I'll start off with a really nice one which I saw in another forum. A group of 10 logical people are stranded on an island. They all each have either blue, green or brown eyes, but the numbers of each are not known. The people are unable to speak or communicate with each other in any way. The only way to get off the island is on a ship which comes every night. The only way to get on the ship is if you can tell the captain with certainty what colour eyes you have; if anyone gets the answer wrong or says anything else other than, "My eye colour is ____", everyone is killed (the Capt. is bit of a douche). The people have all been trapped on the island for some months until one day, the captain gets bored and shouts, "I CAN SEE SOMEBODY WITH BLUE EYES" before he sails away for the day. Everyone on the island hears him. Assume (i) everyone wants to leave the island, (ii) no one is suicidal, (iii) there are no mirrors/reflections, (iv) killing people/eye gouging is a no-no. The question is: WHO LEAVES WHEN? Hint 1: Spoiler There is at least 1 person that gets off the island.
Wow, that took me by surprise! I had the correct answer rather quickly, but thinking it far too simplistic, I looked up answers to similar questions thinking it was a lateral-thinking puzzle! However, this is the answer: Spoiler Assuming only one person has blue eyes, he would leave on the first night, as he would just look around and see that nobody else has blue eyes. Assuming that two people have blue eyes, neither of them would leave on the first night, however they would each see one person among the group with blue eyes, but realise their mistake the second night, assuming that they each have blue eyes themselves, as the only other blue-eyed person didn't leave. Hope that works for ya! Though the other versions of this puzzle I found were much more clear in their explanation of the situation.
I think this part is throwing a bit of a curve to the answer. I believe Malfoleo is essentially correct, but I think the answer can be expanded to the following: Spoiler All people with blue eyes leave the island, and it takes a number of days equal to the number of blue-eyed people (e.g., only 1 person = 1st day; 2 people = second day; 3 people = third day; etc.) Either way, it seems that the answer is dependent upon the islanders making assumptions about the scenario.
Your logic is on the right lines. However, what if there are 3, 4, 5, blue eyed people? Almost there though. Correct! You win sir!
Heres a good one: How do you measure 45 minutes using two ropes that each burn for 1 hour and a lighter? The only problem is that the ropes have variable diameter so they burn at variable rates (e.g in 10 minutes half of the rope may burn and in the next 10 minutes only a 12th might burn). Pretty big hint, use as last resort: Spoiler Think about burning the ropes at both ends
Light rope A at 1 end and rope B at both ends. After 30 mins, rope B is consumed. At that point light the other end of rope A. Once it has been consumed 45 mins has elapsed.
Well I guess if it obvious your answer is correct, you can go for it And I don't see why not your given answer is wrong this case? That's pretty clever!
Ok then, here's one I expect the bright minds here to be able to work out very quickly: A single-cell organism in a test tube divides itself once every five minutes. If you placed one organism in a test tube, it would take 46 hours to fill the test tube through division. If you placed two of the organisms in an identical test tube, how long would it take to fill the tube?
Doing teh maths: Spoiler Using 1 cell at the start results in 2^(46*12) = 2^(552) cells in 46 hours Equating this number of cells to using 2 cells at the start: 2^(552) = 2*2^(y*12) 2^(552) = 2^(12y+1) 552 = 12y + 1 y = 551/12 = 45 and 11/12th hours Thus, 45 hours and 55 mins