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rookie1ja

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Everything posted by rookie1ja

  1. this forum is currently just a beta 5 version - phpbb 3.0.B5 I would like to update this beta to new stable version phpbb 3.0.0 (which would bring more stability and new cool features) do you know someone who could help me with that? btw, there is no official support for update from beta versions and all information I found was quite complicated anyone skilled enough to face the challenge?
  2. good point ... and that's exactly what happens in the next puzzle called Masters of Logic Puzzles II. (hats)
  3. You're right, of course, rookie. The original answer was indeed the optimal solution. Sorry. no worries ... if you know some other good brain teasers, you can post it in the new puzzles section
  4. rookie1ja

    riddle

    a true classic ... you can check my version of the puzzle with 3 ways to ask the question that secures your survival - puzzle called Honestants and Swindlecants III.
  5. simple as that ... I guess the alt tag gave it away
  6. reminds me of this years Independence Day
  7. This answer gives a total required of 0 + 1 + 2 + 4 + 7 + 13 + 24 = 51 bears. This is not a minimum, however, so it is wrong. Since we know that there are exactly four of the seven boxes that contain fake bears, we can use this fact to get a better number sequence. 0 I don't think so ... what about 15 8+4+2+1=15 8+4+3+0=15
  8. checked it quickly and the following should be corrected ... good - evil ... should be both nice usage of flash though
  9. what condition in particular is wrong?
  10. I have not mentioned anything about enough space ... less does not mean not enough ... btw, have you seen a rabbit that would fit into your or my image? ps: have you noticed that your solution is not made of matchsticks of the same size? check the hypotenuses
  11. so you claim that my solution: 1. does not offer the same (equally big) area for 6 rabbits 2. made of 12 matchsticks ???
  12. actually, it was edited long time ago as follows: "... guess (Edited: predict the fate = guess correctly) ..." I don't think editing paradoxes/ sophisms / whatnot really counts.... the crocodile didn't get a chance to reword it once he was tricked, did he? If you want to reword all these famous brainteasers you are really cutting down the use of creative thinking... instead say, "you could do that, you got it right. now try this new but similar paradox, it has new wording so you have to think up a new answer" the sophism was just incorrectly translated into English language ... I know all the brain teasers/paradoxes in Slovak language and try to translate them into English language the best I can
  13. I have no problem with that activity ... let's hope there will be enough people ... unfortunately, I don't have time to be a part of it
  14. sum of the 3 numbers can not be unique ... so to use your "solution" ... 2+3+6=11 ... you can not split 11 into other 3 numbers that would make a product of 36, can you? so if I knew that we drank 11 beers and the product was 36, then I would know the ages ... however, I did not know the ages (because we did not drink 11 beers)
  15. that's great ... I am not a native speaker but I like the riddle a lot ... btw, nice title
  16. actually, it was edited long time ago as follows: "... guess (Edited: predict the fate = guess correctly) ..."
  17. the goal is to push the brown boxes on the yellow dots ... the game can be downloaded from my brain games page
  18. good point ... that's one of the things I am working on ... the rss feed you are referring to is actually a java script showing random puzzle (from a certain number) and I intend to add new puzzles in the script I am waiting for stable phpbb 3 release, so that I could get some mod to automatically generate rss feed for new puzzles in this forum patience
  19. nope, if "B is a lying monkey" is a lie ... then the truth is that: 1. B is a truth telling monkey OR 2. B is a lying human (anything but monkey) OR 3. B is a truth telling human ... any of the 3 options can be true eg. when you lie that you drank 2 beers last night, actually, you could still have had 1 (or 10)
  20. as already mentioned above by savagegamer90 and Boiling Oil ... the sum of the remaining numerals after 4 round-offs is indeed 24 after 1st rounding - 473816950 after 2nd rounding - 473817000 after 3rd rounding - 473817000 after 4th rounding - 473820000 so 4+7+3+8+2=24
  21. yes - I repeated the solution, since I assumed that it is clear ... I will try to take it a step further ... so let's assume that I am a wife and I know that there are 3 UH's 1st night Let's see if all 3 UH's will be shot ... but nothing happens ... I could have known that ... I guess that each of the 3 wives expected 2 shots ... hmmm, maybe the next night 2nd night let's wait what happens ... I guess that each of the 3 wives is anxiously awaiting the 2 shots now ... I imagine that one of them might think as follows: "I know there are 2 UH's. They were not shot the first night, since both of the 2 wives thought that there is just 1 UH. So this 2nd night both of them must kill their UH's. What the ...? I hear no shots. And now I know why - not only those 2, but also I do have an UH. We'll shoot them the 3rd night" ... I think that is the way one of the 3 wives might have thought 3rd night I'll open the window ... time is ticking away and then I hear something ... nope, was just some cat outside ... what is going on here? no shots? it is morning already ... the wives must have thought the way I imagined it ... unless ... wait a minute ... each of them does not know 2 UH's but 3 of them ... so there are 4 UH's? ... but who is the 4th one??? oh noooooo ... where is my gun ...
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