Will post logical puzzles for a beta key.

"3 prisoners (lets call them X, Y & Z) who are being punished are buried neck deep-riddle"

The middle (y) prisoner will guess the right way.
Why?
cause he does know, that Z will see his and x'ses Hats, and Z will see a black and a red hat, so Z cannot draw the conclusion what hat he is wearing.
so Y must wear another colour than X, and since he can see X' hat he can for sure tell, what hat he is wearing.
"
Bash76again wrote:
"3 prisoners (lets call them X, Y & Z) who are being punished are buried neck deep-riddle"

The middle (y) prisoner will guess the right way.
Why?
cause he does know, that Z will see his and x'ses Hats, and Z will see a black and a red hat, so Z cannot draw the conclusion what hat he is wearing.
so Y must wear another colour than X, and since he can see X' hat he can for sure tell, what hat he is wearing.


Darn, I just thought about it in the shower ^^ Didn't make it in time xD gz
@Bash76again
Yep that is correct

In this particular case, Y can correctly guess the color of his own hat.
1. If X and Y had white hats. Then Z would have answered that his own hat was black.
2. Since Z remains silent. This implies that either both X & Y are black or X & Y are one black, one white.
3. Now Y looks at X and sees that he is wearing a white hat. Thus the only explanation for Z's silence is that Y must be wearing a black hat.
Here's another one

There are 100 coins on a table, of which 26 show heads and the 74 show tails. You are now blindfolded and asked to separate the coins into 2 stacks such that each stack contains equal number heads. How do you do this?

Conditions:
1. While blindfolded you can flip any number of coins, ie change heads to tails or tails to heads.
2. You cannot feel and find out whether it is heads or tails.
3. You cannot place coins on their side.

"
arjunzv8 wrote:
Here's another one

There are 100 coins on a table, of which 26 show heads and the 74 show tails. You are now blindfolded and asked to separate the coins into 2 stacks such that each stack contains equal number heads. How do you do this?

Conditions:
1. While blindfolded you can flip any number of coins, ie change heads to tails or tails to heads.
2. You cannot feel and find out whether it is heads or tails.
3. You cannot place coins on their side.



You first divide all the coins into 2 stashes, 50 in each.
Then you take half the coins in each stash and flip them and voila =)

Or I'm I being a fool now o.O let me think one sec about it...

It's wrong...hmmmmmm
Last edited by Zemeter on Apr 11, 2012, 8:04:03 AM
"
VideoGeemer wrote:


-snip-

And I'm not sure if it's because this is the internet, or because he had prior knowledge, but rtbchef actually was right. The reason given is that the middle sheep is on a hill in the middle of the meadow (must be a pretty steep one), but a tree stump could also work.

I ask this question a lot, because it seems so obvious as soon as you think of it geometrically, and yet I've never had anyone in person answer it correctly. There's one guy I should still try, because he's very intelligent, but you'd be surprised how intelligent minds can somehow not think about it in that way. :)


I was rehearsing hydrocarbons and for some reason CH4 popped into my head when i read that puzzle :p
This is a BETA people, it's for SUGGESTIONS and REPORTING BUGS
"
arjunzv8 wrote:
Here's another one

There are 100 coins on a table, of which 26 show heads and the 74 show tails. You are now blindfolded and asked to separate the coins into 2 stacks such that each stack contains equal number heads. How do you do this?

Conditions:
1. While blindfolded you can flip any number of coins, ie change heads to tails or tails to heads.
2. You cannot feel and find out whether it is heads or tails.
3. You cannot place coins on their side.



Ok so first you need to split them into 2 groups. One group of 26 and the other of 74. Then you flip the whole group of 26 onto the other side. You will get the same amount in both groups but you wont know what that number is.

Eg. If in the group of 26 you have 10 heads and 16 tails that means that in the other group you must have 16 heads and 58 tails. If you flip all of the ones in the first group you will get 16 heads in both groups.
@Neebz
Correct :-)
Last edited by arjunzv8 on Apr 11, 2012, 2:00:22 PM
"
Neebz wrote:

Ok so first you need to split them into 2 groups. One group of 26 and the other of 74. Then you flip the whole group of 26 onto the other side. You will get the same amount in both groups but you wont know what that number is.

Eg. If in the group of 26 you have 10 heads and 16 tails that means that in the other group you must have 16 heads and 58 tails. If you flip all of the ones in the first group you will get 16 heads in both groups.



O.O

That's amazing. Took me a minute to figure out how that was really possible, even after reading it. :)
Invited to Beta 2012-03-18 / Supporter since 2012-04-08
"
Sixth Puzzle (By arjunzv8):
"3 prisoners (lets call them X, Y & Z) who are being punished are buried neck deep in sand in a straight line such that
1. X cannot see anyone.
2. Y can see only X.
3. Z can see both X & Y.

A hat is placed on each of their head but they cannot see the hat on their head and they cannot speak to each other.
They are informed there are 5 hats of which 3 are Black and 2 are White. Out of these, 3 are selected and placed on each of their head. They have one try to guess the color of the hat on their head, if they guess correctly, they will be set free.

In this particular case the hats were placed as stated below
1. X has white hat
2. Y has black hat
3. Z has black hat

One person guesses correctly. Who and how?
Do assume all the prisoners are smart, this is important."


The answer depends only on the color of the first prisoner's hat.
We will label the prisoners like this:
C > B > A
So C can see everyone ,B only A, and A no one.

1) Let's assume A wears a white hat.
If B wears a white hat also, then C will say he is wearing a Black hat and will be Correct.
If B wears a Black hat, then C won't answer and, seeing that A is wearing a White hat, he will know he is wearing a Black hat and will be set free.

2) Let's assume A wears a black hat.
If B wears a black hat as well, then C will not know what to say and B, seeing A is wearing a black hat, will not know what to say either, so A will know he's wearing a Black hat, since this can only happen if he has the Black hat.
If B wears a white hat, then it's the same situation as above.

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