[OZPruductions]

Well, the point of this thread is that someone posts riddles and everybody needs to answer them, and the one who answers the riddle can write one of his own and etc....



And now for the first one:

1. Only people who know hebrew can answer that question:

What is special about this sentence:

Yes, we can see.

Clue: Think outside the box.

[NeatNit]

I reject your riddle for excluding most of the users! I'd also like to add another rule - the person who solves the riddle gives the next one.


Since your riddle is disqualified, I'll give a different one:
How much wood would a woodchuck chuck if a woodchuck could chuck wood?






(by the way, we can change this game into "disqualify the riddle above you" xD)

[noobay]

NeatNit, on 06 November 2009 - 08:57 PM, said:

How much wood would a woodchuck chuck if a woodchuck could chuck wood?

Well that's a sneezy, 100% more then the amount a woodchuck would chuck if a woodchuck couldn't chuck wood.

now what do I win?

[NeatNit]

You win telling the next riddle.

[noobay]

Blue Godzilla with a perl in it.
what am I referring to?

[VerdeFlash]

Blue Öyster Cult?

Blue Öyster Cult! I just saw them in concert a couple weeks back.

[.eXeC!]

No idea. Ima cheat here and set my own riddle:

Your house has 4 walls. All 4 of them are southfacing. A bear walks past. What colour is the bear?

[noobay]

Quote

Your house has 4 walls. All 4 of them are southfacing. A bear walks past. What colour is the bear?

All the walls face south because you are in the south!
The bear is white because you're in the north(pole).
It's a polar bear...

[noobay]

VerdeFlash, on 06 November 2009 - 11:09 PM, said:

Blue Öyster Cult?

Blue Öyster Cult! I just saw them in concert a couple weeks back.

You are correct, they have a song called Godzilla, awesome song.

[OZPruductions]

Now for a HARD one:

You got 100 coins, in EVERY coin there are two sides, black and white.

The coins are everywhere inside a room, you can't see anything in that room.

There are 10 coins turned on the white side up and 90 turned on the black side up.

How can you divide the coins into 2 groups, and in every group there will be the same amount of coins with their white side up?

Good luck.

[NeatNit]

Cut every coin in half and put half of it in one group and half in the other group, and NEVER LET ANYTHING FLIP.

Or just take it outside the room and arrange it.

[.eXeC!]

OZPruductions, on 06 November 2009 - 09:38 PM, said:

Now for a HARD one:

You got 100 coins, in EVERY coin there are two sides, black and white.

The coins are everywhere inside a room, you can't see anything in that room.

There are 10 coins turned on the white side up and 90 turned on the black side up.

How can you divide the coins into 2 groups, and in every group there will be the same amount of coins with their white side up?

Good luck.

Turn the 10 white ones over. That way you have 2 groups of 50 coins black side up and 0 coins white side up. Which is equal.


Riddle:

There is a crown in the middle of a room, on top of a carpet. The carpet is 50 meters wide and 50 meters long. Without using ANYTHING but your bare hands (no tools, planes, wires etc) how do you go to the crown and wear it WITHOUT treading on the carpet?

[NeatNit]

What's the problem? Just take it from the corner of the carpet and wear it

Seriously, though, you can just turn the carpet over and stuff until the crown falls away from it.


Also, your solution to OZ's question is wrong. He said you can't see anything in that room (although he didn't make it very clear...)

[.eXeC!]

You feel around for the coins then flip them and put them into a group at random. If probability prevails you will end up with even amount of white coins to black coins.

[OZPruductions]

No, you are all wrong, do you want me to give you the answer?

BTW: You can only make 2 groups of coins, and you can't cut any coin to a half

[NeatNit]

Take all the coins outside the room where you can see. My final answer. If not please tell me the answer.

[OZPruductions]

Give-upper, ok.

If you give out an answer, make a delete line on it so who doesn't wanna see the answer yet can ignore it.


You make two groups of coins, one group with 10 coins in it, and the other with 90 coins. You just flip all the coins in the group with 10 coins, that way in each group there is the same amout of coins like the other group. Example: if in the 10 coin group there are 7 white coins, and in the 90 coin group there are 3 white coins up, you flip all the coins in the 10 group, and you get the same amount as in the group with the 90 coins, because 10-7=3. Now it's again my turn: You got 10 sacks full of coins, in 9 sacks there are real coins, and in 1 sack there are fake coins. Real coins weight 4 grams, and a fake coin weight 3 grams, In 1 weighing, figure out which of the sacks has the fake coins in it. Additional info: You can get out as many coins from the sacks as you want.


That's it, and Ido, or Hen, please delete all the useless posts.

[NeatNit]

I'd also like to know the answer to this one:

OZPruductions, on 06 November 2009 - 08:29 PM, said:

Only people who know hebrew can answer that question:

What is special about this sentence:

Yes, we can see.

What useless posts? I see none.

[.eXeC!]

I'd assume it's todo with the rhyme: "Yeswe cansee"

[VerdeFlash]

Since I won one a while back, what 9 letter English word can you continuously remove one letter from and have another word every time?

Also, for OZ's, if there's no light then the coins would not be white or black because without light there is no color WAITER, CHECK PLEASE.