Brain Teasers Q&A Game - Answer the First Question Now!

[Hootenanny]

And I can do nothing but say that you are correct.

Thank you

At what time immediately prior to Six O'clock the hands of the clock are exactly opposite to each other?

Give the exact time in hours, minutes and seconds (nearest second).

 

[Jimmy Snyder]

He always thinks of new and exotic ways of doing so and there are no two alike

He's not all that bad. It seems that he has killed no more than two people.

 

[Moridin]

He's not all that bad. It seems that he has killed no more than two people.

The ideas for execution method the emperor has is different, the actual deaths are of course very similar.

The current question seems interesting. I'm pretty sure about the position of the hour hand, but not much more.

 

[davee123]

At what time immediately prior to Six O'clock the hands of the clock are exactly opposite to each other?

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

 

[Gokul43201]

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.

 

[davee123]

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

 

[dontdisturbmycircles]

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

 

[Hootenanny]

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?

 

[dontdisturbmycircles]

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 :-)

 

[Hootenanny]

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

 

[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 no one 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. ;-).

 

[davee123]

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.

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 no one 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:
"PLEASE TRY AGAIN"

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

DaveE

 

[dontdisturbmycircles]

Very nicely done DaveE! :) Correct.

 

[cristo]

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

 

[dontdisturbmycircles]

Thanks, I logged off and went to bed thinking that I made it too hard. But I guess not :)

 

[davee123]

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 anyone 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

 

[dontdisturbmycircles]

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

 

[davee123]

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

 

[dontdisturbmycircles]

ok, thanks :)

 

[dontdisturbmycircles]

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

 

[Your.Master]

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.

 

[dontdisturbmycircles]

We have 4 guys, art history, biochemistry, law, and mathematics.

Let me rename them please because I named them differently on my piece of paper and I don't want to convert the names

I named them Math, science, biology, and chemistry.

The main thing that I was going for is to ensure that they all sit at different tests while being flexible.

The plan is this. The Math guy goes in, and flips over paper #1.


When Math guy sits at paper #1

If it is math, he sits down and writes his test

If it is science, he goes to #2

If it is bioogy, he goes to # 3

If it is Chemisty, he goes to #4

When the Science guy sits down, he goes to paper #2

If it is math, he goes to #1

If it is science, he sits and writes his test

If it is chemistry, he goes to #4

If it is Biology, He goes to #3

When the Biology guy goes to write his test, he goes to table #3

If it is math, he goes to #1

If it is science, he goes to #2

If it is biology, he sits and writes his test

If it is Chemitry, he goes to #4

When the Chemistry guy goes, he looks at paper #4 first

If it is math, he goes to #1

If it is science, he goes to #2

If it is Bio, he goes to #3

If it is Chemistry, he sits down and writes his test.

I think that solving the probability for this strategy in itself is a brain teaser... lol, I know that they for sure have a 1/6 chance of a guaranteed win (math and science compose first 2 booklets) I can also see that if I can get the first one seated correctly, and that the second one sits at a different desk, then his chances of selecting the right desk is higher because when he first picks up the paper and finds out where he should sit, if the first guy is seated correctly, then there is only 2 seats to choose from. I am going to bed, goodnight, good luck to anyone who is trying to solve this :P.

Edit: Thought of a way to test probability... doing it now.

 

[dontdisturbmycircles]

I am correct, the probability of success using my system is 41.66666%. I will post proof in a second.

EDIT:

Proof(Not the most elegant/simple proof but it is a 'proof' none the less) You can either work through it or trust me! :-)):

Please note that the B with the erasing marks should be a C.

http://img207.imageshack.us/img207/2351/math01002sj1.jpg

http://img207.imageshack.us/img207/2171/math01003yk4.jpg

The underlined ones are what actually happens in the given situations. I tally up the successes in the right side and divide the successes by the total number in the sample space. Proving that my system will work 10 out of 24 of the possible times. Hurrah! That was one of the best puzzles I have ever seen DaveE, seriously.

Oh, and excuse the lack of a % sign after the 41 2/3 :P.

I don't know why it took me so long to think of writing out the entire sample space...

M=Math
S=Science
B=Biology
C=Chemistry

I started to use different names and then never changed. But obviously it is the same solution. :P

 

[davee123]

the probability of success using my system is 41.66666%.
...
Proving that my system will work 10 out of 24 of the possible times. Hurrah!

Correct!

Now, the thing that totally blew *my* mind about this problem was that even if the remaining students are allowed to watch *everything* that's going on in the test room (so they can see which tests are which as the current exam-taker reveals them), they STILL have a 41.666% chance of success! The only thing that it saves them is the number of wrong guesses! Explanation:

Possible situations (with other students watching):
A) 1st student picks wrong, then wrong = 3/4 * 2/3 = 1/2
nothing else matters, they fail

B) 1st student picks wrong, then correct = 3/4 * 1/3 = 1/4
now, the remaining students know the identity of one of the remaining tests, and that student can go straight for it without guessing. The two other students know that they can go for the other two options, and are guaranteed success.

C) 1st student picks correctly on his first try = 1/4
now, the remaining students have learned very little. New student goes in:
C1) 2nd student picks correctly on his first try = 1/3 (victory!)
C2) 2nd student picks correctly on his second try = 2/3 * 1/2 = 1/3 (victory!)
C3) 2nd student picks incorrectly on both = 2/3 * 1/2 = 1/3 (failure!)

So overall chance of success is the chance of B (1/4) plus the chance of success in C (1/4 * (1/3 + 1/3)). So 1/4 + 1/6 = 10/24 = 41.666%!

DaveE

 

[dontdisturbmycircles]

Thanks DaveE. I am not very good at probability so I just solved it from a logical standpoint. I knew that even if the test writers can't tell the others what they did, the future test writers can tell what they did by looking at the tests. For example, when the second guy goes in and looks at the second test. If it is not math then he knows that either they already failed or that the first desk was math. So he wants to go to a different desk obviously because if two of them write the same test, they fail.

Then for the third and fourth guys, they also know what happened. If third guy walks in and the 1st guy's test is on his desk, he knows that either they screwed it up, or the 3rd guy's test is sitting on the first desk. So yea, two different ways to solve it I guess. The only difficult part for me was trying to prove the probability... I need to improve my probability skills.

I'll think of another problem in a bit.

 

[dontdisturbmycircles]

Here is my main puzzle

Q: Place the same two letters in the exact centre of each word so that five longer words are formed. Which two letters must be used?

CARS EARS LAST PANT WARS

 

[Jimmy Snyder]

Q: Place the same two letters in the exact centre of each word so that five longer words are formed. Which two letters must be used?

CARS EARS LAST PANT WARS

Answer (hilite to see) ET works. eom.

 

[dontdisturbmycircles]

Nicely done JimmySnyder. Your turn to ask a question.

 

[vamfun]

Hi all, I wanted to post this comment re the second teasor. Is there a way to have it appear in a relavent place rather than here?

"I agree with this answer... But it must allow 0 hairs. I would reason a little differently.
To get colony started 1st man must have 0 hair to meet condition 3.
The second man must have 1 hair to meet conditions 1 and 3
The third man must have 2 hairs to meet conditions 1 and 3 etc
So there is always one more man than the max number of hairs on a head and the number of hairs increases by one with each new man added.

Since no man can have exactly 483207 hairs then the population stops at 483207 with the last man having exactly 483207-1 hairs."