90909, so the hands are 180^ apart at 4 hours and .9090909x60 minutes = 54.5454 minutes and .5454*60 seconds = ~33s So the hands are exactly opposite at 4:54:33 :-)
 Blog Entries: 1 Recognitions: Gold Member Science Advisor Staff Emeritus Well done 'circles! An alternative method would be as follows; The hour hand moves 1 degree for every 12 degrees that the minute hand moves. Let the hour hand be X degree away from 5 O'clock. Therefore the minute hand is 12X degree away from 12 O'clock. Therefore solving for X Angle between minute hand and 12 O'clock + Angle between 12 O'clock and 4 O'clock + Angle between 4 O'clock and hour hand = 180 12X + 120 + (30-X) = 180 11X = 30 Hence X = 30/11 degrees (hour hand is X degree away from 5 O'clock) Now each degree the hour hand moves is 2 minutes. Therefore minutes are = 2 * 30/11 = 60/11 = 5.45 (means 5 minutes 27.16 seconds) Therefore the exact time at which the hands are opposite to each other is = 4 hrs. 54 min. 32.74 seconds Technically, its your turn 'circles since davee didn't offer an explanation. However, if you wish to nominate davee to take your turn, that's you prerogative
 Thanks Hootenanny. Well, since I am here. Here is a code I just made up :-) Try to decode it. JFYUMY NLS UAUCH edit: I don't know how hard this will be, so if noone has got it by tommorow, I will add a hint... For now, I will say that it is a very simple cipher, and is very famous as well. ;-).

Dang. Forgot my explanation. I effectively solved it by the "bash-away" method. I looked right away and saw it should be roughly 4:50-something, or thereabouts. I calculated the angles by the number of seconds that had passed total since 12:00:

Angle of minute hand:
(seconds % 3600)/10

Angle of hour hand:
(seconds*360)/43200

And a bash-at-the-answer computer program said it was at 4:54:33. So I manually verified 4:54:31, 4:54:32, 4:54:33, 4:54:34, and 4:54:35, just to double check.

 Quote by dontdisturbmycircles Thanks Hootenanny. Well, since I am here. Here is a code I just made up :-) Try to decode it. JFYUMY NLS UAUCH edit: I don't know how hard this will be, so if noone has got it by tommorow, I will add a hint... For now, I will say that it is a very simple cipher, and is very famous as well. ;-).
Hm. I get:

Each of the letters is rotated by 6 places. So an "A" becomes "G", a "B" becomes an "H", etc.

DaveE
 Very nicely done DaveE! :) Correct.

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 Quote by davee123 Hm. I get: "PLEASE TRY AGAIN" Each of the letters is rotated by 6 places. So an "A" becomes "G", a "B" becomes an "H", etc. DaveE
Well done Dave! I was sat for ages staring at that this morning! Nice question circles
 Thanks, I logged off and went to bed thinking that I made it too hard. But I guess not :)

 Quote by dontdisturbmycircles Very nicely done DaveE! :) Correct.
Cool :) Ok, next question. Hm. Alright, here's a variant of one I've heard, completely reworded and rather silly:

Four college students major in different subjects-- art history, biochemistry, law, and mathematics. In order to graduate, the dean has given them a ridiculously unfair and stupid challenge. But they're crazy college kids, so they accepted.

Together, the four students must pass four different exams: an art history exam, a biochem exam, a law exam, and a math exam, each of equal length. A room is prepared for the students with 4 desks in it, all in a row. On each desk is a copy of a different one of the exams with a blank cover sheet (the students have no idea which is which). One at a time, the students are sent into the room to take an exam. When a student enters the room, he must sit down at the desk of his choice and take one of the exams, without peeking underneath the cover sheet. When a student selects an exam, he may remove the cover sheet and look at which exam he has chosen. However, after a student selects an exam, he gets ONE chance to switch exams. If he decides to switch, he may choose another desk (again without looking under the cover sheet of the exams), and take the corresponding exam instead.

After the student takes the exam he has chosen, he is escorted out of the room into another room. He thus cannot contact any of his companions who have not already taken an exam. His exam is submitted for grading, and removed from the room. Then, any exam, desk, chair, etc. that the student altered in any way is replaced or adjusted so that it is exactly the way it was before the student entered. Precisely 2 hours after the student entered the exam room (each is given 1 hour to complete the exam), the next student is led into the room and must perform the same task.

If any one of the students fails, they all fail. Each student knows that they can pass the exam of their own major. However, they also know that none of them has any chance of passing an exam of a different topic than their own major.

While preparing for the challenge, the students first decide that their chances are slim to none. Each student has a 50% chance of choosing an exam that they can pass. So they figure that their chances are 0.5 * 0.5 * 0.5 * 0.5 = 0.0625. Pretty low. However, being brilliant young college kids, they happen upon a plan that will give them a success rate of just over 41%. Not great, but not bad, either.

What's their plan?

DaveE
 Are all methods of communication out of the question? Such as leaving a scuff mark on the way to the exam room? Because at that point I could get them to have a 50% chance of passing. This one is hard! edit: I am close... lol

 Quote by dontdisturbmycircles Are all methods of communication out of the question? Such as leaving a scuff mark on the way to the exam room?
Yep! Flat out!

For all intents and purposes, you could say that each student is led simultaneously to identically laid out rooms.

DaveE
 ok, thanks :)
 This problem is seriously inhibiting my ability to get work done today! lol. My strategy gets so complex by the end that accurately determining the probability is hard. I think I have the right idea, there is a kink somewhere though. Very good riddle DaveE. Edit, I thought of something else, I have to solve this soon darnit! :P
 I can quickly up their odds to 12.5% Let's call Art History guy A, Biochemistry guy B, Law guy L, and Math guy M. They decide to label the desk closest to the door as 1 and the furthest is 4. Alternatively, if row is aligned such that no desk is closest the door, then you choose the rightmost as 1 and the leftmost as 4. Or we could do something with cardinal directions. Anyway, we just need them labelled. A chooses between 1 and 2. 50% he's wrong, and then we've failed. Everybody else operates under the assumption he's right in his choice. B chooses between 1 and 2. Since we're assuming A was right (otherwise, we're boned anyway), he has a 25% chance of being right. 50% times 25% is 12.5%. L and M each choose between 3 and 4. A and B have to already be right, or else it doesn't matter. If they are right, then 1 and 2 are already accounted for, L and M are both guaranteed to get the right exam. This isn't good enough yet, clearly, just my start.

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