know_buddy_kares Posted June 15, 2008 Report Share Posted June 15, 2008 a porthole appeard and sucked 2 of them into a vortex!!! Link to comment Share on other sites More sharing options...
Angel of Death Posted June 15, 2008 Report Share Posted June 15, 2008 Still love the avatar BTW thank you... and uhh...hmmm...ya that's the best I got...dammit Link to comment Share on other sites More sharing options...
know_buddy_kares Posted June 15, 2008 Report Share Posted June 15, 2008 i think i got it! Cher was standing infront of rev, with 2 already in hand, which technically would make those in front of rev.. then she took one, and walked off... hence taking 3 (2 of which she already had in possession of) wich left rev with only 9 Link to comment Share on other sites More sharing options...
phee Posted June 15, 2008 Author Report Share Posted June 15, 2008 i think i got it! Cher was standing infront of rev, with 2 already in hand, which technically would make those in front of rev.. then she took one, and walked off... hence taking 3 (2 of which she already had in possession of) wich left rev with only 9 I like that answer... not what I was looking for... but very good thoughts... *CLICK HERE FOR HINT* Link to comment Share on other sites More sharing options...
AstralCrux Posted June 15, 2008 Report Share Posted June 15, 2008 Nice hint. Link to comment Share on other sites More sharing options...
phee Posted June 15, 2008 Author Report Share Posted June 15, 2008 Nice hint. I thought so... Link to comment Share on other sites More sharing options...
know_buddy_kares Posted June 15, 2008 Report Share Posted June 15, 2008 hm.. roman numerals probably XII equals 12.. take one away XI would be 11 still.. but flipped around it would be IX making 9... so he could have 9 or 11, depending on the persective of view... am i at least on the right path here?? Link to comment Share on other sites More sharing options...
phee Posted June 15, 2008 Author Report Share Posted June 15, 2008 hm.. roman numerals probably XII equals 12.. take one away XI would be 11 still.. but flipped around it would be IX making 9... so he could have 9 or 11, depending on the persective of view... am i at least on the right path here?? Sort of... Link to comment Share on other sites More sharing options...
know_buddy_kares Posted June 15, 2008 Report Share Posted June 15, 2008 Sort of... ok ok,, alphabet time.. A=1 B=2 C=3 and so forth.. so I=9 and L=12 copsticks are at an angle like this.. |_ making an L=12, cher takes the _ one leaving the | making 9! Link to comment Share on other sites More sharing options...
phee Posted June 15, 2008 Author Report Share Posted June 15, 2008 ok ok,, alphabet time..A=1 B=2 C=3 and so forth.. so I=9 and L=12 copsticks are at an angle like this.. |_ making an L=12, cher takes the _ one leaving the | making 9! Thinking too hard... Link to comment Share on other sites More sharing options...
know_buddy_kares Posted June 15, 2008 Report Share Posted June 15, 2008 Thinking too hard... *picks the table up and throws it off a cliff* damnit, now there's no chopsticks... wut?? seriously, i give up... let someone else figure this out. Link to comment Share on other sites More sharing options...
Angel of Death Posted June 15, 2008 Report Share Posted June 15, 2008 did rev maybe move so that 2 were no longer in front of him??? Link to comment Share on other sites More sharing options...
phee Posted June 15, 2008 Author Report Share Posted June 15, 2008 did rev maybe move so that 2 were no longer in front of him??? Nope... Cher removes one and NINE remain Link to comment Share on other sites More sharing options...
know_buddy_kares Posted June 15, 2008 Report Share Posted June 15, 2008 |\| |\ |\| E E making 4 by it's self... counting the red one, totaling 12.. remove the red one, and it spells out "NINE" Link to comment Share on other sites More sharing options...
phee Posted June 16, 2008 Author Report Share Posted June 16, 2008 |\||\ |\| E E making 4 by it's self... counting the red one, totaling 12.. remove the red one, and it spells out "NINE" WE HAVE A WINNER!!! An easer one: know_buddy_cares is condemned to death. He must choose between three rooms. The first is full of raging fires, the second is full of assassins with loaded guns, and the third is full of lions that haven't eaten in 3 years. Which room is safest for him? Link to comment Share on other sites More sharing options...
AstralCrux Posted June 16, 2008 Report Share Posted June 16, 2008 The lions- they haven't eaten in 3 years.... They're dead. Link to comment Share on other sites More sharing options...
know_buddy_kares Posted June 16, 2008 Report Share Posted June 16, 2008 The lions- they haven't eaten in 3 years.... They're dead. I agree, but I'm sure most the mods would choose the room with fire... so long as they have front row seats to the show. Link to comment Share on other sites More sharing options...
phee Posted June 16, 2008 Author Report Share Posted June 16, 2008 Correct.... I am not sure if KBK is correct because I am not sure what he means.... Suppose Rayne wants to send in the mail a valuable thing to AstralCrux. She has a box which is big enough to hold the thing. The box has a locking ring which is large enough to have a lock attached and Rayne has several locks with keys. However, AstralCrux does not have the key to any lock that Rayne has. Rayne cannot send the key in an unlocked box since it may be stolen or copied. How does Rayne send the valuable object, locked, to AstralCrux - so it may be opened by her? Link to comment Share on other sites More sharing options...
Angel of Death Posted June 16, 2008 Report Share Posted June 16, 2008 use a combination lock? Link to comment Share on other sites More sharing options...
phee Posted June 16, 2008 Author Report Share Posted June 16, 2008 use a combination lock? I think that is a better answer then the actual one to be honest.... but "Rayne has several locks with keys" She does not have, or have means of getting... (because of some stupid reason) a combination lock. Link to comment Share on other sites More sharing options...
Angel of Death Posted June 16, 2008 Report Share Posted June 16, 2008 OK maybe... you take the key from the first box and put it in a second box, and put the key for the second box in the lock of the first box?? Link to comment Share on other sites More sharing options...
phee Posted June 16, 2008 Author Report Share Posted June 16, 2008 Nope..... one could still get into the boxes..... Link to comment Share on other sites More sharing options...
Angel of Death Posted June 16, 2008 Report Share Posted June 16, 2008 Nope..... one could still get into the boxes..... How's the key for a different lock going to unlock the 1st lock...can't you send them separately? Link to comment Share on other sites More sharing options...
Angel of Death Posted June 16, 2008 Report Share Posted June 16, 2008 Rayne sends the box locked, AstralCrux attaches her own lock and sends it back, then you take off your lock and send it back...??!! Link to comment Share on other sites More sharing options...
phee Posted June 16, 2008 Author Report Share Posted June 16, 2008 How's the key for a different lock going to unlock the 1st lock...can't you send them separately? I think you have defeated this one... The actual answer is you send the box with your lock... they receive the box and send it back with a lock on their own, you receive the box unlock yours, send it back, they remove their lock and viola.... I think your answers were both right though.... ahem.... Moving on: Angel of Death and DeadBurgerKing play five games of checkers. Each of them wins the same number of games. There are no ties. Explain this. Link to comment Share on other sites More sharing options...
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