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| Posted by humoeba | 06/24/04 |
| I like it!!! Cool :D |
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| Posted by jimbo | 06/26/04 |
| Interesting logic but I had in mind something that would take less time. Hope the scorekeeper doesn't die before the strategy carries through. :D |
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| Posted by hersheykiss8908 | 06/27/04 |
| that was really hard, and took a little too much time. i liked it though, and i would have never thought of the answer |
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| Posted by crystalstar | 06/27/04 |
| Hmm...It's okaaaaaaaay, I guess. But yeah, it's basically impossible for anyone to think of the answer to that, and also I didn't know what we had to figure out, so I think you should have made that clearer! :wink: |
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| Posted by Varthen | 07/01/04 |
| I dont like people who give me headaches... |
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| Posted by rashad | 07/04/04 |
| Ok i liked this 1 but the problem that there was no question...there is no point...the answer may be 15ducj=k and i have the right to say that cuz there is no question make it clearer the next time |
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| Posted by kuya_kei | 07/13/04 |
| This is very good. I liked the way it made my head ache. |
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| Posted by sakirski | 07/13/04 |
| Excellent Teaser in my opinion. It's perfectly logical, and it's perfectly solvable, though certainly not very easy.
The question though not stated in the given is implied, so I don't think it to be a big deal that it's missing. |
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| Posted by sakirski | 07/15/04 |
| OK so I just found a problem with the solution (provided the current statement of the given) For the solution to work the prisoners will have to keep visiting the jail cell even after all of them visited it an equal number of times.
You see nowhere in the problem does it say that the warden will keep pulling prisoners into the switch room indefinitely. Problem states that he will pick them randomly, and that everyone will visit the room as many times as everyone else. Well what if the luck of the draw happens that randomly they will all be picked only once and all 23 will visit the room only once? Then the counter can only do one flip of the switch, and then they will never be freed.
So basically the problem needs an extra statement: “Assume that this exercise continues indefinitely until one of the prisoners speaks out.†|
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| Posted by swordflame | 07/30/04 |
| It seems to me that the answer is slightly incorrect. In part 3 of the answer, the second ON should read OFF. |
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| Posted by biztycl | 07/30/04 |
| Thanks Sakirski, the pre-condition you mentioned is stated in the hint section. I like those 3 teasers you posted too. :D |
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| Posted by valeriy | 08/24/04 |
| Assuming non of the prisoners died in the jail. Then they would stay there 4ever. |
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| Posted by stormtrooper | 02/13/05 |
| wha...wha...wha? |
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| Posted by jacstop | 05/11/05 |
| As a matter of fact or fiction, this scenario's countless improbabilities bequeath indistinct objectives to its foreseen populace. This is neither bad nor good :roll: :roll: |
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| Posted by netgoof | 05/13/05 |
| If the warden picked one random prisoner every day, it would take about 596 days on average. |
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| Posted by sweetime | 05/16/05 |
| I wasn't sure what the question was asking - i was trying to work out how the warden would know if they were lying or not...(for some reason i had a nice double blind going on, and the warden didn't know who he was taking out either...) |
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| Posted by smarty_blondy | 11/07/05 |
| What exactly are we supposed to find out? What does this teaser ask? I only see plain text with an explanation fallen from the sky. :-? |
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| Posted by shenqiang | 11/15/05 |
| Where is Scenario B? |
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| Posted by HibsMax | 02/28/06 |
| Unless I don't understand the problem correctly, I think there is a flaw. For ease let's say that the warden has already decided that he is going to pick each prisoner 5 times. What if the scorekeeper is picked to make the first 5 visits? Who keeps score then? |
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| Posted by HibsMax | 02/28/06 |
| Sorry, I'm new here. :) I guess I should have looked at the date of the problem first. |
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| Posted by mercenary007 | 04/01/06 |
| I think this is a nice try... for those of you who say you didn't know what to do then something is wrong with you cuz anyone could figure out that the prisoners should devise a plan to get out... DUH!!!! OBVIOUSLY... now as for the answer... if the scorekeeper counts 1 for every time that he flicks off the switch B that would be incorrect. If 1 prisoner goes before him and flicks B on and then the scorekeeper goes and flicks it off and they alternate like that for 3 turns then the scorekeeper would count 3 when in fact only 1 of his fellow prisoners has been to the room. For this reason the answer is wrong in my mind. 8) |
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| Posted by safire2191 | 04/14/06 |
| what was the question? |
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| Posted by calmsavior | 05/16/06 |
| that solution makes a lot of sense... :oops: |
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| Posted by Creshosk | 06/24/06 |
| I have to agree with mercenary007. alternating between the scorekeeper and one other person multiple times would ruin the solution.
As it stands the prisoners are alligator chow. |
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| Posted by ztodd | 10/16/06 |
| Read the problem again mercenary and cresh. Each transmitter only flips B to on if it's off one time, then only flips A after that, no matter whether they find B off again or not in later visits.
I agree though that it's not totally clear if the warden is guaranteed to keep bring prisoners back or if he will stop after they've all had the same number of turns. |
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| Posted by senther7 | 12/30/06 |
| I think THINK that i foud a flaw.
What happened if 2 transmitters go to the room, and neither of them have gone to the room, how does the reciver know that 2 people has gone, not one?Unless someone explaines the tatic, i think the answer is wrong :roll: |
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| Posted by senther7 | 12/30/06 |
| O wait, never mind :oops: |
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| Posted by jesdexter | 03/25/07 |
| i never knew the question |
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| Posted by Odessius | 04/20/07 |
| (sigh) well people, its to the alligators. Theres no way; we could have been in there for a hundred billion years. Still, yeah sure, the answer works out, and I was beat by the puzle... ih. :-? I tried to split em into 2's and 4's with 4 possible combinations, and clubbing the warden, nope wasn't gona solve that one. a transmitter... don't feel bad, a transmitter, dude! okay so what if we're in the planning stage, and someone says what if the transmitter dies. hmm... :roll: |
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| Posted by notsosmart111 | 06/11/07 |
| OW!!! My head hurts how do you come up with this stuff!!! It's so brilliant!!! |
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| Posted by spinnercat | 07/23/07 |
| brilliant!!!!! My math teacher gave this to us, but never told us the answer. Thanks!!!! |
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| Posted by 974999gec | 10/26/07 |
| How do you come up with this stuff? I mean really! Does it just hit you? :-? :-? :roll: :wink: |
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| Posted by plokolplok | 05/14/08 |
| how 'bout flicking switch B on everytime a prisoner goes into the switch room for the first time, and flicking switch A if he/she already got there...then it's everybody's role to count the ons and offs of light B, because if they appoint only one 'counter', just one missed count could lead to dying old in prison...the counter would never reach 23...or is this what the solution suggests?
ow.
head achy achy... |
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| Posted by Sorena888 | 11/06/08 |
| Yes it works. Every one could count. This is a better way. |
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| Posted by MikeG | 10/31/11 |
| I love this teaser.
And I know it's 3 years later, but what plokolplok suggests doesn't work. If you have more than one person who counts, who turns off switch B? Only one person can turn off switch B, otherwise they could be staying in there a LOT longer than they would with the poster's solution. And if one person turns it off, you can't have other people counting; a counter could come in five times in a row, see the switch ON each time and add 5, even though nobody else came in. |
