Riddles - come in and see if you can solve them!

Posted: Wed Apr 23, 2008 5:12 pm

If you have the answer to the second one, I shouldn't need to confirm it.Congrats . Feel free to post or you can let others sit on it a bit longer.

Do people like these more math based riddles, if so I can post many more, or branch off into more logic based or word oriented.

HINT BELOW:


HINT for the second riddle:
If the infinite expanse is throwing you off, it works the same if you limit the size to 1000 (or any number for that matter) total quarters.

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Posted: Fri Apr 25, 2008 1:05 pm

One more:

this is a magic trick performed by two magicians, A and B, with one regular, shuffled deck of 52 cards. A asks a member of the audience to randomly select 5 cards out of a deck. the audience member -- who we will refer to as C from here on -- then hands the 5 cards back to magician A. after looking at the 5 cards, A picks one of the 5 cards and gives it back to C. A then arranges the other four cards in some way, and gives those 4 cards face down, in a neat pile, to B. B looks at these 4 cards and then determines what card is in C's hand (the missing 5th card). how is this trick done?


NOTES:

Note 1: There's no secretive message communication in the solution, like encoded speech or ninja hand signals or ESP or whatever ... the only communication between the two magicians is in the logic of the 4 cards transferred from A to B. Think of these magicians as mathematicians.

Note 2: This magic trick is originally credited to magician and mathematician Fitch Cheney.
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riddle 1

Posted: Sat Apr 26, 2008 4:08 pm

for the second riddle.,
you take twenty (20) quarters and flip them all.
this way., any combination of tails you get., after fliping the selected 20 quarters, you have an equal amount of tails.

example.. you flip pile 1

pile 1 | pile 2(the onesl you left)
20 tails | 0 tails
you flip;
0 tails | 0 tails

19 tails/ 1 head | 1 tail
you flip;
19 heads/ 1 tail | 1 tail

18 tails/ 2 heads | 2 tails
you flip;
18 heads/ 2 tails | 2 tails
and so on and so forth.....

always equal in number of tails... ^^

[edit]
and to the magician riddle, b took a look at the remaning 47 cards and knew which were taken out by c.,
so he knows d 5 cards c took out., when he saw the 4 cards a gave him., he ofcourse know what card c is holding... ryt.?
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Posted: Mon Apr 28, 2008 9:23 pm

If only the card trick were that easy

Only need the 4 he handed you though

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Posted: Tue Apr 29, 2008 3:12 pm

This is rather tricky.

Hmmm..
I'm thinking that the 4 cards returned are arranged in some manner that determines the value of the returned card?
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Posted: Tue Apr 29, 2008 9:29 pm

If people would like a hint message me and I'll be happy to oblige. I prefer to keep as many hints as I can off the thread for those who would like to try without.

The card trick took me a few days to work out, but it's actually a fun trick to do to your friends. You just need an accomplice.

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Posted: Wed Apr 30, 2008 5:01 am

That much I figured. Ahahahaha...
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Posted: Wed Apr 30, 2008 5:38 am

hmmm

it maybe based on the order/arrangement of the cards
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Posted: Wed Apr 30, 2008 3:17 pm

Hmm..
First, the magician would have to figure out the suit. Then he has know the value of the card.

Since it's five cards, there is always a chance that two of them will have the same suit [there are only 4 suits]. The assistant MUST give back a card having the same suit as the one in the remaining 4 cards. The 4 cards will be arranged in an order such that the magician will know the suit of the returned card.

I'm pretty sure about on what I said above. I'm trying to figure out how the magician will know the face value.
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Posted: Sun May 04, 2008 3:10 pm

Maybe it has a sequence like an arithmetic sequence in a second year intermediate algebra.

Let me guess... Is it one card that the 3rd magician get to his trick. Isn't it?

This is tricky. Is it a sphinx's riddle?
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Posted: Mon May 05, 2008 2:11 am

I think I know the answer already.. One that involves arranging the three cards to add 6 to the card that shares the same suit.
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Posted: Mon May 05, 2008 6:08 pm

Sounds like your very close, if you don't already have it Yukino.

Just checking from work so it's a quick post. Let me know if you want me to keep posting these riddles, or I can leave the floor open for others.
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Posted: Thu May 15, 2008 10:09 pm

Here's another for ya:


You are trapped in a small phone booth shaped room. In the middle of each side of the room there is a hole. In each hole there is a push button that can be in either an off or on setting. You can't see in the holes but you can reach your hands in them and push the buttons. You can't tell by feel whether they are in the on or off position. You may stick your hands in any two holes at the same time and push neither, either, or both of the buttons as you please. Nothing will happen until you remove both hands from the holes. You succeed if you get all the buttons into the same position, after which time you will immediately be released from the room. Unless you escape, after removing your hands the room will spin around, disorienting you so you can't tell which side is which. How can you escape?

Now generalize. You are in a room with N sides, each side having a hole with a push button. What is the minimum number of hands you need to escape the trap room?




side note: no I'm not reciting these from memory or making them up. They are mostly taken from Willy Wu Tang's (don't know his real name)site at http://www.ocf.berkeley.edu/~wwu/riddles/hard.shtml

I do not post riddles I haven't already solved though.
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Posted: Mon May 19, 2008 5:07 am

a little hard riddle..how about calling someone
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Posted: Wed May 21, 2008 9:26 pm

The two button case is fairly straight forward, it's progressing into the n sided room that gets tricky.

Oddly enough, calling someone is not the answer. Even if you could how would they find you? (There is phone tracing i suppose... but certainly not the answer)
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