Can You Solve the Mystery of the Tape Recorder Riddle?

[JamesU]

I've been gone for too long...

 

[Aditya]

As has already been concluded from the other posts, there are only two possibilities: 2 red marks and 1 blue mark, or three red marks. (Although it doesn't take a person that long to realize that) But, if there are 2 red marks and 1 blue mark, then each of the scientists other than the one with the blue mark could easily conclude that they have a red mark since all three raised their hands and if one has a blue mark then the other person knows that he must have a red mark since the other person with the red mark also raised his hand. And the scientist who did answer the question answered "red" since each of the "smart" scientists took a while to answer and the scientist who did answer the question figured that if there was 1 blue mark and 2 red marks, then the scientist with the other red mark could easily conclude that he has a red mark on his head due to the fact that one of the scientists has a blue mark. Thus, the scientist who answered the question just figured that all three must be "red" since it would be easy if there were 2 blue and 1 red. Sorry about no commas and really long sentences.

 

[Rahmuss]

I'll give the answer to Huck. He answsered it first and gave a good enough explanation. Balls is in your court Huck.

 

[Huckleberry]

You throw it out when you want to use it. You bring it in when you don't want to use it. What is it?

 

[Berislav]

You throw it out when you want to use it. You bring it in when you don't want to use it. What is it?

In white:
Is it maybe an anchor?

 

[Huckleberry]

Yup, your up.

 

[Berislav]

Here's a visual one:
Below is a picture of a square with sides of unit lenght. Inside it is a smaller square. Divide the shaded area into four areas of same shape and size.

 

[JamesU]

Can you give a link to the picture? (it's not working)

 

[theCandyman]

I do not have permission to view it. Why not?

 

[cerberus]

i don't think anyone can view it

 

[Icebreaker]

Is it this?

As for attachments, log out and you will be able to see. Weird, and hope PF fixes it someday.

 

[cerberus]

thanks icebreaker loging out worked, and no the image u provided is not the same as the one we tryed to originally access

 

[Icebreaker]

No, I mean I think I found the solution. The one I attached is my solution.

 

[JamesU]

ICEBREAKER! *slaps icebreaker* other people were supposed to solve the riddle, not you!

 

[Rahmuss]

Am I too late to post the picture he was talking about?

http://img287.echo.cx/my.php?image=squareproblem2pz.png

 

[Berislav]

Sorry it took me this long to reply.

Good work, Icebreaker. Your turn.

 

[Icebreaker]

ICEBREAKER! *slaps icebreaker* other people were supposed to solve the riddle, not you!

Well that was certainly odd.

Good work, Icebreaker. Your turn.

Find a counterexample to the Riemann hypothesis.

Just kidding. Here's an easy one:

The length of the perimeter of a right triangle is 60 and the length of its altitude perpenticular to the hypotenuse is 12. Find the sides and show your work.

Good luck.

 

[Moo Of Doom]

The length of the perimeter of a right triangle is 60 and the length of its altitude perpenticular to the hypotenuse is 12. Find the sides and show your work.

Let a and b denote the lengths of the legs of the triangle, and let c denote the length of the hypotenuse.

60=a+b+c
a^2+b^2=c^2
ab/2=12c/2

Solving this system of equations, we get

a=15
b=20
c=25

 

[Icebreaker]

Yup. Your turn.

 

[JamesU]

Okay, moo of doom, your turn

 

[Moo Of Doom]

Yikes, I don't really have anything...

Here goes...
Find the next row in this sequence:

1
2, 1
3, 3, 2, 1
4, 4, 4, 3, 2, 1
5, 9, 10, 11, 12, 12, 12, 12, 12, 12, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1
?

 

[JamesU]

6, 5, 9, 10, 11, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1


I'm really not sure, I took a huge guess

 

[Moo Of Doom]

Nope.

I'll give you a hint: it's more than twice as long as the last one. (sorry :P)

 

[JamesU]

Can you give a bigger hint?

 

[Moo Of Doom]

There are 62 numbers in the sequence, and the highest number in the sequence is 32, which is repeated 17 times.

There's another big hint I'm willing to give if it's absolutely necessary.

 

[JamesU]

does it involve adding, subtraction, mult... you get it, so does it?

 

[Moo Of Doom]

It involves subtraction, but not in the way you think. It also involves addition, but in an even weirder way than the subtraction.

(Basically, if you think the pattern is anything akin to add one, subtract two, etc. you're on the wrong track. There is no such rule, really.)

 

[Moo Of Doom]

I have the feeling this puzzle is a bit too freakishly difficult, so here comes a big hint:

The sequence is very closely related to this one:

1
10, 1
11, 10, 2, 1
100, 11, 10, 3, 2, 1
...
?

If you're still stuck, I'll give the fifth row of the above sequence, although it kind of gives it away.

 

[JamesU]

30, 31, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

 

[Moo Of Doom]

You have part of it, but you're missing the first 12 numbers, and one of the 32s.

 

[JamesU]

18, 19, 20, 21, 22, 23, 24, 25,, 26, 27, 28, 29, 30, 31, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

 

[Moo Of Doom]

You have it right from 24 on, but the bit before that is wrong.
Think about how the first sequence might relate to the second sequence, and see if that helps. Tell me if you want the next row for the second sequence.

 

[JamesU]

20, 21, 22, 23, 24, 25, 24, 25,, 26, 27, 28, 29, 30, 31, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

 

[Moo Of Doom]

Nope, colder.

 

[JamesU]

I give up, give another.