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

[Hootenanny]

The aim of the game is simple (similar to the Math Q&A game). I will post a brain teaser, the first person to give the correct answer gets to set the next question, the first person to then answer this question correctly gets to set the following question... ad infinitum. A few rules before we begin;

1. The person who sets the brain teaser has the final and only decision as to who answered the question correctly.
2. No new question may be posted until the person who set the current question has indicated who has answered the question correctly first.
3. The person who was identified as answering the brain teaser correctly, by the person who set the current question, may then post one of their own brain teasers.
4. Each answer must be accompanied by an explanation. Any answer, correct or otherwise, which is not qualified with an explanation shall be disqualified.
5. The person whom set the question may at his/her discretion post the solution to the brain teaser which he/she has set and nominate another person to set the next question.

Let the games begin ...

First Question

A bear walks south for one kilometer, then it walks west for one kilometer, then it walks north for one kilometer and ends up at the same point from which it started. What color was the bear?

 

[dontdisturbmycircles]

He must have been standing at the north pole in order to arrive at the same point. Thus he was white because he was a polar bear? lol.

 

[Hootenanny]

He must have been standing at the north pole in order to arrive at the same point. Thus he was white because he was a polar bear? lol.

Correct! Your turn...

 

[neutrino]

>>White - It's a polar bear, and it probably started walking from the north pole. [/color]<<

I don't have any interesting brain teasers right now . Can I let someone else post a question? (in the case that I've posted the correct answer before anyone else does)

(Just can't resist answering questions, you know )

 

[neutrino]

Great! I have the right answer and I don't have to post a questions!

Edit: Why can't it be a brown bear, or black bear that was transferred to the Arctic for some experiment?

 

[dontdisturbmycircles]

I am going to need to be solving this as you are... I was caught unprepared lol, but I found one that looks sort of interesting, so here goes.

In the city of Funkytown, the following facts are true:

No two inhabitants have exactly the same number of hairs.
No inhabitant has exactly 483,207 hairs.
There are more inhabitants than there are hairs on the head of anyone inhabitant.
What is the largest possible number of inhabitants of Funkytown?

 

[dontdisturbmycircles]

I think this riddle is a dud, I don't see a limit? lol am I just tired?

 

[Hootenanny]

I think I have it. The riddle states that there is no one with 483,207 hairs, therefore we can say that there may be 483,207 inhabitants with between [0, 483 206] hairs on their head. However, if there were more than 483,207 inhabitants this would violate the third condition. Therefore, the maximum number of inhabitants is 483,207.

What say you?

 

[dontdisturbmycircles]

I see the answer as being 0. In the minimum case there is 1 man with 1 hair, that causes there to also be another man with more than one hair, which causes there to be another man with more than 2 hairs, add infinitum. So 0 is the only plausible answer.

edit: hold on let me read your post, didn't see it.

 

[neutrino]

I see the answer as being 0. In the minimum case there is 1 man with 1 hair, that causes there to also be another man with more than one hair, which causes there to be another man with more than 2 hairs, add infinitum. So 0 is the only plausible answer.

That was nice.

edit: How about this case, one man with one hair, and another, who is bald. Conditions satisfied, no?

edit2: Is it specified that there are no bald inhabitants?

 

[dontdisturbmycircles]

edit:double post

 

[dontdisturbmycircles]

Makes sense to me, your turn. Thank god, I can go to sleep now. Good night! :-)

I didn't even think of including a bald guy. lol

If every man had at least 1 hair, then 0 would be right. :P

 

[Hootenanny]

I didn't even think of including a bald guy. lol

I very nearly didn't; I was about to submit my answer when I re-read the conditions.

Anyway the next teaser;

Find the remainder if we divide {\color{blue}1!+2!+3!+\ldots +100!} by 7

 

[neutrino]

Thankfully, there is a weekly that we subscribe to which has this little teaser in every issue. Most of them are simple enough.

Fill in the blank so that line 4 is consistent with the first three.

Triangle, Pentagon, Square

Square, Hexagon, Hexagon, Square

Pentagon, Hexagon, Hexagon, Hexagon, Square, Triangle

Hexagon, Octagon, Octagon, Octagon, Octagon, __________

 

[Hootenanny]

edit: How about this case, one man with one hair, and another, who is bald. Conditions satisfied, no?

Except we are to find the maximum number of inhabitants.

 

[neutrino]

Oops! Didn't notice Hoot's post. I though 'circles was referring to me.

Anyway, I'm not planning to delete the stuff I typed. You can solve this, too, "outside" the game.

 

[davee123]

Find the remainder if we divide {\color{blue}1!+2!+3!+\ldots +100!} by 7

Since all the terms from 7! to 100! are evenly dividisble by 7, we only need to find the remainder of (1!+2!+3!+4!+5!+6!)/7, which is 5 :)

DaveE

 

[dontdisturbmycircles]

edit:woops I did this wrong, lol. I really need to go to bed. goodnight

 

[Hootenanny]

Since all the terms from 7! to 100! are evenly dividisble by 7, we only need to find the remainder of (1!+2!+3!+4!+5!+6!)/7, which is 5 :)

DaveE

Correct! Your turn davee...

I am just now learning about factorial stuff, but I think I can answer this. Noting that all numbers above 7! are divisible by 7, we have 1! 2!... 6! to deal with

\frac{1!+2!+3!...6!}{7}=124.7

edit:woops derr I did this wrong, I was supposed to take the remainder of each term. lol. I really need to go to bed. goodnight

Yeah, you nearly had it...till davee stole your thunder!

 

[davee123]

Correct! Your turn davee...

Alright... What letter comes next?

o, e, o, e, r, e, x, ?

DaveE

 

[Moridin]

Alright... What letter comes next?

o, e, o, e, r, e, x, ?

DaveE

Well, the last six letters are the last letters of numbers in a row eg. x is the last letter of six, e is the last letter of five, r is the last letter of four and e is the last letter of three, o is the last letter of two, e is the last letter of e. However, I'm not sure what the first o stands for.

If I continue this series, then I would get an n since that is the last letter in the number seven?

Edit: o is the last letter in zero

 

[davee123]

Well, the last six letters are the last letters of numbers in a row eg. x is the last letter of six, e is the last letter of five, r is the last letter of four and e is the last letter of three, o is the last letter of two, e is the last letter of e. However, I'm not sure what the first o stands for.

If I continue this series, then I would get an n since that is the last letter in the number seven?

Edit: o is the last letter in zero

Exactly! I hoped the "zero" might throw people off. I was actually tempted to start the sequence at "negative three", yielding:

e, o, e, o, e, o, e, r, e, x

But I figured that might be a bit too confusing and arbitrary.

DaveE

 

[dontdisturbmycircles]

Thankfully, there is a weekly that we subscribe to which has this little teaser in every issue. Most of them are simple enough.

Fill in the blank so that line 4 is consistent with the first three.

Triangle, Pentagon, Square

Square, Hexagon, Hexagon, Square

Pentagon, Hexagon, Hexagon, Hexagon, Square, Triangle

Hexagon, Octagon, Octagon, Octagon, Octagon, __________

This one was a pain in the a-- :-). When I first saw it I thought that it would have to do with Pascal's triangle, but within 30 seconds it became apparent that it didn't. Eventually I noticed the first shape always has one more side than in the sentence before it. So perhaps it is just a number used to make a list. Then I noticed that 3 was dividible by 9, and then it was easy to tie it all together.

Square.

3^2 [5+4]

4^2 [6+6+4]

5^2 [6+6+6+4+3]

6^2 [8+8+8+8+4]

 

[dontdisturbmycircles]

I have a feeling that Moridin won't be back.

 

[neutrino]

Square.

3^2 [5+4]

4^2 [6+6+4]

5^2 [6+6+6+4+3]

6^2 [8+8+8+8+4]

That's a correct answer. It could also be any other quadrilateral.

 

[Gokul43201]

A bear walks south for one kilometer, then it walks west for one kilometer, then it walks north for one kilometer and ends up at the same point from which it started. What color was the bear?

I thought this might be Bernhard (from German for "brave bear") Jensen, the captain of the "Southern Cross" Antarctic Expedition. I also thought that after a 3 mile walk in the cold, the captain would be all blue!

 

[Hootenanny]

I thought this might be Bernhard (from German for "brave bear") Jensen, the captain of the "Southern Cross" Antarctic Expedition. I also thought that after a 3 mile walk in the cold, the captain would be all blue!

Your capacity for remembering inane trivia amazes me...

 

[Moridin]

I have a feeling that Moridin won't be back.

I live for contests. You shouldn't be trusting your instinct

I've spend the last day trying to come up with a suitable and interesting brainteaser.

"On a deserted island, an evil emperor lives alone. He hates people who come to the islands and kills them. He always thinks of new and exotic ways of doing so and there are no two alike, although he has a strange sense of humor - he gives them one chance of survival.

He allows them to guess once on the execution style he has in mind for them. If the person guess correctly, they will be hung. If they guess incorrectly, they will be shot."

If you get caught by the emperor and you wish to survive, what style of execution shall you guess on?

 

[Hootenanny]

"On a deserted island, an evil emperor lives alone. He hates people who come to the islands and kills them. He always thinks of new and exotic ways of doing so and there are no two alike, although he has a strange sense of humor - he gives them one chance of survival.

He allows them to guess once on the execution style he has in mind for them. If the person guess correctly, they will be hung. If they guess incorrectly, they will be shot."

If you get caught by the emperor and you wish to survive, what style of execution shall you guess on?

You should choose to be shot. This would create a paradox, since the statement is true implying you should be hung, however, if you are hung the statement is false implying you should be shot. The emperor can do nothing but let you go.

 

[Moridin]

You should choose to be shot. This would create a paradox, since the statement is true implying you should be hung, however, if you are hung the statement is false implying you should be shot. The emperor can do nothing but let you go.

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

 

[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